NCERT Exemplar Class 12 Maths Solutions Chapter 13 Probability

Edited By Ravindra Pindel | Updated on Sep 15, 2022 - 5:25 p.m. IST | #CBSE Class 12th

NCERT exemplar Class 12 Maths solutions chapter 13 deals with the possibility of occurrence of any event or explains how likely it is for an event to occur. The possibility of occurring or not occurring of any event cannot be predicted but can be displayed in the form of probability with the absolute level of certainty. Probability of any event is a number between 0 to 1 where 0 shows the absolute impossibility of an event, and 1 shows the maximum chances of happening of an event certainly. NCERT exemplar Class 12 Maths solutions chapter 13 help students to interrelate such knowledge with real-life problems and display the application of such knowledge in different lives scenarios and situations along with enhancing decision-making at an efficient level.

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Question:1

For a loaded die, the probabilities of outcomes are given as under:
P (1) = P (2) = 0.2, P (3) = P (5) = P (6) = 0.1 and P (4) = 0.3.
The die is thrown two times. Let A and B be the events, same number each time and a total score is 10 or more respectively. Determine whether or not A and B are independent.

Answer:

Given
For a loaded die –
P (1) = P (2) = 0.2, P (3) = P (5) = P (6) = 0.1 and P (4) = 0.3
The die is thrown twice and

Event of the same number each time
Event of total score of 10 or more.

Therefore, for A,
$\\A = \{ $ (1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$ \} $ \\ P(A)= [P (1,1) +P (2,2) +P (3,3) + P (4,4) + P (5,5) + P (6,6)] \\ P(A)= [P (1) $ \times $ P (1) + P (2) $ \times $ P (2) + P (3) $ \times $ P (3) + P (4) $ \times $ P (4) +P (5) $ \times $ P (5) + P (6) $ \times $ P (6)] \\ P(A)= [0.2$ \times $ 0.2+ 0.2$ \times $ 0.2+ 0.1$ \times $ 0.1+0.3$ \times $ 0.3+ 0.1$ \times $ 0.1+ 0.1 $ \times $ 0.1] \\ P(A)= [0.04+ 0.04+ 0.01+ 0.09+ 0.01+0.01] \\ P(A)= [0.20] \\$
For B,
B= EVENT OF TOTAL SCORE IS 10 OR MORE
B= {(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)}
$\\ P(B)= [P (4,6) + P (5,5) + P (5,6) + P (6,4) + P (6,5) + P (6,6)] \\ \\ P(B)= [P (4) $ \times $ P (6) + P (5) $ \times $ P (5) + P (5) $ \times $ P (6) + P (6) $ \times $ P (4) + P (6) $ \times $ P (5) + P (6) $ \times $ P (6)] \\ \\ P(B)= [0.3$ \times $ 0.1+ 0.1$ \times $ 0.1+ 0.1$ \times $ 0.1+ 0.1$ \times $ 0.3+ 0.1$ \times $ 0.1+ 0.1$ \times $ 0.1] \\ \\ P(B)= [0.03+ 0.01+ 0.01+ 0.03+ 0.01+ 0.01] \\ \\ P(B)= [0.10] \\$

For the probability of intersection B i.e. both the events occur simultaneously,
$A \cap B = {(5,5), (6,6)}$
Therefore,
P (A $\cap$ B) = P (5,5) + P (6,6)
P (A $\cap$ B) = P (5) × P (5) + P (6) ×P (6)
P (A $\cap$ B) = 0.1×0.1+ 0.1×0.1
P (A $\cap$ B) = 0.01+0.01
P (A $\cap$ B) = 0.02
Knowing that if two events are independent, then,
$\\P (A \cap B) = P(A) P(B) \\ P(A). P(B)= 0.20$ \times $ 0.10 = 0.02 \\$
Therefore,
$P (A \cap B) = P(A) P(B) \\$
Hence, A and B are independent events.

Question:2

Refer to Exercise 1 above. If the die were fair, determine whether or not the events A and B are independent.

Answer:

Given
$\\A=\{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\}$ \\$\mathrm{So}, \mathrm{n}(\mathrm{A})=6, \mathrm{n}(\mathrm{S})=(6)^{2}=36$$
Therefore,
$\mathrm{P}(\mathrm{A})=\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{6}{36}=\frac{1}{6}$$
And $B=\{(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)\}$$
$\Rightarrow \mathrm{n}(\mathrm{B})=6$$
Therefore,
$\\ \mathrm{P}(\mathrm{B})=\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{6}{36}=\frac{1}{6}$\\ \\$A \cap B=[(5,5),(6,6)]$
Therefore, $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{2}{36}=\frac{1}{18}$$
Therefore,
P (A $\cap$ B) ≠ P(A). P(B)
Hence, A and B are not independent events.

Question:3

The probability that at least one of the two events A and B occurs is 0.6. If A and B occur simultaneously with probability 0.3, evaluate $P(\bar{A})+P(\bar{B})$ .

Answer:

Given-
At least one of the two events A and B occurs is 0.6 i.e. P(A $\cup$ B) = 0.6
If A and B occur simultaneously, the probability is 0.3 i.e. P(A $\cap$ B) = 0.3
It is known to us that
P(A $\cup$ B) = P(A)+ P(B) – P(A $\cap$ B)
$\Rightarrow$ 0.6 = P(A)+ P(B) – 0.3
$\Rightarrow$ P(A)+ P(B) = 0.6+ 0.3 = 0.9
To find- $P(\bar{A})+P(\bar{B})$
Therefore,
$\\ P(\bar{A})+P(\bar{B})=[1-P(A)+1-P(B)] \\ \Rightarrow P(\bar{A})+P(\bar{B})=2-[P(A)+P(B)] \\ \Rightarrow P(\bar{A})+P(\bar{B})=2-0.9 \\ P(\bar{A})+P(\bar{B})=1.1$

Question:4

A bag contains 5 red marbles and 3 black marbles. Three marbles are drawn one by one without replacement. What is the probability that at least one of the three marbles drawn be black, if the first marble is red?

Answer:

Given
A bag contains 5 red marbles and 3 black marbles
If the first marble is red, the following conditions have to be followed for at least one marble to be black.

Second marble is black and third is red = $E_1$
Second and third, both marbles are black = $E_2$

(iii) Second marble is red and third marble is black = $E_3$
Let event $R_n$ = drawing red marble in $n^{th}$ draw
And event $R_n$ = drawing black marble in $n^{th}$ draw
$\\\therefore P(E_ 1)=P\left(R_{1}\right) \cdot P\left(B_{1} / R_{1}\right) \cdot P\left(R_{2} / B_{1} R_{1}\right)=\frac{5}{8} \times \frac{3}{7} \times \frac{4}{6}=\frac{5}{28}$ \\$\therefore \mathrm{P}\left(\mathrm{E}_{2}\right)=\mathrm{P}\left(\mathrm{R}_{1}\right) \cdot \mathrm{P}\left(\mathrm{B}_{1} / \mathrm{R}_{1}\right) \cdot \mathrm{P}\left(\mathrm{B}_{2} / \mathrm{B}_{1} \mathrm{R}_{1}\right)=\frac{5}{8} \times \frac{3}{7} \times \frac{2}{6}=\frac{5}{56}$ \\$\therefore \mathrm{P}\left(\mathrm{E}_{3}\right)=\mathrm{P}\left(\mathrm{R}_{1}\right) \cdot \mathrm{P}\left(\mathrm{R}_{2} / \mathrm{R}_{1}\right) \cdot \mathrm{P}\left(\mathrm{B}_{1} / \mathrm{R}_{1} \mathrm{R}_{2}\right)=\frac{5}{8} \times \frac{4}{7} \times \frac{3}{6}=\frac{5}{28}$$
Hence, the for the probability for at least one marble to be black is P
$(\mathrm{E})=\mathrm{P}\left(\mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right)+\mathrm{P}\left(\mathrm{E}_{3}\right)\\$ Therefore, $P(E)=\frac{5}{28}+\frac{5}{56}+\frac{5}{28}=\frac{25}{56}$$

Question:5

Two dice are thrown together and the total score is noted. The events E, F and G are ‘a total of 4’, ‘a total of 9 or more’, and ‘a total divisible by 5’, respectively. Calculate P(E), P(F) and P(G) and decide which pairs of events, if any, are independent.

Answer:

Given-
Two dice are drawn together i.e. n(S)= 36
S is the sample space
E = a of total 4
F= a total of 9 or more
G= a total divisible by 5
Therefore, for E,
E = a of total 4
∴E = {(2,2), (3,1), (1,3)}

∴n(E) = 3
For F,
F= a total of 9 or more
∴ F = {(3,6), (6,3), (4,5), (5,4), (6,4), (4,6), (6,5), (6,6), (5,5), (5,6)}
∴n(F)=10
For G,
G = a total divisible by 5
∴ G = {(1,4), (4,1), (2,3), (3,2), (4,6), (6,4), (5,5)}
∴ n(G) = 7
Here, (E $\cap$ F) = φ AND (E $\cap$ G) = φ
Also, (F $\cap$ G) = {(4,6), (6,4), (5,5)}
$\Rightarrow$ n (F $\cap$ G) = 3 and (E $\cap$ F $\cap$ G) = φ
$\\ P(E)=\frac{n(E)}{n(S)}=\frac{3}{36}=\frac{1}{12} \\ P(F)=\frac{n(F)}{n(S)}=\frac{10}{36}=\frac{5}{18} \\ P(G)=\frac{n(G)}{n(S)}=\frac{7}{36} \\ P(F \cap G)=\frac{3}{36}=\frac{1}{12} \\ P(F) \cdot P(G)=\frac{5}{18} \times \frac{7}{36}=\frac{35}{648}$
Therefore,
P (F $\cap$ G) ≠ P(F). P(G)
Hence, there is no pair which is independent

Question:6

Explain why the experiment of tossing a coin three times is said to have binomial distribution.

Answer:

Let p= events of failure and q=events of success
It is known to us that,
A random variable X (=0,1, 2,…., n) is said to have Binomial parameters n and p, if its probability distribution is given by
$\mathrm{P}(\mathrm{X}=\mathrm{r})=\mathrm{n}_{\mathrm{c}_{\mathrm{r}}} \mathrm{p}^{\mathrm{r}} \mathrm{q}^{\mathrm{n}-\mathrm{r}}$ \\Where, $q=1-p$ and $r=0,1,2, \ldots . . n$ \\In the experiment of a coin being tossed three times $\mathrm{n}=3$ and random variable $\mathrm{X}$ can take $\mathrm{q}=\frac{1}{2}$ values $r=0,1,2$ and 3 with $p=\frac{1}{2}$ and $q = \frac{1 }{2}$
$\begin{array}{|l|l|l|l|l|} \hline \mathrm{X} & 0 & 1 & 2 & 3 \\ \hline \mathrm{P}(\mathrm{X}) & 3_{\mathrm{c}_{0}} \mathrm{q}^{3} & 3_{\mathrm{c}_{1}} \mathrm{pq}^{2} & 3_{\mathrm{c}_{2}} \mathrm{p}^{2} \mathrm{q} & 3_{\mathrm{c}_{3}} \mathrm{p}^{3} \\ \hline \end{array}$
Therefore, in the experiment of a coin being tossed three times,
we have random variable X which can take values 0,1,2 and 3 with parameters n=3 and $p=\frac{1}{2}$
Hence, tossing of a coin 3 times is a Binomial distribution.

Question:7

A and B are two events such that $P(A)=\frac{1}{2},P(B)=\frac{1}{2}\: \: $and $P(A \cap B)= \frac{1}{4}$
Find:
(i) P(A|B) (ii) P(B|A) (iii) P(A’|B) (iv) P(A’|B’)

Answer:

Given-
$P(A)=\frac{1}{2},P(B)=\frac{1}{2}\: \: $and $P(A \cap B)= \frac{1}{4}$
$\\(i) P(A \mid B)=\frac{P(A \cap B)}{P(B)}$ \\$=\frac{\frac{\frac{1}{4}}{1}}{3}$ \\$=\frac{3}{4}$ \\(ii) $P(B \mid A)=\frac{P(A \cap B)}{P(A)}$ \\$=\frac{\frac{1}{4}}{2}$ \\$=\frac{1}{2}$$
$\begin{aligned} &\text { (iii) }\\ &\mathrm{P}\left(\mathrm{A}^{\prime} \mid \mathrm{B}\right)=\mathrm{P}\left(\frac{\mathrm{A}^{\prime} \cap \mathrm{B}}{\mathrm{P}(\mathrm{A})}\right)\\ &=\frac{\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \mathrm{U} \mathrm{B})}{\mathrm{P}(\mathrm{B})}\\ &=\frac{\frac{1}{3}-\frac{1}{4}}{\frac{1}{3}}\\ &=\frac{\frac{1}{12}}{\frac{1}{3}}\\ &=\frac{1}{4} \end{aligned}$
iv)
$\\ P\left(A^{\prime} \mid B'\right)=P\left(\frac{A^{\prime} \cap B^{\prime}}{P\left(B^{\prime}\right)}\right)=\frac{1-P(A U B)}{1-P(B)} \\\\ =\frac{1-[P(A)+P(B)-P(A \cap B)]}{1-P(B)} \\\\ =\frac{1-\left[\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right]}{1-\frac{1}{3}} \\\\ =\frac{1-\left[\frac{5}{6}-\frac{1}{4}\right]}{\frac{2}{3}} \\\\ =\frac{1-\frac{14}{24}}{\frac{2}{3}} \\\\ =\frac{\frac{10}{24}}{\frac{2}{3}} \\\\ =\frac{30}{48} \\\\ =\frac{5}{8}$

Question:8

Three events A, B and C have probabilities $\frac{2}{5},\frac{1}{3}$ and $ \frac{1}{2}$ , respectively. Given that $P(A \cap C)=\frac{1}{5} $ and $ \quad P(B \cap C)=\frac{1}{4}$ , find the values of P (C | B) and P (A'∩ C').

Answer:

Given-
$\begin{aligned} &P(A)=\frac{2}{5}\\ &P(B)=\frac{1}{3} , P(C)=\frac{1}{2}\\ &P(A \cap C)=\frac{1}{5} , P(B \cap C)=\frac{1}{4}\\ &\therefore P(C \mid B)=\frac{P(B \cap C)}{P(B)}\\ &=\frac{\frac{1}{\frac{4}{1}}}{3}\\ &=\frac{3}{4}\\ &\text { By De Morgan's laws: }\\ &(A \cup B)^{\prime}=A' \cap B^{\prime}\\ &(\mathrm{A} \cap \mathrm{B})^{\prime}=\mathrm{A}^{\prime} \cup \mathrm{B}^{\prime} \end{aligned}$
$\\ P\left(A^{\prime} \cap C^{\prime}\right)=P(A \cup C)^{\prime} \\ =1-P(A \cup C) \\ =1-[P(A)+P(C)-P(A \cap C)] \\ =1-\left[\frac{2}{5}+\frac{1}{2}-\frac{1}{5}\right] \\ =1-\left[\frac{4+5-2}{10}\right] \\ =1-\frac{7}{10} \\ =\frac{3}{10}$

Question:9

Let $E_1$ and $E_2$ be two independent events such that P( $E_1$ ) = $p_1$ and P( $E_2$ ) = $p_2$ .
Describe in words of the events whose probabilities are
$\\(i) $P_{1} P_{2}$ \\(ii) $\left(1-P_{1}\right) P_{2}$ \\(iii) $1-\left(1-P_{1}\right)\left(1-P_{2}\right)$ \\$(\mathrm{iv}) \mathrm{P}_{1}+\mathrm{P}_{2}-2 \mathrm{P}_{1} \mathrm{P}_{2}$$
Answer:

$\\$Given $ \mathrm{P}\left(\mathrm{E}_{1}\right)=\mathrm{p}_{1}$ and $\mathrm{P}\left(\mathrm{E}_{2}\right)=\mathrm{p}_{2}$ $\\\\(\mathrm{i}) \mathrm{p}_{1} \mathrm{p}_{2}$ $=P\left(E_{1}\right) \cdot P\left(E_{2}\right)=P\left(E_{1} \cap E_{2}\right) \\$Hence, $E_{1}$ and $E_{2}$ occur simultaneously. \\\\(ii)(1-p_1) $\mathrm{p}_{2}$ $=\mathrm{P}\left(\mathrm{E'}_{1}\right)\cdot \mathrm{P}\left(\mathrm{E}_{2}\right)$ $=P\left(E_{1}^{\prime} \cap E_{2}\right)$\\ Hence, $\mathrm{E}_{1}$ does not occur but $\mathrm{E}_{2}$ occur. \\\\(iii) $1-\left(1-p_{1}\right) \left(1-p_{2}\right)$ $=1-P\left(E'_{1}\right) P\left(E'_{2}\right)$ \\$=1-P\left(E_{1}^{\prime} \cap E_{2}^{\prime}\right)$$
By De Morgan's laws
$\begin{array}{l} (\mathrm{A} \cup \mathrm{B})^{\prime}=\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime} \\ (\mathrm{A} \cap \mathrm{B})^{\prime}=\mathrm{A}^{\prime} \cup \mathrm{B}^{\prime} \\ \end{array}$
Knowing that
$\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{A})^{\prime}=1$$
$\\=1-\left[1-P\left(E_{1} \cup E_{2}\right)\right]$ \\=P\left(E_{1} \cup E_{2}\right)$$
Hence, either $E_{1}$ or $E_{2}$ or both $E_{1}$ and $E_{2}$ occurs.$
(iv)
$\\\mathrm{p}_{1}+\mathrm{p}_{2}-2 \mathrm{p}_{1} \mathrm{p}_{2}$ \\$=\mathrm{P}\left(\mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right)-2 \mathrm{P}\left(\mathrm{E}_{1}\right) \mathrm{P}\left(\mathrm{E}_{2}\right)$ \\$=\mathrm{P}\left(\mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right)-2 \mathrm{P}\left(\mathrm{E}_{1} \cap \mathrm{E}_{2}\right)$ \\$=\mathrm{P}\left(\mathrm{E}_{1} \cup \mathrm{E}_{2}\right)- \mathrm{P}\left(\mathrm{E}_{1} \cap \mathrm{E}_{2}\right)$$
Hence, either $E_{1}$ or $E_{2}$ occurs but not both.$

Question:10

A discrete random variable X has the probability distribution given as below:
$\begin{array}{|l|l|l|l|l|} \hline \mathrm{x} & 0.5 & 1 & 1.5 & 2 \\ \hline \mathrm{P}(\mathrm{x}) & \mathrm{k} & \mathrm{k}^{2} & 2 \mathrm{k}^{2} & \mathrm{k} \\ \hline \end{array}$
(i) Find the value of k
(ii) Determine the mean of the distribution.

Answer:

Given-
$\begin{array}{|l|l|l|l|l|} \hline \mathrm{x} & 0.5 & 1 & 1.5 & 2 \\ \hline \mathrm{P}(\mathrm{x}) & \mathrm{k} & \mathrm{k}^{2} & 2 \mathrm{k}^{2} & \mathrm{k} \\ \hline \end{array}$
$\begin{aligned} &\text { (i) As we know that } \sum_{i=1}^{n} P_{i}=1, \text { where } p i \geq 0\\ &\Rightarrow \mathrm{P}_{1}+\mathrm{P}_{2}+\mathrm{P}_{3}+\mathrm{P}_{4}=1\\ &\Rightarrow k+k^{2}+2 k^{2}+k=1\\ &\Rightarrow 3 k^{2}+2 k-1=0\\ &\Rightarrow 3 k^{2}+3 k-k-1=0 \end{aligned}$
$\\\Rightarrow 3 \mathrm{k}(\mathrm{k}+1)-1(\mathrm{k}+1)=0$ \\$\Rightarrow \mathrm{k}=\frac{1}{3}$ and $\mathrm{k}=-1$ \\since, $k \geq 0,$ we take $k=\frac{1}{3}$ \\(ii) Mean of distribution $(\mu)=E(X)=\sum_{i=1}^{n} x_{i} P_{i}=1$ \\$=0.5(\mathrm{k})+1\left(\mathrm{k}^{2}\right)+1.5\left(2 \mathrm{k}^{2}\right)+2(\mathrm{k})$ \\$=4 k^{2}+2.5 k \quad\left[\because k=\frac{1}{3}\right]$ \\$=4 \times \frac{1}{9}+2.5 \times \frac{1}{3}$ \\$=\frac{4+7.5}{9}=\frac{23}{18}$$

Question:11

Prove that
$(i) P(A)= P(A\cap B) + P(A\cap \bar{B})$
(ii) $P(A\cup B) = P(A\cap B) + P(A\cap \bar{B} ) + P( \bar{A}\cap B)$

Answer:

It has to be proven that $P(A)=P(A \cap B)+P\left(A \cap {\bar{B}}\right)$$
As we know, $A= A \cap S$
Therefore,
$\\\mathrm{s}=\mathrm{A} \cup \mathrm{A}^{\prime}$ and \\$\mathrm{s}=\mathrm{B} \cup \mathrm{B'}$$
$\\A=A \cap\left(B \cup B^{\prime}\right)$ \\$=(A \cap B) \cup\left(A \cap B^{\prime}\right)$$
When two events cannot occur at the same time, they are called mutually exclusive or disjoint events.
Since ( $A \cap B$ ) means A and B both occurring at the same time and
( $A \cap B \textsuperscript{'}$ ) means A and $B\textsuperscript{'}$ both occurring at the same time.
Therefore, it is not possible that $(A \cap B) and (A \cap B\textsuperscript{'})$ occur at the same time.
Hence, $(A \cap B) and (A \cap B\textsuperscript{'})$ are mutually exclusive.
∴ P[(A $\cap$ B) $\cap$ (A $\cap$ B’)] = 0 ….. (1)
Therefore, A = A $\cap$ (B $\cup$ B’)
Knowing that P (A $\cup$ B) = P (A) + P(B) - P (A $\cap$ B)
$\begin{aligned} &\mathrm{P}(\mathrm{A})=\mathrm{P}\left[(\mathrm{A} \cap \mathrm{B}) \cup\left(\mathrm{A} \cap \mathrm{B}^{\prime}\right)\right]\\ &=P(A \cap B)+P\left(A \cap B^{\prime}\right)-P\left[(A \cap B) \cap\left(A \cap B^{\prime}\right)\right]\\ &\text { From }(1)\\ &P(A)=P(A \cap B)+P(A \cap \bar{B}) \end{aligned}$
Hence proved.
ii) It is to be proven that, $P(A \cup B)=P(A \cap B)+P(A \cap \bar{B})+P(\bar{A} \cap B)$

A $\cup$ B means the all the possible outcomes of both A and B.
From the above Venn diagram,
$\mathrm{AUB}=(\mathrm{A} \cap \mathrm{B}) \cup(\mathrm{A} \cap \overline{\mathrm{B}}) \cup(\overline{\mathrm{A}} \cap \mathrm{B})$
When two events cannot occur at the same time, they are called mutually exclusive or disjoint events.
Since (A $\cap$ B) means A and B both occurring at the same time and
(A $\cap$ B’) means A and B’ both occurring at the same time.
(A' $\cap$ B) means A’ and B both occurring at the same time.
Therefore, it is not possible that (A $\cap$ B), (A $\cap$ B’) and ( $\bar{A}$ B) occur at the same time.
Hence these events are mutually exclusive.
$\\ P\left[(A \cap B) \cap\left(A \cap B^{\prime}\right)\right]=0 \ldots . .(1)$ \\$P\left[\left(A \cap B^{\prime}\right) \cap P(\bar{A} \cap B)\right]=0 \ldots . .(2)$ \\$P[(A \cap B) \cap P(\bar{A} \cap B)]=0 \ldots . .(3)$ \\$P\left[(A \cap B) \cap\left(A \cap B^{\prime}\right) \cap P(\bar{A} \cap B)\right]=0 \ldots .(4)$ \\$P(A \cup B)=P[(A \cap B) \cup(A \cap \bar{B}) \cup(\bar{A} \cap B)]$$
It is known that,
$\mathrm{P}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{C})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{C})-\mathrm{P}(\mathrm{B} \cap \mathrm{C})+\mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})$$
Therefore,
$\\P(A U B)=P[(A \cap B) \cup(A \cap \bar{B}) \cup(\bar{A} \cap B)]=P(A \cap B)+P(A \cap \bar{B})+P(\bar{A} \cap B)-P[(A \cap B) \cap(A \cap \bar{B})]-$$P[(A \cap B) \cap(\bar{A} \cap B)]-P[(A \cap \bar{B}) \cap(\bar{A} \cap B)]+P\left[(A \cap B) \cap\left(A \cap B^{\prime}\right) \cap P(\bar{A} \cap B)\right]$$
From (1),(2),(3) and (4) we get,
$P(A U B)=P(A \cap B)+P\left(A \cap {\bar{B}}\right)+P\left(A \cap{ B}\right)$$
Hence Proved

Question:12

If X is the number of tails in three tosses of a coin, determine the standard deviation of X.

Answer:

Given-
Radom variable X is the member of tails in three tosses of a coin
Therefore, X= 0,1,2,3
$\\ \Rightarrow \mathrm{P}(\mathrm{X}=\mathrm{x})=^\mathrm{n}C_{\mathrm{x}}(\mathrm{p})^{\mathrm{n}} \mathrm{q}^{\mathrm{n}-\mathrm{x}} \\ \text { Where } \mathrm{n}=3, \mathrm{p}=\frac{1}{2}, \mathrm{q}=\frac{1}{2} \text { and } \mathrm{x}=0,1,2,3$
$\begin{array}{|l|c|c|l|l|} \hline \mathrm{X} & 0 & 1 & 2 & 3 \\ \hline \mathrm{P}(\mathrm{X}) & \frac{1}{8} & \frac{3}{8} & \frac{3}{8} & \frac{1}{8} \\ \hline \mathrm{XP}(\mathrm{X}) & 0 & \frac{3}{8} & \frac{3}{4} & \frac{3}{8} \\ \hline \mathrm{X}^{2} \mathrm{P}(\mathrm{X}) & 0 & \frac{3}{8} & \frac{3}{2} & \frac{9}{8} \\ \hline \end{array}$
$\begin{aligned} &\text { As we know, Var }(X)=E\left(X^{2}\right)-[E(X)]^{2} \ldots \ldots \text { (i) }\\ &\text { Where, } E\left(X^{2}\right)=\sum_{i=1}^{n} x_{i}^{2} P(x) \text { and } E(X)=\sum_{i=1}^{n} x_{i} P\left(x_{i}\right)\\ &\therefore\left[\mathrm{E}\left(\mathrm{X}^{2}\right)\right]=\sum_{i=1}^{\mathrm{n}} \mathrm{x}_{i}^{2} \mathrm{P}\left(\mathrm{X}_{i}\right)\\ &=0+\frac{3}{8}+\frac{3}{2}+\frac{9}{8}=\frac{24}{8}\\ &=3\\ &\text { And }[\mathrm{E}(\mathrm{X})]^{2}=\left[\sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{x}_{i} \mathrm{P}\left(\mathrm{X}_{\mathrm{i}}\right)\right]^{2} \end{aligned}$
$\\=\left[0+\frac{3}{8}+\frac{3}{4}+\frac{3}{8}\right]^{2}$ \\$=\left[\frac{12}{8}\right]^{2}=\frac{144}{64}$ \\$=\frac{9}{4}$$
Substituting the values of $E(X)^{2}$ and $[E(X)]^{2}$$ in equation (i), we get:
$\\\therefore \operatorname{Var}(\mathrm{X})=3-\frac{9}{4}=\frac{3}{4}$$
And standard deviation of $\\ \mathrm{X}=\sqrt{\operatorname{Var}(\mathrm{X})} =\sqrt{\frac{3}{4}}$$ $=\frac{\sqrt{3}}{2}$$

Question:13

In a dice game, a player pays a stake of Re1 for each throw of a die. She receives Rs 5 if the die shows a 3, Rs 2 if the die shows a 1 or 6, and nothing otherwise. What is the player’s expected profit per throw over a long series of throws?

Answer:

Let X = the random variable of profit per throw
Probability of getting any number on dice is $\frac{1}{6}$ .
Since, she loses Rs 1 on getting any of 2, 4 or 5.
Therefore, at X= -1,
P(X) = P (2) +P(4) +P(5)
$\\P(X)=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}$ \\$=\frac{3}{6}$ \\$=\frac{1}{2}$$
In the same way, =1 if dice shows of either 1 or 6 .
$\mathrm{P}(\mathrm{X})=\mathrm{P}(1)+\mathrm{P}(6)$ \\$P(X)=\frac{1}{6}+\frac{1}{6}$ \\$=\frac{1}{3}$$
and at X=4 if die shows a 3
$\mathrm{P}(\mathrm{X})=\mathrm{P}(3)$ \\$\mathrm{P}(\mathrm{X})=\frac{1}{6}$$
$\begin{aligned} &\begin{array}{|l|c|l|l|} \hline \mathrm{X} & -1 & 1 & 4 \\ \hline \mathrm{P}(\mathrm{X}) & \frac{1}{2} & \frac{1}{3} & \frac{1}{6} \\ \hline \end{array}\\ &\therefore \text { Player's expected profit }=E(X)=X P(X)\\ &=-1 \times \frac{1}{2}+1 \times \frac{1}{3}+4 \times \frac{1}{6}\\ &=\frac{-3+2+4}{6}\\ &=\frac{3}{6}\\ &=\frac{1}{2}=\text { Rs } 0.50 \end{aligned}$

Question:14

Three dice are thrown at the same time. Find the probability of getting three twos’, if it is known that the sum of the numbers on the dice was six.

Answer:

Since three dice are thrown at the same time, the sample space is [n(S)] = 6³= 216.
Let E₁ be the event when the sum of numbers on the dice was six and
E₂ be the event when three twos occur.
$\\ \Rightarrow \quad E_{1}=\{(1,1,4),(1,2,3),(1,3,2),(1,4,1),(2,1,3),(2,2,2,),(2,3,1),(3,1,2),(3,2,1),(4,1,1)\} \\ \Rightarrow n\left(E_{1}\right)=10 \text { and } E_{2}\{2,2,2\} \\ \Rightarrow n\left(E_{2}\right)=1 \\ \text { And }\left(E_{1} \cap E_{2}\right)=1 \\ P\left(E_{1}\right)=\frac{10}{216} \\ P\left(E_{1} \cap E_{2}\right)=\frac{1}{216} \\ \therefore P\left(E_{2} \mid E_{1}\right)=\frac{P\left(E_{1} \cap E_{2}\right)}{P\left(E_{1}\right)} \\ \frac{\frac{1}{216}}{\frac{10}{216}}=\frac{1}{10}$

Question:15

Suppose 10,000 tickets are sold in a lottery each for Re 1. First prize is of Rs 3000 and the second prize is of Rs. 2000. There are three third prizes of Rs. 500 each. If you buy one ticket, what is your expectation.

Answer:

Let X be the variable for the prize
The possibility is of winning nothing, Rs 500, Rs 2000 and Rs 3000.
So, X will take these values.
Since there are 3 third prizes of 500, the probability of winning third prize is $\frac{3}{10000}$ .
1 first prize of 3000, so probability of winning third prize is $\frac{1}{10000}$ .
1 second prize of 2000, so probability of winning third prize is $\frac{1}{10000}$ .
$\begin{aligned} &\begin{array}{|l|l|l|l|l|} \hline \mathrm{X} & 0 & 500 & 2000 & 3000 \\ \hline \mathrm{P}(\mathrm{X}) & \frac{9995}{10000} & \frac{3}{10000} & \frac{1}{10000} & \frac{1}{10000} \\ \hline \end{array}\\ &\text { since, } E(X)=X(P X)\\ &\text { Therefore, }\\ &\mathrm{E}(\mathrm{X})=0 \times \frac{9995}{10000}+\frac{1500}{10000}+\frac{2000}{10000}+\frac{3000}{10000}\\ &=\frac{1500+2000+3000}{10000}\\ &=\frac{6500}{10000}\\ &=\frac{13}{20}=\mathrm{Rs} 0.65 \end{aligned}$

Question:16

A bag contains 4 white and 5 black balls. Another bag contains 9 white and 7 black balls. A ball is transferred from the first bag to the second and then a ball is drawn at random from the second bag. Find the probability that the ball drawn is white.

Answer:

Given-
$W_1$ = [4 white balls] and $B_1$ = [5 black balls]
$W_2$ = [9 white balls] and $B_2$ = [7 black balls]
Let $E_1$ be the event that the ball transferred from the first bag is white and
$E_2$ be the event that the ball transferred from the bag is black.
E is the event that the ball drawn from the second bag is white.
$\begin{aligned} &\therefore \mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{1}\right)=\frac{0}{17}, \mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{2}\right)=\frac{9}{17}\\ &\text { And }\\ &\mathrm{P}\left(\mathrm{E}_{1}\right)=\frac{4}{9} \text { and } \mathrm{P}\left(\mathrm{E}_{2}\right)=\frac{5}{9}\\ &\therefore P(E)=P\left(E_{1}\right) \cdot P\left(E \mid E_{1}\right)+P\left(E_{2}\right) \cdot P\left(E \mid E_{2}\right)\\ &=\frac{4}{9} \times \frac{10}{17}+\frac{5}{9} \times \frac{9}{17}\\ &=\frac{40+45}{153}\\ &=\frac{85}{153}\\ &=\frac{5}{9} \end{aligned}$

Question:17

Bag I contains 3 black and 2 white balls, Bag II contains 2 black and 4 white balls. A bag and a ball are selected at random. Determine the probability of selecting a black ball.

Answer:

Given-
Bag I= [3Black, 2White], Bag II= [2 black, 4 white]
Let E₁ be the event that bag I is selected
E₂ be the event that bag II is selected
E₃ be the event that a black ball is selected
Therefore,
$\\ \Rightarrow P(E 1)=P(E 2)=\frac{1}{2} \\ P\left(E \mid E_{2}\right)=\frac{3}{5} \\ P\left(E \mid E_{2}\right)=\frac{2}{6}=\frac{1}{3} \\ \therefore P(E)=P\left(E_{1}\right) \cdot P\left(E \mid E_{1}\right)+P\left(E_{2}\right) \cdot P\left(E \mid E_{2}\right) \\ =\frac{1}{2} \times \frac{3}{5}+\frac{1}{2} \times \frac{2}{6} \\ =\frac{3}{10}+\frac{2}{12} \\ =\frac{18+10}{60} \\ =\frac{28}{60} \\ =\frac{7}{15}$

Question:18

A box has 5 blue and 4 red balls. One ball is drawn at random and not replaced. Its colour is also not noted. Then another ball is drawn at random. What is the probability of second ball being blue?

Answer:

Given-
The box has 5 blue and 4 red balls.
Let E₁ be the event that first ball drawn is blue
E₂be the event that the first ball drawn is red and
E be the event that second ball drawn is blue.
$\\ P\left(E_{1}\right)=\frac{5}{9} \\ P\left(E_{2}\right)=\frac{4}{9} \\ P\left(E \mid E_{1}\right)=\frac{4}{8} \\ P\left(E \mid E_{2}\right)=\frac{5}{8} \\ \therefore P(E)=P\left(E_{1}\right) . P\left(E \mid E_{1}\right)+P\left(E_{2}\right) \cdot P\left(E \mid E_{2}\right) \\ =\frac{5}{9} \times \frac{4}{8}+\frac{4}{9} \times \frac{5}{8} \\ =2\left(\frac{20}{72}\right) \\ =\frac{40}{72} \\ =\frac{5}{9}$

Question:19

Four cards are successively drawn without replacement from a deck of 52 playing cards. What is the probability that all the four cards are kings?

Answer:

Let E₁, E₂, E₃and E₄ be the events that the first, second, third and fourth card is king respectively.
As we know, there are 4 kings,
$\mathrm{P}\left(\mathrm{E}_{1}\right)=\frac{4}{52}$
when 1 king is taken out, there are 3 kings and total 51 cards left.
Therefore, probability of drawing a king when one king has been taken out is:
$\mathrm{So}, \mathrm{P}\left(\mathrm{E}_{2} \mid \mathrm{E}_{1}\right)=\frac{3}{51}$$
When 2 kings are taken out, there are 2 kings and 50 cards left. Therefore, probability of drawing a king when two kings have been taken out is:
$P\left(E_{3} \mid E_{1} \cap E_{2}\right)=\frac{2}{50}$$
When 3 kings are taken out, 1 king and 49 cards are left.
Therefore, probability of drawing a king when three kings have been taken out is:
$\mathrm{P}\left[\mathrm{E}_{4} \mid\left(\mathrm{E}_{1} \cap \mathrm{E}_{2} \cap \mathrm{E}_{3} \cap \mathrm{E}_{4}\right)\right]=\frac{1}{49}$$
Probability that all 4 cards are king is:
$\\\left.\mathrm{P}\left(\mathrm{E}_{1} \cap \mathrm{E}_{2} \cap \mathrm{E}_{3} \cap \mathrm{E}_{4}\right)=\mathrm{p}\left(\mathrm{E}_{1}\right) . \mathrm{P}\left(\mathrm{E}_{2} \mid \mathrm{E}_{1}\right) . \mathrm{P}\left(\mathrm{E}_{3} \mid \mathrm{E}_{1} \cap \mathrm{E} 2\right) . \mathrm{P}\left[\mathrm{E}_{4}\right]\left(\mathrm{E}_{1} \mathrm{NE}_{2} \cap \mathrm{E}_{3} \mathrm{NE}_{4}\right)\right]$ $=\frac{4}{52} \times \frac{3}{51} \times \frac{2}{50} \times \frac{1}{49}=\frac{24}{52 \times 51 \times 50 \times 49}$ \\$=\frac{1}{13 \times 14 \times 25 \times 49}$ \\$=\frac{1}{270725}$$

Question:20

A die is thrown 5 times. Find the probability that an odd number will come up exactly three times.

Answer:

Given-
n=5, Odd numbers = 1,3,5
$\\p($ odd number $)=P(1)+P(3)+P(5)$ \\$=\left(\frac{1}{6}+\frac{1}{6}+\frac{1}{6}\right)$ \\$=\frac{3}{6}=\frac{1}{2}$$
and $\mathrm{q}=1-\mathrm{p}=1-\frac{1}{2}=\frac{1}{2}$$
also, r=3
$\\ \therefore \mathrm{P}(\mathrm{X}=\mathrm{r})=\mathrm{n}_{\mathrm{c}_{\mathrm{r}}}(\mathrm{p})^{\mathrm{r}} \mathrm{q}^{\mathrm{r}} \mathrm{q}^{\mathrm{n}-\mathrm{r}}$ \\$=^5C_{3}\left(\frac{1}{2}\right)^{3}\left(\frac{1}{2}\right)^{5-3}$ \\$=\frac{5 !}{3 ! 2 !} \times \frac{1}{8} \times \frac{1}{4}$ \\$=\frac{10}{32}$ \\$=\frac{5}{16}$$

Question:21

Ten coins are tossed. What is the probability of getting at least 8 heads?

Answer:

Let X = the random variable for getting a head.
Here, n=10, r≥8
r=8,9,10
$\begin{aligned} &\mathrm{p}=\frac{1}{2}, \mathrm{q}=\frac{1}{2}\\ &\text { It is known to us that, }\\ &\mathrm{P}(\mathrm{X}=\mathrm{r})=\mathrm{n}_{\mathrm{c}_{\mathrm{r}}}(\mathrm{p})^{\mathrm{r}} \mathrm{q}^{\mathrm{r}} \mathrm{q}^{\mathrm{n}-\mathrm{r}}\\ &\therefore P(X=r)=P(r=8)+P(r=9) + P(r=10)\\ &=^{10}C_{8}\left(\frac{1}{2}\right)^{10} (\frac{1}{2})^{10-8}+^{10}{c_{9}}\left(\frac{1}{2}\right)^{9}\left(\frac{1}{2}\right)^{10-9}+^{10}c_{10}\left(\frac{1}{2}\right)^{10} \frac{1}{2}^{10-10}\\ &=\frac{10 !}{8 ! 2 !}\left(\frac{1}{2}\right)^{10}+\frac{10 !}{9 ! 1 !}\left(\frac{1}{2}\right)^{10}+\frac{10 !}{0 ! 10 !}\left(\frac{1}{2}\right)^{10}\\ &=\left(\frac{1}{2}\right)^{10}\left[\frac{10 \times 9}{2}+10+1\right]\\ &=\left(\frac{1}{2}\right)^{10} \times 56=\frac{1}{2^{7} \times 2^{3}} \times 56\\ &=\frac{7}{128} \end{aligned}$

Question:22

The probability of a man hitting a target is 0.25. He shoots 7 times. What is the probability of his hitting at least twice?

Answer:

Given-
The man shoots 7 times, therefore, n=7
And probability of hitting target is
$\\ \mathrm{P}=0.25 \mathrm{or } \frac{1}{4} \\ \mathrm{q}=1-\frac{1}{4}=\frac{3}{4}, \mathrm{r} \geq 2 \\ \text { Where, } \mathrm{P}(\mathrm{X})=^\mathrm{n}{\mathrm{c_r}}(\mathrm{p})^{\mathrm{r}} \mathrm{q}^{\mathrm{n}-\mathrm{r}} \\ \therefore \mathrm{P}(\mathrm{X}=\mathrm{r} \geq 2)=1-[\mathrm{p}(\mathrm{r}=0)+\mathrm{P}(\mathrm{r}=1)]$
$\\ \Rightarrow 1-\left[7 c_{0}\left(\frac{1}{4}\right)^{0}\left(\frac{3}{4}\right)^{7-0}+7 c_{1}\left(\frac{1}{4}\right)^{1}\left(\frac{3}{4}\right)^{7-1}\right] \\ =1-\left[\left(\frac{3}{4}\right)^{7}+{7}\left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^{6}\right] \\ =1-\left[\left(\frac{3}{4}\right)^{6}\left(\frac{3}{4}+\frac{7}{4}\right)\right] \\ =1-\left[\frac{3^{6}}{4^{6}} \times \frac{10}{4}\right] \\ =\quad \\ =1-\left[\frac{7290}{16384}\right] \\ =\frac{4547}{8192}$

Question:23

A lot of 100 watches is known to have 10 defective watches. If 8 watches are selected (one by one with replacement) at random, what is the probability that there will be at least one defective watch?

Answer:

Given-

There are 10 defective watches in 100 watches

The probability of defective watch from a lot of 100 watch $=\frac{10}{100}=\frac{1}{10}$$

$P=\frac{1}{10},q=\frac{9}{10}, n=8$ and $r \geq 1$$
As we know that

$\\\mathrm{P}(\mathrm{X})=^\mathrm{n}{\mathrm{c}_{\mathrm{r}}}(\mathrm{p})^{\mathrm{r}} \mathrm{q}^{\mathrm{n}-\mathrm{r}}$ \\$\therefore \mathrm{P}=(\mathrm{r} \geq 1)=1-\mathrm{P}(\mathrm{r}=0)$ \\$=1-^8{c_{0}}\left(\frac{1}{10}\right)^{0}\left(\frac{9}{10}\right)^{8-0}$ \\$=1-\frac{8 !}{0 ! 8 !} \times\left(\frac{9}{10}\right)^{8}$ \\$=1-\left(\frac{9}{10}\right)^{8}$$

Question:24

Consider the probability distribution of a random variable X:
$\begin{array}{|l|l|l|l|l|l|} \hline \mathrm{X} & 0 & 1 & 2 & 3 & 4 \\ \hline \mathrm{P}(\mathrm{X}) & 0.1 & 0.25 & 0.3 & 0.2 & 0.15 \\ \hline \end{array}$
Calculate (i) $V\left ( \frac{X}{2} \right )$ (ii) Variance of X.

Answer:

Given-
$\begin{aligned} &\begin{array}{|l|l|l|l|l|l|} \hline \mathrm{X} & 0 & 1 & 2 & 3 & 4 \\ \hline \mathrm{P}(\mathrm{X}) & 0.1 & 0.25 & 0.3 & 0.2 & 0.15 \\ \hline \mathrm{XP}(\mathrm{X}) & 0 & 0.25 & 0.6 & 0.6 & 0.60 \\ \hline \mathrm{X}^{2} \mathrm{P}(\mathrm{X}) & 0 & 0.25 & 1.2 & 1.8 & 2.40 \\ \hline \end{array}\\ &\operatorname{Var}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^{2}\right) \cdot-[\mathrm{E}(\mathrm{X})]^{2}\\ &\text { Where, }\\ &E(X)=\mu=\sum_{i=1}^{n} x_{i} P\left(x_{i}\right) \end{aligned}$
$\begin{aligned} &\mathrm{E}(\mathrm{X})^{2}=\sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{x}_{\mathrm{i}}^{2} \mathrm{P}\left(\mathrm{x}_{\mathrm{i}}\right)\\ &\therefore E(x)=0+0.25+0.6+0.6+0.60=2.05\\ &\text { And } E(X)^{2}=0+0.25+1.2+1.8+2.40=5.65\\ &\text { (i) } V\left[\frac{x}{2}\right]\\ &\text { It is known that } \operatorname{Var}(a x)=a^{2} \operatorname{yar}(x)\\ &\Rightarrow \mathrm{V}\left[\frac{\mathrm{X}}{2}\right]=\frac{1}{4} \mathrm{~V}(\mathrm{X})\\ &=\frac{1}{4}\left[5.65-(2.05)^{2}\right] \end{aligned}$
$\begin{aligned} &=\frac{1}{4}[5.65-4.2025]\\ &=\frac{1}{4} \times 1.4475\\ &=0.361875\\ &\text { (ii) } \mathrm{V}(\mathrm{X})\\ &\operatorname{Var}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^{2}\right)-[\mathrm{E}(\mathrm{X})]^{2}\\ &=5.65-(2.05)^{2}\\ &=5.65-4.2025\\ &=1.4475 \end{aligned}$

Question:25

The probability distribution of a random variable X is given below:
$\begin{array}{|l|l|l|l|l|} \hline \mathrm{X} & 0 & 1 & 2 & 3 \\ \hline \mathrm{P}(\mathrm{X}) & \mathrm{K} & \frac{\mathrm{k}}{2} & \frac{\mathrm{k}}{4} & \frac{\mathrm{k}}{8} \\ \hline \end{array}$
(i) Determine the value of k.

(ii) Determine P (X ≤ 2) and P (X > 2)
(iii) Find P (X ≤ 2) + P (X > 2).

Answer:

Given-

$\begin{array}{|l|l|l|l|l|} \hline \mathrm{X} & 0 & 1 & 2 & 3 \\ \hline \mathrm{P}(\mathrm{X}) & \mathrm{K} & \frac{\mathrm{k}}{2} & \frac{\mathrm{k}}{4} & \frac{\mathrm{k}}{8} \\ \hline \end{array}$
(i) since,
$\\\sum_{i=1}^{n} P_{i}=1,2, \ldots, n$ and $P_{i} \geq 0$ \\$\therefore \mathrm{k}+\frac{\mathrm{k}}{2}+\frac{\mathrm{k}}{4}+\frac{\mathrm{k}}{8}=1$ \\$\Rightarrow 8 k+4 k+2 k+k=8$ \\$\therefore 15 k=8$ \\$k=\frac{8}{15}$$
(ii)
$\\\mathrm{P}(\mathrm{X} \leq 2)=\mathrm{P}(0)+\mathrm{P}(1)+\mathrm{P}(2)$ \\$=\mathrm{k}+\frac{\mathrm{k}}{2}+\frac{\mathrm{k}}{4}$ \\$=\left(\frac{4 \mathrm{k}+2 \mathrm{k}+\mathrm{k}}{4}\right)$$
$\\ =\frac{7 \mathrm{k}}{4}=\frac{7}{4} \times \frac{8}{15}=\frac{14}{15} \\ \text { and } \mathrm{P}(\mathrm{X}>2)=\mathrm{P}(3) \\ =\frac{\mathrm{k}}{8} \\ =\frac{1}{8} \times \frac{8}{15} \\ =\frac{1}{15} \\ \text { (iii) } \mathrm{P}(\mathrm{X} \leq 2)+\mathrm{P}(\mathrm{X}>2)=\frac{14}{15}+\frac{1}{15}=1$

Question:26

For the following probability distribution determine standard deviation of the random variable X.
$\begin{array}{|l|l|l|l|} \hline \mathrm{X} & 2 & 3 & 4 \\ \hline \mathrm{P}(\mathrm{X}) & 0.2 & 0.5 & 0.3 \\ \hline \end{array}$

Answer:

Given-

$\begin{aligned} &\begin{array}{|l|l|l|l|} \hline \mathrm{X} & 2 & 3 & 4 \\ \hline \mathrm{P}(\mathrm{X}) & 0.2 & 0.5 & 0.3 \\ \hline \mathrm{XP}(\mathrm{X}) & 0.4 & 1.5 & 1.2 \\ \hline \mathrm{X}^{2} \mathrm{P}(\mathrm{X}) & (4 \times 0.2) & (9 \times 0.5) & 16 \times 0.3 \\ & =0.8 & =4.5 & =4.8 \\ \hline \end{array}\\ &\text { It is known that standard deviation of }\\ &\mathrm{X}=\sqrt{\operatorname{VarX}}\\ &\text { Where, Var }=E\left(X^{2}\right)-[E(X)]^{2}\\ &=\sum_{i=1}^{n} x_{i}^{2} P\left(x_{i}\right)-\left[\sum_{i=1}^{n} x_{i} P_{i}\right]^{2} \end{aligned}$
$\begin{aligned} &\therefore \operatorname{Var} \mathrm{X}=[0.8+4.5+4.8]-[0.5+1.5+1.2]^{2}\\ &=10.1-(3.1)^{2}\\ &=10.1-9.61\\ &=0.49\\ &\text { Hence, standard deviation of }\\ &\mathrm{X}=\sqrt{\operatorname{Var} \mathrm{X}}=\sqrt{0.49}=0.7 \end{aligned}$

Question:27

A biased die is such that P (4) = 1/10 and other scores being equally likely. The die is tossed twice. If X is the ‘number of fours seen’, find the variance of the random variable X.

Answer:

Given-
X= number of four seen
On tossing to die, X=0,1,2
$\mathrm{P}_{4}=\frac{1}{10}\ \ , \mathrm{P}_{\mathrm{not} 4}=\frac{9}{10}$$
Therefore, $\mathrm{P}(\mathrm{X}=0)=\mathrm{P}_{\mathrm{not} 4} \cdot \mathrm{P}_{\mathrm{not} 4}=\frac{9}{10} \times \frac{9}{10}=\frac{81}{100}$$
$\\ \mathrm{P}(\mathrm{X}=1)=\mathrm{P}_{\mathrm{not} 4} \cdot \mathrm{p}_{4} +P_4\cdot \mathrm{P}_{\mathrm{not} 4}=\frac{9}{10} \times \frac{1}{10}+\frac{1}{10} \times \frac{9}{10}=\frac{18}{100}$ \\$\mathrm{P}(\mathrm{X}=2)=\mathrm{P}_{4} \cdot \mathrm{P}_{4} =\frac{1}{10} \times \frac{1}{10}=\frac{1}{100}$$
Thus, the table is derived
$\begin{aligned} &\begin{array}{|l|l|l|l|} \hline \mathrm{X} & 0 & 1 & 2 \\ \hline \mathrm{P}(\mathrm{X}) & \frac{81}{100} & \frac{18}{100} & \frac{1}{100} \\ \hline \mathrm{XP}(\mathrm{X}) & 0 & \frac{18}{100} & \frac{2}{100} \\ \hline \mathrm{X}^{2} \mathrm{p}(\mathrm{x}) & 0 & \frac{18}{100} & \frac{4}{100} \\ \hline \end{array}\\ &\therefore \operatorname{Var}(\mathrm{X})=\mathrm{E}(\mathrm{X})^{2}-\left[\mathrm{E}(\mathrm{X})^{2}\right]=\mathrm{X}^{2} \mathrm{P}(\mathrm{x})-[\mathrm{XP}(\mathrm{X})]^{2}\\ &\Rightarrow\left[0+\frac{18}{100}+\frac{4}{100}\right]-\left[0+\frac{18}{100}+\frac{2}{100}\right]^{2}\\ &=\frac{22}{100}-\left(\frac{20}{100}\right)^{2}=\frac{11}{50}-\frac{1}{25}\\ &=\frac{11-2}{50}=\frac{9}{50}=\frac{18}{100}=0.18 \end{aligned}$

Question:28

A die is thrown three times. Let X be ‘the number of twos seen’. Find the expectation of X.

Answer:

Given-
X= no. of twos seen
Therefore, on throwing a die three times, we will have X=0,1,2,3
$\begin{aligned} &\therefore P(X=0)=P_{n o t 2} \cdot P_{n o t2} \cdot P_{n o t 2}=\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6}=\frac{125}{216}\\ &P(X=1)=\left(\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}\right)+\left(\frac{5}{6} \times \frac{1}{6} \times \frac{5}{6}\right)+\left(\frac{1}{6} \times \frac{5}{6} \times \frac{5}{6}\right)\\ &=\frac{25}{36} \times \frac{3}{6}\\ &=\frac{25}{72}\\\end{aligned}$
$\\P(X=2) =\left(\frac{5}{6} \times \frac{1}{6} \times \frac{1}{6}\right)+\left(\frac{1}{6} \times \frac{1}{6} \times \frac{5}{6}\right)+\left(\frac{1}{6} \times \frac{5}{6} \times \frac{1}{6}\right) \\ =3\left(\frac{5}{6} \times \frac{1}{6} \times \frac{1}{6}\right) \\ =\frac{5}{72} \\ P(X=3)=P_{2} \cdot P_{2} \cdot P_{2} \\ =\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} \\ =\frac{1}{216}$
$\begin{aligned} &\text { As it is known that, }\\ &\mathrm{E}(\mathrm{X})=\Sigma \mathrm{XP}(\mathrm{X})=0 \times \frac{125}{216}+1 \times \frac{25}{72}+2 \times \frac{15}{216}+3 \times \frac{1}{216}\\ &=\frac{75+30+3}{216}\\ &=\frac{108}{216}\\ &=\frac{1}{2} \end{aligned}$

Question:29

Two biased dice are thrown together. For the first die P (6) = 1/2. the other scores being equally likely while for the second die, P (1) = 2/5 and the other scores are equally likely. Find the probability distribution of ‘the number of ones seen’.

Answer:

Given-
$\begin{aligned} &\text { For first die, }\\ &\mathrm{P}(6)=\frac{1}{2} \text { and } \mathrm{P}\left(6^{\prime}\right)=\frac{1}{2}\\ &\Rightarrow P(1)+P(2)+P(3)+P(4)+P(5)=\frac{1}{2}\\ &[\because \mathrm{P}(1)=\mathrm{P}(2)=\mathrm{P}(3)=\mathrm{P}(4)=\mathrm{P}(5)]\\ &\Rightarrow 5 \mathrm{P}(1)=\frac{1}{2}\\ &\Rightarrow \mathrm{P}(1)=\frac{1}{10}\\ &\Rightarrow \mathrm{P}\left(1^{\prime}\right)=1-\mathrm{P}(1)\\ &=1-\frac{1}{10}=\frac{9}{10} \end{aligned}$
For the second die
$P(1)=\frac{2}{5}$ and $P\left(1^{\prime}\right)=1-\frac{2}{5}=\frac{3}{5}$$
Let x be the number of ones seen
For X=0,
$\\\mathrm{P}(\mathrm{X}=0)=\mathrm{P}\left(1^{\prime}\right) \cdot\mathrm{P}\left(1^{\prime}\right)=\frac{9}{10} \times \frac{3}{5}=\frac{27}{50}=0.54$ \\$\mathrm{P}(\mathrm{X}=1)=\frac{9}{10} \times \frac{2}{5}+\frac{1}{10} \times \frac{3}{5}$ \\$=\frac{18}{50}+\frac{3}{50}=\frac{21}{50}$ \\$=0.42$ \\$\mathrm{P}(\mathrm{X}=2)=\mathrm{P}(1) \cdot \mathrm{P}(1)$$
$\begin{aligned} &=\frac{1}{10} \times \frac{2}{5}\\ &=0.04\\ &\text { Hence, the required probability distribution is as follows }\\ &\begin{array}{|l|l|l|l|} \hline \mathrm{X} & 0 & 1 & 2 \\ \hline \mathrm{P}(\mathrm{X}) & 0.54 & 0.42 & 0.04 \\ \hline \end{array} \end{aligned}$

Question:30

Two probability distributions of the discrete random variable X and Y are given below.
$\begin{array}{|l|l|l|l|l|} \hline \mathrm{X} & 0 & 1 & 2 & 3 \\ \hline \mathrm{P}(\mathrm{X}) & \frac{1}{5} & \frac{2}{5} & \frac{1}{5} & \frac{1}{5} \\ \hline \mathrm{Y} & 0 & 1 & 2 & 3 \\ \hline \mathrm{P}(\mathrm{y}) & \frac{1}{5} & \frac{3}{10} & \frac{2}{5} & \frac{1}{10} \\ \hline \end{array}$
Prove that $E(Y^2)=2E(X)$ .

Answer:

To prove that- $E(Y)^2=2E(X)$
Taking LHS of equation (i), we have:
$E(Y^{2})=Y^{2} P(Y)$$
$\\=0 \times \frac{1}{5}+1 \times \frac{3}{10}+4 \times \frac{2}{5}+9 \times \frac{1}{10}$ \\$=\frac{3}{10}+\frac{8}{5}+\frac{9}{10}=\frac{28}{10}=\frac{14}{5}$ \\$\Rightarrow \mathrm{E}(\mathrm{Y}^2)=\frac{14}{5}$$
Taking RHS of equation (i) we get:
$E(X)=X P(X)$$
$\\ =0 \times \frac{1}{5}+1 \times \frac{2}{5}+3 \times \frac{1}{5}=\frac{7}{5}$ \\$2 \mathrm{E}(\mathrm{X})=2\left(\frac{7}{5}\right)$ \\$=\frac{14}{5}$$
Thus, from equations (ii) and (iii), we get:
$E\left(Y^{2}\right)=2 E(X)$$
Hence proved.

Question:31

A factory produces bulbs. The probability that any one bulb is defective is 1/50 and they are packed in boxes of 10. From a single box, find the probability that
(i) none of the bulbs is defective
(ii) exactly two bulbs are defective
(iii) more than 8 bulbs work properly

Answer:

Let X be the random variable which denotes that the bulb is defective.
And $n=10, p=\frac{1}{50}$ and $P(X=r)=n_{c_{r}}(p)^{r} q^{n-r}$$
(j) None of the bulbs is defective i.e., r=0
$\therefore P(X=r)=P_{0}=10_{c_{0}}\left(\frac{1}{50}\right)^{0}\left(\frac{49}{50}\right)^{10-0}=\left(\frac{49}{50}\right)^{10}$$
(ii)Exactly two bulbs are defective i.e., r=2
$\mathrm{P}(\mathrm{X}=\mathrm{r})=\mathrm{P}_{2}=10 \mathrm{c}_{2}\left(\frac{1}{50}\right)^{2}\left(\frac{49}{50}\right)^{10-2}$
$\begin{aligned} &=\frac{10 !}{8 ! 2 !}\left(\frac{1}{50}\right)^{2} \cdot\left(\frac{49}{50}\right)^{8}\\ &=45 \times\left(\frac{1}{50}\right)^{10} \times(49)^{8}\\ &\text { (iii)More than } 8 \text { bulbs work properly i.e., there are less than } 2 \text { bulbs that are defective. }\\ &\text { Therefore, } r<2 \Rightarrow r=0,1\\ &\therefore P(X=r)=P(r<2)=P(0)+P(1)\\ &=10_{c_{0}}\left(\frac{1}{50}\right)^{0}\left(\frac{49}{50}\right)^{10-0}+10_{c_{1}}\left(\frac{1}{50}\right)^{1}\left(\frac{49}{50}\right)^{10-1}\\&=\left(\frac{49}{50}\right)^{10}+\frac{1}{5} \times\left(\frac{49}{50}\right)^{9}\\ &=\left(\frac{49}{50}+\frac{10}{50}\right)\left(\frac{49}{50}\right)^{9}\\ &=\frac{59(49)^{9}}{(50)^{10}} \end{aligned}$

Question:32

Suppose you have two coins which appear identical in your pocket. You know that one is fair and one is 2-headed. If you take one out, toss it and get a head, what is the probability that it was a fair coin?

Answer:

Let E₁ be the event that a fair coin is drawn
E₂ be the event that two headed coin is drawn
E be the event that tossed coin get a head
$\begin{aligned} & \mathrm{P}\left(\mathrm{E}_{1}\right)=\frac{1}{2}, \mathrm{P}\left(\mathrm{E}_{2}\right)=\frac{1}{2}, \mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{1}\right)=\frac{1}{2}, \text { And } \mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{2}\right)=1\\ &\text { Using Bayes' theorem, we have }\\ &\mathrm{P}\left(\mathrm{E}_{1} \mid \mathrm{E}\right)=\frac{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{1}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \cdot \mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{2}\right)}\\ &=\frac{\frac{1}{2} \times \frac{1}{2}}{\frac{1}{2} \times \frac{1}{2}+\frac{1}{2} \times 1}\\ &=\frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{2}}\\ &=\frac{\frac{1}{4}}{\frac{3}{4}}\\ &=\frac{1}{3} \end{aligned}$

Question:33

Suppose that 6% of the people with blood group O are left handed and 10% of those with other blood groups are left handed 30% of the people have blood group O. If a left-handed person is selected at random, what is the probability that he/she will have blood group O?

Answer:

Given-
$\begin{array}{|l|l|l|} \hline & \begin{array}{l} \text { Blood group } \\ \text { 'O } \end{array} & \begin{array}{l} \text { Other than } \\ \text { blood group } \\ \text { 'O' } \end{array} \\ \hline \begin{array}{l} \text { I. Number of } \\ \text { people } \end{array} & 30 \% & 70 \% \\ \hline \begin{array}{l} \text { II. Percentage } \\ \text { of left-handed } \\ \text { people } \end{array} & 6 \% & 10 \% \\ \hline \end{array}$
Let E₁ be the event that the person selected is of group O
E₂ be the event that the person selected is of other than blood group O
And E₃ be the event that the person selected is left handed
∴P(E₁) =0.30, P(E₂) =0.70
P(E₃|E₁) = 0.060 And P(E₃|E₂) =0.10
Using bayes’ theorem, we have:
$\\ P\left(E_{1} \mid E_{3}\right)=\frac{P\left(E_{1}\right) \cdot P\left(E_{3} \mid E_{1}\right)}{P\left(E_{1}\right) \cdot P\left(E_{3} \mid E_{1}\right)+P\left(E_{2}\right) \cdot P\left(E_{3} \mid E_{2}\right)} \\ =\frac{0.30 \times 0.06}{0.30 \times 0.06+0.70 \times 0.10} \\ =\frac{0.0180}{0.0180+0.0700} \\ =\frac{0.0180}{0.0880} \\ =\frac{180}{880} \\ =\frac{9}{44}$

Question:34

Two natural numbers r, s are drawn one at a time, without replacement from the set S= {1, 2, 3, ...., n}. Find P [r ≤ p|s ≤ p], where p ∈ S.

Answer:

The notation P [r ≤ p| s ≤ p] means that
P (r ≤p) *given that s ≤ p
Since we know s ≤ p , then it means that s is drawn first.
Let us have n numbers before s is drawn:
(1 . . s …. . p . . .. n)
After s is drawn,
[ 1 ... p] has one element missing, so there are (p-1) elements.
Also, there is one element missing from the entire set, so there are (n-1) altogether.

$p( r is among p-1 elements $)=\frac{p-1}{n-1}$
Among (1 . . s …. . p) the probability of drawing s is $\frac{p}{p}$ .

$P[r \leq p l s \leq p]$$ is the probability that $r \leq p$ when $s \leq p$.$
Therefore, $P[r \leq p \mid s \leq p]=\frac{(p-1)}{(n-1)} \times \frac{p}{p}$$
$=\frac{(p-1)}{(n-1)}$$

Question:35

Find the probability distribution of the maximum of the two scores obtained when a die is thrown twice. Determine also the mean of the distribution.

Answer:

Let X be the random variable score obtained when a die is thrown twice.
∴ X= 1,2,3,4,5,6
The sample space is
$\\S= {(1,1), (1,2)... (2,1) (2,2).. (3,1) (3,2) (3,3),..... (6,6)} \\\therefore P(X=1)=\frac{1}{6} \times \frac{1}{6}=\frac{1}{36}$ \\$P(X=2)=\frac{1}{6} \times \frac{1}{6}+\frac{1}{6} \times \frac{1}{6}+\frac{1}{6} \times \frac{1}{6}=\frac{3}{36}$ \\$\mathrm{P}(\mathrm{X}=3)=\frac{5}{36}$$
In the same way,
$\\P(X=4)=\frac{7}{36} \\ P(X=5)=\frac{9}{36}$ \\$P(X=6)=\frac{11}{36}$$
Therefore, the required distribution is,
$\begin{aligned} &\begin{array}{|l|l|l|l|l|l|l|} \hline \mathrm{X} & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \mathrm{P}(\mathrm{X}) & \frac{1}{36} & \frac{3}{36} & \frac{5}{36} & \frac{7}{36} & \frac{9}{36} & \frac{11}{36} \\ \hline \end{array}\\ &\text { Also, it is known that, }\\ &\operatorname{Mean}\left\{\mathrm{E}(\mathrm{X})=\Sigma \mathrm{XP}(\mathrm{X})=\frac{1}{36}+\frac{6}{36}+\frac{15}{36}+\frac{28}{36}+\frac{45}{36}+\frac{66}{36}\right.\\ &=\frac{161}{36} \end{aligned}$

Question:36

The random variable X can take only the values 0, 1, 2. Given that P (X = 0) = P (X = 1) = p and that $E(X^2) = E[X]$ , find the value of p.

Answer:

Given-
X=0,1,2 and P(X) at X=0 and 1,
Let X=2, P(X) is x.
$\\\Rightarrow p+p+x=1$ \\$\Rightarrow x=1-2 p$$
The following distribution is obtained
$\begin{array}{|l|l|l|l|} \hline \mathrm{X} & 0 & 1 & 2 \\ \hline \mathrm{P}(\mathrm{X}) & \mathrm{P} & \mathrm{P} & 1-2 \mathrm{p} \\ \hline \end{array}$
$\\\therefore E(X)=X P(X)$ \\$=0 \times \mathrm{P}+1 \times \mathrm{P}+2(1-2 \mathrm{P})$ \\$=P+2-4 P=2-3 P$$
And $E(X)^{2}=X^{2} P(X)$$
$\\=0 \times p+1 \times P+4 \times(1-2 P)$ \\$=\mathrm{P}+4-8 \mathrm{P}=4-7 \mathrm{P}$$
And, as we know that $E\left(X^{2}\right)=E(X)$$
$\\\Rightarrow 4-7 p=2-3 p$ \\$\Rightarrow 4 p=2$ \\$\Rightarrow p=\frac{1}{2}$$

Question:37

Find the variance of the distribution:
$\begin{array}{|l|l|l|l|l|l|l|} \hline \mathrm{X} & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline \mathrm{P}(\mathrm{X}) & \frac{1}{6} & \frac{5}{18} & \frac{2}{9} & \frac{1}{6} & \frac{1}{9} & \frac{1}{18} \\ \hline \end{array}$

Answer:

Given-
$\begin{aligned} &\begin{array}{|l|c|c|c|c|c|c|} \hline \mathrm{X} & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline \mathrm{P}(\mathrm{X}) & \frac{1}{6} & \frac{5}{18} & \frac{2}{9} & \frac{1}{6} & \frac{1}{9} & \frac{1}{18} \\ \hline \mathrm{X} \mathrm{P}(\mathrm{X}) & 0 & \frac{5}{18} & \frac{4}{9} & \frac{1}{2} & \frac{4}{9} & \frac{5}{18} \\ \hline \mathrm{X}^{2} \mathrm{P}(\mathrm{X}) & 0 & \frac{5}{18} & \frac{8}{9} & \frac{3}{2} & \frac{16}{9} & \frac{25}{18} \\ \hline \end{array}\\ &\therefore \text { Variance }=E(X)^{2}-[E(X)]^{2}=X^{2} P(X)-[X P(X)]^{2} \end{aligned}$
$\\ =\left[0+\frac{5}{18}+\frac{8}{9}+\frac{3}{2}+\frac{16}{9}+\frac{25}{18}\right]-\left[0+\frac{5}{18}+\frac{4}{9}+\frac{1}{2}+\frac{4}{9}+\frac{5}{18}\right]^{2} \\ =\left[\frac{5+16+27+32+25}{18}\right]-\left[\frac{5+8+9+8+5}{18}\right] \\ =\frac{105}{18}-\frac{35 \times 35}{18 \times 18} \\ =\frac{18 \times 105-35 \times 35}{18 \times 18} \\ =\frac{35}{18 \times 18}[54-35] \\ =\frac{19 \times 35}{324} \\ =\frac{665}{324}$

Question:38

A and B throw a pair of dice alternately. A wins the game if he gets a total of 6 and B wins if she gets a total of 7. It A starts the game, find the probability of winning the game by A in third throw of the pair of dice.

Answer:

Given-
A and B throw a pair of dice alternately.
A wins if he gets a total of 6
And B wins if she gets a total of 7
Therefore,
A = {(2,4), (1,5), (5,1), (4,2), (3,3)} and
B = {(2,5), (1,6), (6,1), (5,2), (3,4), (4,3)}
Let P(B) be the probability that A wins in a throw $\Rightarrow P(A)=\frac{5}{36}$$
And P(B) be the probability that B wins in a throw $\Rightarrow P(B)=\frac{1}{6}$$
$\therefore$$ The probability of A winning the game in the third row
$\rightarrow$ $P(\bar{A}) \cdot P(\bar{B}) \cdot P(A)=\frac{31}{36} \times \frac{5}{6} \times \frac{5}{36}=\frac{775}{216 \times 36}$$
$=\frac{775}{7776}$$

Question:39

Two dice are tossed. Find whether the following two events A and B are independent:
A = {(x, y): x+y = 11} and B = {(x, y): x ≠ 5}
where (x, y) denotes a typical sample point.

Answer:

Given-
A= {(x, y):x+y=11}
And B= {(x, y): x≠5}
∴ A = {(5,6), (6,5)}
B= {(1,1), (1,2), (1,3), (1,4), ((1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}
$\begin{aligned} &\Rightarrow \mathrm{n}(\mathrm{A})=2, \mathrm{n}(\mathrm{B})=30, \mathrm{n}(\mathrm{A} \cap \mathrm{B})=1\\ &P(A)=\frac{2}{36}=\frac{1}{18} \text { And } \mathrm{P}(\mathrm{B})=\frac{30}{36}=\frac{5}{6}\\ &\Rightarrow P(A) \cdot P(B)=\frac{5}{108} \text { And } P(A \cap B)=\frac{1}{18} \neq P(A) \cdot P(B)\\ &\text { Hence, } \mathrm{A} \text { and } \mathrm{B} \text { are not independent. } \end{aligned}$

Question:40

An urn contains m white and n black balls. A ball is drawn at random and is put back into the urn along with k additional balls of the same colour as that of the ball drawn. A ball is again drawn at random. Show that the probability of drawing a white ball now does not depend on k.

Answer:

Given-
An urn contains m white and n black balls.
Let E₁ be the first ball drawn of white colour
E₂ be the first ball drawn of black colour
And E₃ be the second ball drawn of white colour
$\begin{aligned} &P=\left(E_{1}\right)=\frac{m}{m+n} \text { and } P\left(E_{2}\right)=\frac{n}{m+n}\\ &\text { And, }\\ &\mathrm{P}\left(\mathrm{E}_{3} \mid \mathrm{E}_{1}\right)=\frac{\mathrm{m}+\mathrm{k}}{\mathrm{m}+\mathrm{n}+\mathrm{k}} \text { and } \mathrm{P}\left(\mathrm{E}_{3} \mid \mathrm{E}_{2}\right)=\frac{\mathrm{m}}{\mathrm{m}+\mathrm{n}+\mathrm{k}} \end{aligned}$
Using the probability theorem, we get,
$\\ \therefore P\left(E_{3}\right)=P\left(E_{1}\right) \cdot P\left(E_{3} \mid E_{1)}+P\left(E_{2}\right) \cdot P\left(E_{3} \mid E_{2}\right)\right.$ \\$=\frac{\mathrm{m}}{\mathrm{m}+\mathrm{n}} \times \frac{\mathrm{m}+\mathrm{k}}{\mathrm{m}+\mathrm{n}+\mathrm{k}}+\frac{\mathrm{n}}{\mathrm{m}+\mathrm{n}} \times \frac{\mathrm{m}}{\mathrm{m}+\mathrm{n}+\mathrm{k}}$ \\$=\frac{m(m+k)+n m}{(m+n+k)(m+n)}=\frac{m^{2}+m k+n m}{(m+n+k)(m+n)}$ \\$=\frac{m(m+k+n)}{(m+n+k)(m+n)}$ \\$=\frac{\mathrm{m}}{\mathrm{m}+\mathrm{n}}$$
Hence, the probability of drawing a white ball does not depend on k.

Question:41

Three bags contain a number of red and white balls as follows:
Bag 1: 3 red balls, Bag 2 : 2 red balls and 1 white ball
Bag 3: 3 white balls.
The probability that bag i will be chosen and a ball is selected from it is i/6, i = 1, 2, 3. What is the probability that
(i) a red ball will be selected? (ii) a white ball is selected?

Answer:

Let E₁, E₂, and E₃ be the events that Bag 1, Bag 2 and Bag 3 are selected, and a ball is chosen from it.
Bag 1: 3 red balls,
Bag 2: 2 red balls and 1 white ball
Bag 3: 3 white balls.
The probability that bag i will be chosen and a ball is selected from it is i|6.
$\\ \mathrm{P}\left(\mathrm{E}_{1}\right)=\frac{1}{6}, \mathrm{P}\left(\mathrm{E}_{2}\right)=\frac{2}{6}$ and \\$\mathrm{P}\left(\mathrm{E}_{3}\right)=\frac{3}{6}$$
The Law of Total Probability:
In a sample space S, let E₁,E₂,E₃…….E_n be n mutually exclusive and exhaustive events associated with a random experiment. If A is any event which occurs with E₁,E₂,E₃…….E_n, then
$\mathrm{P}(\mathrm{A})=\mathrm{P}\left(\mathrm{E}_{1}\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{2}\right)+\ldots \ldots \mathrm{P}\left(\mathrm{E}_{n}\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{n}\right)$
(i) Let “E” be the event that a red ball is selected.
P(E|E1) is the probability that red ball is chosen from the bag 1.
P(E|E2) is the probability that red ball is chosen from the bag 2.
P(E|E3) is the probability that red ball is chosen from the bag 3.
Therefore,
$\mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{1}\right)=\frac{3}{3}, \mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{2}\right)=\frac{2}{3}, \mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{3}\right)=0$
As red ball can be selected from Bag 1, Bag 2 and Bag 3.
Therefore, probability of choosing a red ball is the sum of individual probabilities of choosing the red from the given bags.
From the law of total probability,
$\\ P(E)=P\left(E_{1}\right) \times P\left(E \mid E_{1}\right)+P\left(E_{2}\right) \times P\left(E \mid E_{2}\right)+P\left(E_{3}\right) \times P\left(E \mid E_{3}\right) \\ =\frac{1}{6} \times \frac{3}{3}+\frac{2}{6} \times \frac{2}{3}+\frac{3}{6} \times 0 \\ =\frac{1}{6}+\frac{2}{9} \\ =\frac{3+4}{18} \\ =\frac{7}{18}$
Let F be the event that a white ball is selected.
Therefore, P(F|E1) is the probability that white ball is chosen from the bag 1.
P(F|E2) is the probability that white ball is chosen from the bag 2.
P(F|E3) is the probability that white ball is chosen from the bag 2.
P(F|E1) = 0
$\\ \mathrm{P}\left(\mathrm{F} \mid \mathrm{E}_{2}\right)=\frac{1}{3}$ \\$\mathrm{P}\left(\mathrm{F} \mid \mathrm{E}_{3}\right)=\frac{3}{3}=1$$
As white ball can be selected from Bag 1, Bag 2 and Bag 3
Therefore, sum of individual probabilities of choosing the red from the given bags is the probability of choosing a white ball.
$\\ \mathrm{P}(\mathrm{F})=\mathrm{P}\left(\mathrm{E}_{1}\right) \times \mathrm{P}\left(\mathrm{F} \mid \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \times \mathrm{P}\left(\mathrm{F} \mid \mathrm{E}_{2}\right)+\mathrm{P}\left(\mathrm{E}_{3}\right) \times \mathrm{P}\left(\mathrm{F} \mid \mathrm{E}_{3}\right)$ \\$=\frac{1}{6} \times 0+\frac{2}{6} \times \frac{1}{3}+\frac{3}{6} \times 1$ \\$=\frac{2}{18}+\frac{3}{6}$ \\$=\frac{2+9}{18}$ \\$=\frac{11}{18}$$

Question:42

Refer to Question 41 above. If a white ball is selected, what is the probability that it came from
(i) Bag 2 (ii) Bag 3

Answer:

Referring to the previous question, using Bayes theorem, we get
Let E₁, E₂, and E₃ be the events that Bag 1, Bag 2 and Bag 3 is selected, and a ball is chosen from it.
Bag 1: 3 red balls,
Bag 2: 2 red balls and 1 white ball
Bag 3: 3 white balls.
The probability that bag i will be chosen and a ball is selected from it is i|6.
$\\ P\left(E_{1}\right)=\frac{1}{6}, P\left(E_{2}\right)=\frac{2}{6}$ and \\$P\left(E_{3}\right)=\frac{3}{6}$$
Let F be the event that a white ball is selected. Therefore, $\mathrm{P}\left(\mathrm{F} \mid \mathrm{E}_{1}\right)$$ is the probability that white ball is chosen from the bag 1 .
$\mathrm{P}\left(\mathrm{F} \mid \mathrm{E}_{2}\right)$$ is the probability that white ball is chosen from the bag 2 .
$\\ \mathrm{P}\left(\mathrm{F} \mid \mathrm{E}_{3}\right)$ is the probability that white ball is chosen from the bag 2 . \\$\mathrm{P}\left(\mathrm{F} \mid \mathrm{E}_{1}\right)=0$ \\$\mathrm{P}\left(\mathrm{F} \mid \mathrm{E}_{2}\right)=\frac{1}{3}$ \\$\mathrm{P}\left(\mathrm{F} \mid \mathrm{E}_{3}\right)=\frac{3}{3}=1$$
To find: the probability that if white ball is selected, it is selected from:
(i) Bag 2
Using Bayes’ theorem to find the probability of occurrence of an event A when event B has already occurred.
$\begin{aligned} &\therefore \mathbf{P}(\mathbf{A} \mid \mathbf{B})=\frac{\mathbf{P}(\mathbf{A}) \mathbf{P}(\mathbf{B} \mid \mathbf{A})}{\mathrm{P}(\mathrm{B})}\\ &\mathrm{P}\left(\mathrm{E}_{2} \mid \mathrm{F}\right) \text { is the probability that white ball is selected from bag } 2 \text { . }\\ &\mathrm{P}\left(\mathrm{E}_{2} \mid \mathrm{F}\right)=\frac{\mathrm{P}\left(\mathrm{E}_{2}\right) \times \mathrm{P}\left(\mathrm{F} \mid \mathrm{E}_{2}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \times \mathrm{P}\left(\mathrm{F} \mid \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \times \mathrm{P}\left(\mathrm{F} \mid \mathrm{E}_{2}\right)+\mathrm{P}\left(\mathrm{E}_{3}\right) \times \mathrm{P}\left(\mathrm{F} \mid \mathrm{E}_{3}\right)}\\ &=\frac{\frac{2}{6} \times \frac{1}{3}}{\frac{1}{6} \times 0+\frac{2}{6} \times \frac{1}{3}+\frac{3}{6} \times 1} \end{aligned}$
$=\frac{\frac{1}{9}}{0+\frac{1}{9}+\frac{1}{2}} \\ =\frac{\frac{1}{9}}{\frac{2+9}{18}} \\ =\frac{1}{9} \times \frac{18}{11} \\ =\frac{2}{11}$
(ii)Bag 3
Using Bayes' theorem to find the probability of occurrence of an event A when event B has already occurred.
$\therefore \mathbf{P}(\mathbf{A} \mid \mathbf{B})=\frac{\mathbf{P}(\mathbf{A}) \mathbf{P}(\mathbf{B} \mid \mathbf{A})}{\mathrm{P}(\mathrm{B})}$$
$\mathrm{P}\left(\mathrm{E}_{3} \mid \mathrm{F}\right)$ is the probability that white ball is selected from bag 2$
Using Bayes' theorem, we get the probability of $\mathrm{P}\left(\mathrm{E}_{3} \mid \mathrm{F}\right)$$ as:
$\mathrm{P}\left(\mathrm{E}_{3} \mid \mathrm{F}\right)=\frac{\mathrm{P}\left(\mathrm{E}_{3}\right) \times \mathrm{P}\left(\mathrm{F} \mid \mathrm{E}_{3}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \times \mathrm{P}\left(\mathrm{F} \mid \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \times \mathrm{P}\left(\mathrm{F} \mid \mathrm{E}_{2}\right)+\mathrm{P}\left(\mathrm{E}_{3}\right) \times \mathrm{P}\left(\mathrm{F} \mid \mathrm{E}_{3}\right)}$ $=\frac{\frac{3}{6} \times 1}{\frac{1}{6} \times 0+\frac{2}{6} \times \frac{1}{3}+\frac{3}{6} \times 1}$ $\\=\frac{\frac{1}{2}}{0+\frac{1}{9}+\frac{1}{2}}$$
$\\ =\frac{\frac{1}{2}}{\frac{2+9}{18}} \\ =\frac{1}{2} \times \frac{18}{11} \\ =\frac{9}{11}$

Question:43

A shopkeeper sells three types of flower seeds A1, A2 and A3. They are sold as a mixture where the proportions are 4:4:2 respectively. The germination rates of the three types of seeds are 45%, 60% and 35%. Calculate the probability
(i) of a randomly chosen seed to germinate
(ii) that it will not germinate given that the seed is of type A3,
(iii) that it is of the type A2 given that a randomly chosen seed does not germinate.
Answer:

Given-
A1, A2, and A3 denote the three types of flower seeds and A1: A2: A3 = 4: 4 : 2
Therefore, Total outcomes = 10
Therefore, $\mathrm{P}\left(\mathrm{A}_{1}\right)=\frac{4}{10}, \mathrm{P}\left(\mathrm{A}_{2}\right)=\frac{4}{10} \text { and } \mathrm{P}\left(\mathrm{A}_{3}\right)=\frac{2}{10}$
Let E be the event that a seed germinates and E’ be the event thata seed does not germinate.
P(E|A1) is the probability that seed germinates when it is seed A1.
P(E’|A1) is the probability that seed will not germinate when it is seed A1.
P(E|A2) is the probability that seed germinates when it is seed A2.
P(E’|A2) is the probability that seed will not germinate when it is seed A2.
P(E|A3) is the probability that seed germinates when it is seed A3.
P(E’|A3) is the probability that seed will not germinate when it is seed A3.
$\\ \therefore P\left(E \mid A_{1}\right)=\frac{45}{100}, P\left(E \mid A_{2}\right)=\frac{60}{100} \text { and } P\left(E \mid A_{3}\right)=\frac{35}{100} \quad \text { and } \\ P\left(E^{\prime} \mid A_{1}\right)=\frac{55}{100}, P\left(E^{\prime} \mid A_{2}\right)=\frac{40}{100} \text { and } P\left(E^{\prime} \mid A_{3}\right)=\frac{65}{100}$
The Law of Total Probability:
In a sample space $\mathrm{S},$ let $\mathrm{E}_{1}, \mathrm{E}_{2}, \mathrm{E}_{3} \ldots \ldots .$ En be $\mathrm{n}$$ mutually exclusive and exhaustive events associated with a random experiment. If A is any event which occurs with $\mathrm{E}_{1}, \mathrm{E}_{2}, \mathrm{E}_{3} \ldots \ldots$ $\mathrm{E}_{0,}$$ then
$\left.\mathrm{P}(\mathrm{A})=\mathrm{P}\left(\mathrm{E}_{1}\right) \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{2}\right)+\ldots \ldots \mathrm{P}\left(\mathrm{E}_{n}\right) \mathrm{P}(\mathrm{A}] \mathrm{E}_{n}\right) $$$
(i) Probability of a randomly chosen seed to germinate.
It can be either seed A, B or C, therefore,
From law of total probability,
$\mathrm{P}(\mathrm{E})=\mathrm{P}\left(\mathrm{A}_{1}\right) \times \mathrm{P}\left(\mathrm{E} \mid \mathrm{A}_{1}\right)+\mathrm{P}\left(\mathrm{A}_{2}\right) \times \mathrm{P}\left(\mathrm{E} \mid \mathrm{A}_{2}\right)+\mathrm{P}\left(\mathrm{A}_{3}\right) \times \mathrm{P}\left(\mathrm{E} \mid \mathrm{A}_{3}\right)$ \\$=\frac{4}{10} \times \frac{45}{100}+\frac{4}{10} \times \frac{60}{100}+\frac{2}{10} \times \frac{35}{100}$ \\$=\frac{180}{1000}+\frac{240}{1000}+\frac{70}{1000}$ \\$=\frac{490}{1000}$ \\$=0.49$$
(ii) that it will not germinate given that the seed is of type A3
Knowing that P(A) + P(A’) =1
P(E’|A3) = 1 – P(E|A3)
$\\ =1-\frac{35}{100} \\ =\frac{65}{100}$
(iii) that it is of the type A2 given that a randomly chosen seed does not germinate.
Using Bayes’ theorem to find the probability of occurrence of an event A when event B has already occurred.
$\\ \therefore \mathrm{P}(\mathbf{A} \mid \mathrm{B})=\frac{\mathrm{P}(\mathrm{A}) \mathrm{P}(\mathbf{B} \mid \mathrm{A})}{\mathrm{P}(\mathrm{B})} \\ \mathrm{P}\left(\mathrm{A}_{2} \mid \mathrm{E}^{\prime}\right)=\frac{\mathrm{P}\left(\mathrm{A}_{2}\right) \times \mathrm{P}\left(\mathrm{E}^{\prime} \mid \mathrm{A}_{2}\right)}{\mathrm{P}\left(\mathrm{A}_{1}\right) \times \mathrm{P}\left(\mathrm{E}^{\prime} \mid \mathrm{A}_{1}\right)+\mathrm{P}\left(\mathrm{A}_{2}\right) \times \mathrm{P}\left(\mathrm{E}^{\prime} \mid \mathrm{A}_{2}\right)+\mathrm{P}\left(\mathrm{A}_{3}\right) \times \mathrm{P}\left(\mathrm{E}^{\prime} \mid \mathrm{A}_{3}\right)} \\ =\frac{\frac{4}{10} \times \frac{40}{100}}{\frac{4}{10} \times \frac{55}{100}+\frac{4}{10} \times \frac{40}{100}+\frac{2}{10} \times \frac{65}{100}} \\ =\frac{\frac{160}{1000}}{\frac{220}{1000}+\frac{160}{1000}+\frac{130}{1000}} \\ =\frac{160}{510} \\ =\frac{16}{51}$

Question:44

A letter is known to have come either from TATA NAGAR or from CALCUTTA. On the envelope, just two consecutive letter TA are visible. What is the probability that the letter came from TATA NAGAR.

Answer:

Let events E1, E2 be the following events-
E1 be the event that letter is from TATA NAGAR and E2 be the event that letter is from CALCUTTA
Let E be the event that on the letter, two consecutive letters TA are visible.
Since, the letter has come either from CALCUTTA or TATA NAGAR
$\therefore \mathrm{P}\left(\mathrm{E}_{1}\right)=\frac{1}{2}=\mathrm{P}\left(\mathrm{E}_{2}\right)$
We get the following set of possible consecutive letters when two consecutive letters are visible in the case of TATA NAGAR
{TA, AT, TA, AN, NA, AG, GA, AR}
We get the following set of possible consecutive letters in the case of CALCUTTA,
{CA, AL, LC, CU, UT, TT, TA}
Therefore, P(E|E1) is the probability that two consecutive letters are visible when letter came from TATA NAGAR
P(E|E2) is the probability that two consecutive letters are visible when letter came from CALCUTTA
$\therefore P\left(E \mid E_{1}\right)=\frac{2}{8}, P\left(E \mid E_{2}\right)=\frac{1}{7}$
To find- the probability that the letter came from TATA NAGAR.
Using Bayes’ theorem to find the probability of occurrence of an event A when event B has already occurred.
$\begin{aligned} &\underset{\therefore}{\mathbf{P}}(\mathbf{A} \mid \mathbf{B})=\frac{P(A) P(B \mid A)}{P(B)}\\ &\mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{1}\right) \text { is the probability that the letter came from TATA NAGAR }\\ &\therefore P\left(E \mid E_{1}\right)=\frac{P\left(E_{1}\right) \times P\left(E \mid E_{1}\right)}{P\left(E_{1}\right) \times P\left(E \mid E_{1}\right)+P\left(E_{2}\right) \times P\left(E \mid E_{2}\right)}\\ &=\frac{\frac{1}{2} \times \frac{2}{8}}{\frac{1}{2} \times \frac{2}{8}+\frac{1}{2} \times \frac{1}{7}}\\ &=\frac{\frac{1}{8}}{\frac{1}{8}+\frac{1}{14}}\\ &=\frac{\frac{1}{8}}{\frac{7+4}{56}}=\frac{1}{8} \times \frac{56}{11}=\frac{7}{11} \end{aligned}$

Question:45

There are two bags, one of which contains 3 black and 4 white balls while the other contains 4 black and 3 white balls. A die is thrown. If it shows up 1 or 3, a ball is taken from the Ist bag; but it shows up any other number, a ball is chosen from the second bag. Find the probability of choosing a black ball.

Answer:

Given-
There are 2 bags-
Bag 1: 3 black and 4 white balls
Bag 2: 4 black and 3 white balls
Therefore, Total balls = 7
Let events E1, E2 be the following:
E1 and E2 be the events that bag 1 and bag 2 are selected respectively
We know that a die is thrown.
Therefore, total outcomes = 6
$\therefore P\left(E_{1}\right)=\frac{2}{6}=\frac{1}{3} \text { and } P\left(E_{2}\right)=1-\frac{1}{3}=\frac{2}{3}$
The Law of Total Probability:
In a sample space S, let E1,E2,E3…….En be n mutually exclusive and exhaustive events associated with a random experiment. If A is any event which occurs with E1,E2,E3…….En, then
P(A) = P(E1)P(A/E1)+ P(E2)P(A/E2)+ …… P(En)P(A/En)
Let “E” be the event that black ball is chosen.
P(E|E1) is the probability that black ball is chosen from the bag 1.
P(E|E2) is the probability that black ball is chosen from the bag 2.
Therefore,
$\mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{1}\right)=\frac{3}{7}$ and $\mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{2}\right)=\frac{4}{7}$$
Therefore, probability of choosing a black ball is the sum of individual probabilities of choosing the black from the given bags.
From the law of total probability,
$\\\mathrm{P}(\mathrm{E})=\mathrm{P}\left(\mathrm{E}_{1}\right) \times \mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \times \mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{2}\right)$ \\$=\frac{1}{3} \times \frac{3}{7}+\frac{2}{3} \times \frac{4}{7}$ \\$=\frac{3}{21}+\frac{8}{21}=\frac{11}{21}$$

Question:46

There are three urns containing 2 white and 3 black balls, 3 white and 2 black balls, and 4 white and 1 black balls, respectively. There is an equal probability of each urn being chosen. A ball is drawn at random from the chosen urn and it is found to be white. Find the probability that the ball drawn was from the second urn.

Answer:

Given-
There are 3 urns U1, U2 and U3
Let U1 be 2 white and 3 black balls
U2 be 3 white and 2 black balls
U3 be 4 white and 1 black balls
Therefore, Total balls = 5
As there is an equal probability of each urn being chosen
$\therefore \mathrm{P}\left(\mathrm{U}_{1}\right)=\mathrm{P}\left(\mathrm{U}_{2}\right)=\mathrm{P}\left(\mathrm{U}_{3}\right)=\frac{1}{3}$$
Let $E_{1}, E_{2}$ and $E_{3}$$ be the event that a ball is chosen from an urn $U_{1},$$
$\\\mathrm{U}_{2}$ and $\mathrm{U}_{3}$ respectively. \\$\therefore P\left(E_{1}\right)=P\left(E_{2}\right)=P\left(E_{3}\right)=\frac{1}{3}$$
Let A be the event that white ball is drawn.
P(A|E1) is the probability that white ball is chosen from urn U1
P(A|E2) is the probability that white ball is chosen from urn U2
P(A|E3) is the probability that white ball is chosen from urn U3
$\therefore P\left(A \mid E_{1}\right)=\frac{2}{5}, P\left(A \mid E_{2}\right)=\frac{3}{5}$ and $P\left(A \mid E_{3}\right)=\frac{4}{5}$$
To find- the probability that the ball is drawn was from $\mathrm{U}_{2}$.$
Using Bayes' theorem to find the probability of occurrence of an event A when event B has already occurred.
$\underset{\therefore}{\mathbf{P}}(\mathbf{A} \mid \mathbf{B})=\frac{P(A) P(B \mid A)}{P(B)}$
$\\ \mathrm{P}\left(\mathrm{E}_{2} \mid \mathrm{A}\right)$ is the probability that white ball is selected from urn $\mathrm{U}_{2}$$
$\\ P\left(E_{2} \mid A\right)=\frac{P\left(E_{2}\right) \times P\left(A \mid E_{2}\right)}{P\left(E_{1}\right) \times P\left(A \mid E_{1}\right)+P\left(E_{2}\right) \times P\left(A \mid E_{2}\right)+P\left(E_{3}\right) \times P\left(A \mid E_{3}\right)} \\ =\frac{\frac{1}{3} \times \frac{3}{5}}{\frac{1}{3} \times \frac{2}{5}+\frac{1}{3} \times \frac{3}{5}+\frac{1}{3} \times \frac{4}{5}} \\ =\frac{\frac{1}{5}}{\frac{2}{15}+\frac{3}{15}+\frac{4}{15}} \\ =\frac{\frac{1}{5}}{\frac{9}{15}} \\ =\frac{1}{3}$

Question:47

By examining the chest X ray, the probability that TB is detected when a person is actually suffering is 0.99. The probability of an healthy person diagnosed to have TB is 0.001. In a certain city, 1 in 1000 people suffers from TB. A person is selected at random and is diagnosed to have TB. What is the probability that he actually has TB?

Answer:

Let events E1, E2, E3 be the following events:
E1 - the event that person has TB and E2 - the event that the person does not have TB
Therefore, Total persons = 1000
Therefore, $P\left(E_{1}\right)=\frac{1}{1000}=0.001, P\left(E_{2}\right)=\frac{1000-1}{1000}=\frac{999}{1000}=0.999$$
Let E be the event that the person is diagnosed to have TB
$\mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{1}\right)$ is the probability that $\mathrm{TB}$ is detected when a person is actually suffering$
$\mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{2}\right)$ the probability of an healthy person diagnosed to have TB Therefore, $\mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{1}\right)=0.99$ and $\mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{2}\right)=0.001$$
To find- the probability that the person actually has TB
Using Bayes' theorem to find the probability of occurrence of an event A when event B has already occurred.
$\therefore P(A \mid B)=\frac{P(A) P(B \mid A)}{P(B)}$
$\mathrm{P}\left(\mathrm{E}_{1} \mid \mathrm{E}\right)$$ is the probability that person actually has TB
$\begin{aligned} &\therefore P\left(E_{1} \mid E\right)=\frac{P\left(E_{1}\right) \times P\left(E_{1} \mid E\right)}{P\left(E_{1}\right) \times P\left(E_{1} \mid E\right)+P\left(E_{2}\right) \times P\left(E_{2} \mid E\right)}\\ &=\frac{0.001 \times 0.99}{0.001 \times 0.99+0.999 \times 0.001}\\ &=\frac{\frac{1}{1000} \times \frac{99}{100}}{\frac{1}{1000} \times \frac{99}{100}+\frac{999}{1000} \times \frac{1}{1000}}\\ &=\frac{\frac{99}{1000 \times 100}}{\frac{990+999}{1000 \times 1000}}\\ &=\frac{990}{990+999}\\ &=\frac{990}{1989}\\ &=\frac{110}{221} \quad \text { [Dividing both numerator and denominator by } \left.9\right] \end{aligned}$

Question:48

An item is manufactured by three machines A, B and C. Out of the total number of items manufactured during a specified period, 50% are manufactured on A, 30% on B and 20% on C. 2% of the items produced on A and 2% of items produced on B are defective, and 3% of these produced on C are defective. All the items are stored at one go down. One item is drawn at random and is found to be defective. What is the probability that it was manufactured on machine A?

Answer:

Let E1, E2, E3 be the following events:
E1 -event that item is manufactured by machine A
E2- event that item is manufactured by machine B
E3- event that item is manufactured by machine C
Since, E1, E2 and E3 are mutually exclusive and exhaustive events hence, they represent a partition of sample space.
As we know that
Items manufactured on machine A = 50%
Items manufactured on machine B = 30%
Items manufactured on machine C = 20%
Therefore,
$\begin{aligned} P\left(E_{1}\right)=50 \% &=\frac{50}{100}=\frac{1}{2}, P\left(E_{2}\right)=30 \%=\frac{30}{100}=\frac{3}{10}, P\left(E_{3}\right)=20 \%=\frac{20}{100} \\ =& \frac{1}{5} \end{aligned}$
Let E be the event that ‘an item is defective’.
Therefore, P(E|E1) is the probability of the item drawn is defective given that it is manufactured on machine A = 2%
P(E|E2) is the probability of the item drawn is defective given that it is manufactured on machine B = 2%
P(E|E3) is the probability of the item drawn is defective given that it is manufactured on machine C = 3%
Therefore,
$P\left(E \mid E_{1}\right)=\frac{2}{100}, P\left(E \mid E_{2}\right)=\frac{2}{100}, P\left(E \mid E_{3}\right)=\frac{3}{100}$
To find- the probability that the item which is picked up is defective, it was manufactured on machine A
Using Bayes’ theorem to find the probability of occurrence of an event A when event B has already occurred.
$\underset{\therefore}{P}(A \mid B)=\frac{\mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B} \mid \mathrm{A})}{P(B)}$$
$\mathrm{P}\left(\mathrm{E}_{1} \mid \mathrm{E}\right)$$ is the probability that the item is drawn is defective and it was manufactured on machine A
$\\ \therefore P\left(E_{1} \mid E\right)=\frac{P\left(E_{1}\right) \times P\left(E \mid E_{1}\right)}{P\left(E_{1}\right) \times P\left(E \mid E_{1}\right)+P\left(E_{2}\right) \times P\left(E \mid E_{2}\right)+P\left(E_{3}\right) \times P\left(E \mid E_{3}\right)}$ \\$=\frac{\frac{1}{2} \times \frac{2}{100}}{\frac{1}{2} \times \frac{2}{100}+\frac{3}{10} \times \frac{2}{100}+\frac{1}{5} \times \frac{3}{100}}$ \\$=\frac{\frac{1}{100}}{\frac{1}{100}+\frac{3}{5 \times 100} \times \frac{3}{5 \times 100}}$ \\$=\frac{1}{1+\frac{3}{5}+\frac{3}{5}}$ \\$=\frac{5}{5+6}$ \\$=\frac{5}{11}$$

Question:49

Let X be a discrete random variable whose probability distribution is defined as follows:

$P(X=x)=\left\{\begin{array}{ll} k(x+1) \text { for } x=1,2,3,4 \\ 2 k x & \text { for } x=5,6,7 \\ 0 & \text { otherwise } \end{array}\right.$

where k is a constant. Calculate
(i) the value of k (ii) E (X) (iii) Standard deviation of X.

Answer:

Given-
$P(X=x)=\left\{\begin{array}{ll} k(x+1) \text { for } x=1,2,3,4 \\ 2 k x & \text { for } x=5,6,7 \\ 0 & \text { otherwise } \end{array}\right.$
Therefore, we get the probability distribution of X as
$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \mathrm{X} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & \text { otherwise } \\ \hline \mathrm{P}(\mathrm{X}) & 2 \mathrm{k} & 3 \mathrm{k} & 4 \mathrm{k} & 5 \mathrm{k} & 10 \mathrm{k} & 12 \mathrm{k} & 14 \mathrm{k} & 0 \\ \hline \end{array}$
(i) the value of k
As we know, Sum of the probabilities =1
$i.e. \sum_{i=1}^{n} p_{i}=1$$
$\\\therefore 2 k+3 k+4 k+5 k+10 k+12 k+14 k=1$ \\$\Rightarrow 50 k=1$ \\$\Rightarrow k=\frac{1}{50}$ \\$\Rightarrow \mathrm{k}=0.02$$
To find: E(X)
The probability distribution of X is:
$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \mathrm{X} \text { or } \mathrm{x}_{i} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & \text { Otherwise } \\ \hline \mathrm{P}(\mathrm{X}) \text { or } \mathrm{p}_{\mathrm{i}} & 2 \mathrm{k} & 3 \mathrm{k} & 4 \mathrm{k} & 5 \mathrm{k} & 10 \mathrm{k} & 12 \mathrm{k} & 14 \mathrm{k} & 0 \\ \hline \mathrm{x}_{\mathrm{i}} \mathrm{p}_{\mathrm{i}} & 2 \mathrm{k} & 6 \mathrm{k} & 12 \mathrm{k} & 20 \mathrm{k} & 50 \mathrm{k} & 72 \mathrm{k} & 98 \mathrm{k} & 0 \\ \hline \end{array}$
$\begin{aligned} &\text { Therefore, } \mu=\mathrm{E}(\mathrm{X})\\ &\because E(X)=\sum_{i=1}^{n} x_{i} p_{i}\\ &\therefore E(X)=2 k+6 k+12 k+20 k+50 k+72 k+98 k+0\\ &=260 \mathrm{k}\\ &=260 \times \frac{1}{50}\left[\because k=\frac{1}{50}\right]\\ &=\frac{26}{5}\\ &=5.2 \ldots(i) \end{aligned}$
(iii) To find: Standard deviation of X
$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \mathrm{X} \text { or } \mathrm{x}_{\mathrm{i}} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & \text { Otherwise } \\ \hline \mathrm{P}(\mathrm{X}) \text { or } \mathrm{pi} & 2 \mathrm{k} & 3 \mathrm{k} & 4 \mathrm{k} & 5 \mathrm{k} & 10 \mathrm{k} & 12 \mathrm{k} & 14 \mathrm{k} & 0 \\ \hline \mathrm{XP}(\mathrm{X}) & 2 \mathrm{k} & 6 \mathrm{k} & 12 \mathrm{k} & 20 \mathrm{k} & 50 \mathrm{k} & 72 \mathrm{k} & 98 \mathrm{k} & 0 \\ \hline \mathrm{X}^{2} \mathrm{P}(\mathrm{X}) & 2 \mathrm{k} & 12 \mathrm{k} & 36 \mathrm{k} & 80 \mathrm{k} & 250 \mathrm{k} & 432 \mathrm{k} & 686 \mathrm{k} & 0 \\ \hline \end{array}$
As we know,
$\\\operatorname{Var}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^{2}\right)-[\mathrm{E}(\mathrm{X})]^{2}$ \\$=\Sigma X^{2} P(X)-[\Sigma\{X P(X)]]^{2}$ \\$=[2 k+12 k+36 k+80 k+250 k+432 k+686 k+0]-[5.2]^{2}=1498 k-27.04$ \\$=\left[1498 \times \frac{1}{50}\right]-27.04$ \\$=29.96-27.04$ \\$=2.92$$
As we know,
standard deviation of $\mathrm{X}=\sqrt{\operatorname{Var}(\mathrm{X})}=\sqrt{2.92}=1.7088$ $\cong 1.7$$

Question:50

The probability distribution of a discrete random variable X is given as under:
$\begin{array}{|l|l|l|l|l|l|l|} \hline \mathrm{X} & 1 & 2 & 4 & 2 \mathrm{~A} & 3 \mathrm{~A} & 5 \mathrm{~A} \\ \hline \mathrm{P}(\mathrm{X}) & \frac{1}{2} & \frac{1}{5} & \frac{3}{25} & \frac{1}{10} & \frac{1}{25} & \frac{1}{25} \\ \hline \end{array}$

Calculate :
(i) The value of A if E(X) = 2.94
(ii) Variance of X.

Answer:

i ) Given-
E(X) = 2.94
It is known to us that μ = E(X)
$\\ \because E(X)=\sum_{i=1}^{n} x_{i} p_{i} \\ =1 \times \frac{1}{2}+2 \times \frac{1}{5}+4 \times \frac{3}{25}+2 A \times \frac{1}{10}+3 A \times \frac{1}{25}+5 A \times \frac{1}{25} \\ =\frac{1}{2}+\frac{2}{5}+\frac{12}{25}+\frac{A}{5}+\frac{3 A}{25}+\frac{A}{5} \\ =\frac{25+20+24+10 A+6 A+10 A}{50} \\ =\frac{69+26 A}{50} \\ =2.94=\frac{69+26 A}{50} \quad \text { [given: } \left.E(X)=2.94\right] \\ \Rightarrow 2.94 \times 50=69+26 A \\ \Rightarrow 147-69=26 A \\ \Rightarrow \quad 78=26 A \\ \Rightarrow A=\frac{78}{26} \\$ x1
$\begin{aligned} &\Rightarrow A=3\\ &\text { (ii) As we know that, }\\ &\operatorname{Var}(X)=E\left(X^{2}\right)-[E(X)]^{2}\\ &=\Sigma X^{2} P(X)-[\Sigma\{X P(X)\}]^{2}\\ &=\Sigma X^{2} P(X)-(2.94)^{2}\\ &\begin{array}{l} \text { We first find } \Sigma X^{2} P(X) \\ =1^{2} \times \frac{1}{2}+2^{2} \times \frac{1}{5}+4^{2} \times \frac{3}{25}+(2 A)^{2} \times \frac{1}{10}+(3 A)^{2} \times \frac{1}{25}+(5 A)^{2} \times \frac{1}{25} \end{array}\\ &=\frac{1}{2}+\frac{4}{5}+\frac{48}{25}+\frac{36}{10}+\frac{81}{25}+\frac{225}{25}\\ &=\frac{25+40+96+180+162+450}{50}\\ &=\frac{953}{50}\\ &=19.06\\ &\operatorname{Var}(X)=19.06-(2.94)^{2}\\ &=19.06-8.6436\\ &=10.4164 \end{aligned}$

Question:51

The probability distribution of a random variable x is given as under:

$\mathrm{P}(\mathrm{X}=\mathrm{x})=\left\{\begin{array}{ll} \mathrm{kx}^{2} & \text { for } \mathrm{x}=1,2,3 \\ 2 \mathrm{kx} & \text { for } \mathrm{x}=4,5,6 \\ 0 & \text { otherwise } \end{array}\right.$

where k is a constant. Calculate
(i) E(X) (ii) $E (3X^2)$ (iii) P(X ≥ 4)

Answer:

Given-
$\begin{array}{|c|c|c|c|c|c|c|c|} \hline \mathrm{X} & 1 & 2 & 3 & 4 & 5 & 6 & \text { otherwise } \\ \hline \mathrm{P}(\mathrm{X}) & \mathrm{k} & 4 \mathrm{k} & 9 \mathrm{k} & 8 \mathrm{k} & 10 \mathrm{k} & 12 \mathrm{k} & 0 \\ \hline \end{array}$
As we know, Sum of the probabilities =1
$i, e . \sum_{i=1}^{n} p_{i}=1$ \\$\Rightarrow \mathrm{k}+4 \mathrm{k}+9 \mathrm{k}+8 \mathrm{k}+10 \mathrm{k}+12 \mathrm{k}=1$ \\$\Rightarrow 44 k=1$ \\$\Rightarrow k=\frac{1}{44}$$
(i) To find:
$\mathrm{E}(\mathrm{X})$ As we know, $\mu=E(X)$ $\because E(X)=\sum_{i=1}^{n} x_{i} p_{i}$$
or
$E(X)=\Sigma X P(X)$ \\$=1 \times k+2 \times 4 k+3 \times 9 k+4 \times 8 k+5 \times 10 k+6 \times 12 k$ \\$=k+8 k+27 k+32 k+50 k+72 k$ \\$=190 \mathrm{k}$$
$\\=190 \times \frac{1}{44} \because\left[k=\frac{1}{44}\right]$ \\$=\frac{95}{22}$ \\\\$=4.32$$
(ii) To find: $E\left(3 X^{2}\right)$$
We first find $E\left(X^{2}\right)$$
As we know that,
$\\E\left(X^{2}\right)=\Sigma X^{2} P(X)$ \\$=1^{2} \times \mathrm{k}+2^{2} \times 4 \mathrm{k}+3^{2} \times 9 \mathrm{k}+4^{2} \times 8 \mathrm{k}+5^{2} \times 10 \mathrm{k}+6^{2} \times 12 \mathrm{k}$ \\$=\mathrm{k}+16 \mathrm{k}+81 \mathrm{k}+128 \mathrm{k}+250 \mathrm{k}+432 \mathrm{k}$ \\$=908 \mathrm{k}$ \\$=908 \times \frac{1}{44}$$
$\begin{aligned} &=20.636\\ &\cong 20.64\\ &\therefore E\left(3 X^{2}\right)=3 \times 20.64=61.92\\ &\text { (iii) } P(X \geq 4)=P(X=4)+P(X=5)+P(X=6)\\ &=8 \mathrm{k}+10 \mathrm{k}+12 \mathrm{k}\\ &=30 \mathrm{k}\\ &=30 \times \frac{1}{44}\\ &=\frac{15}{22} \end{aligned}$

Question:52

A bag contains (2n + 1) coins. It is known that n of these coins have a head on both sides whereas the rest of the coins are fair. A coin is picked up at random from the bag and is tossed. If the probability that the toss results in a head is 31/42, determine the value of n.

Answer:

Given-
n coins have head on both the sides and (n + 1) coins are fair coins
Therefore, Total coins = 2n + 1
Let E1, E2 be the following events:
E1 = Event that an unfair coin is selected
E2 = Event that a fair coin is selected
$\therefore P\left(E_{1}\right)=\frac{n}{2 n+1} \text { and } P\left(E_{2}\right)=\frac{n+1}{2 n+1}$
The Law of Total Probability:

In a sample space S, let E1,E2,E3…….En be n mutually exclusive and exhaustive events associated with a random experiment. If A is any event which occurs with E1,E2,E3…….En, then
P(A) = P(E1)P(A/E1)+ P(E2)P(A/E2)+ …… P(En)P(A/En)
Let “E” be the event that the toss result is a head
P(E|E1) is the probability of getting a head when unfair coin is tossed
P(E|E2) is the probability of getting a head when fair coin is tossed
Therefore,
$\begin{aligned} &P\left(E \mid E_{1}\right)=1 \text { and } P\left(E \mid E_{2}\right)=\frac{1}{2}\\ &\text { Erom the law of total probability, }\\ &\mathrm{P}(\mathrm{E})=\mathrm{P}\left(\mathrm{E}_{1}\right) \times \mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \times \mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{2}\right)\\ &\Rightarrow \frac{31}{42}=\frac{n}{2 n+1} \times 1+\frac{n+1}{2 n+1} \times \frac{1}{2} \text { (Given) } \end{aligned}$
$\begin{aligned} &\Rightarrow \frac{31}{42}=\frac{2 n+n+1}{2(2 n+1)}\\ &\Rightarrow 31 \times 2(2 n+1)=42 \times(3 n+1)\\ &\Rightarrow 124 n+62=126 n+42\\ &\Rightarrow 2 n=20\\ &\Rightarrow \mathrm{n}=10\\ &\text { Hence, the value of } \mathrm{n} \text { is } 10 \text { . } \end{aligned}$

Question:53

Two cards are drawn successively without replacement from a well shuffled deck of cards. Find the mean and standard variation of the random variable X where X is the number of aces.

Answer:

Let X be a random variable of number of aces
X can take values 0, 1 or 2 because only two cards are drawn.
Therefore, Total deck of cards = 52
and total no. of ACE cards in a deck of cards = 4
Since the draws are done without replacement, therefore, the two draws are not independent.
Therefore,
P(X = 0) = Probability of no ace being drawn
= P(non – ace and non – ace)
= P(non – ace) × P(non – ace)
$\\=\frac{48}{52} \times \frac{47}{51}$ \\$=\frac{2256}{2652}$ \\$\mathrm{P}(\mathrm{X}=1)=$ Probability that 1 card is an ace \\$=$ P(ace and non - ace or non -ace and ace) \\$=$ P(ace and non $-$ ace $)+P($ non $-$ ace and $a c e)=P(a c e) P($ non $-a c e)+P($ non $-$ ace $)$ P(ace) \\$=\frac{4}{52} \times \frac{48}{51}+\frac{48}{52} \times \frac{4}{51}$ \\$=\frac{384}{2652}$ \\$=\frac{192}{2326}=\frac{96}{1613}$ \\$\mathrm{P}(\mathrm{X}=2)=$ Probability that both cards are ace \\$=$ P(ace and ace) \\$=\mathrm{P}(\mathrm{ace}) \times \mathrm{P}(\mathrm{ace})$ \\$=\frac{4}{52} \times \frac{3}{51}$$
$\begin{aligned} &=\frac{12}{2652}=\frac{6}{1326}=\frac{3}{663}=\frac{1}{221}\\ &\text { As we know that, }\\ &\operatorname{Mean}(\mu)=E(X)=\Sigma X P(X)\\ &=0 \times \frac{2256}{2652}+1 \times \frac{384}{2652}+2 \times \frac{12}{2652}\\ &=\frac{384}{2652}+\frac{24}{2652}\\ &=\frac{408}{2652} \end{aligned}$
$\\ =\frac{2}{13} \\ \text { Also, } \operatorname{Var}(X)=E\left(X^{2}\right)-[E(X)]^{2} \\ =\Sigma X^{2} P(X)-[E(X)]^{2} \\ =\left[0 \times \frac{2256}{2652}+1^{2} \times \frac{384}{2652}+2^{2} \times \frac{12}{2652}\right]-\left(\frac{2}{13}\right)^{2} \\ =\frac{432}{2652}-\frac{4}{169} \\ =0.1629-0.0237 \\ =0.1392 \\ \therefore \text { Standard Deviation }=\sqrt{\operatorname{Var}(X)}=\sqrt{ 0.1392 }\cong 0.373(\text { approx. })$

Question:54

A die is tossed twice. A ‘success’ is getting an even number on a toss. Find the variance of the number of successes.

Answer:

Let X be the random variable for a ‘success’ for getting an even number on a toss.
∴ X = 0, 1, 2
n = 2
Even number on dice = 2, 4, 6
∴ Total possibility of getting an even number = 3
Total number on dice = 6
p = probability of getting an even number on a toss
$\begin{aligned} &=\frac{3}{6}\\ &=\frac{1}{2}\\ &q=1-p\\ &=1-\frac{1}{2}\\ &=\frac{1}{2}\\ &\text { The probability of x successes in n-Bernoulli trials is }{ }^{n} \mathrm{C}_{r} \mathrm{p}^{r} \mathrm{q}^{\text {n-r }}\\ &P(x=0)=^2C_{0}\left(\frac{1}{2}\right)^{0}\left(\frac{1}{2}\right)^{2-0}=1 \times 1 \times \frac{1}{4}=\frac{1}{4}\\ &P(x=1)=^2C_{1}\left(\frac{1}{2}\right)^{1}\left(\frac{1}{2}\right)^{2-1}=2 \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{2}\\ &P(x=2)=^2C_{2}\left(\frac{1}{2}\right)^{2}\left(\frac{1}{2}\right)^{2-2}=1 \times \frac{1}{4} \times 1=\frac{1}{4} \end{aligned}$
$\begin{aligned} &\begin{array}{|c|c|c|c|} \hline \mathrm{X} & 0 & 1 & 2 \\ \hline \mathrm{P}(\mathrm{X}) & 1 / 4 & 1 / 2 & 1 / 4 \\ \hline \end{array}\\ &E(X)=\Sigma X P(X)=0 \times \frac{1}{4}+1 \times \frac{1}{2}+2 \times \frac{1}{4}\\ &=\frac{1}{2}+\frac{1}{2}\\ &=1\\ &\text { And, } \operatorname{Var}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^{2}\right)-[\mathrm{E}(\mathrm{X})]^{2}\\ &=\Sigma X^{2} P(X)-[E(X)]^{2} \end{aligned}$
$\\ =\left[0 \times \frac{1}{4}+1^{2} \times \frac{1}{2}+2^{2} \times \frac{1}{4}\right]-(1)^{2} \\ =\left[\frac{1}{2}+1\right]-1 \\ =\frac{3}{2}-1 \\ =1.5-1 \\ =0.5$

Question:55

There are 5 cards numbered 1 to 5, one number on one card. Two cards are drawn at random without replacement. Let X denote the sum of the numbers on two cards drawn. Find the mean and variance of X.

Answer:

The sample space is
S = { (1,2),(1,3),(1,4),(1,5)
(2,1),(2,3),(2,4),(2,5)
(3,1),(3,2),(3,4),(3,5)
(4,1),(4,2),(4,3),(4,5)
(5,1),(5,2),(5,3),(5,4)}
Total Sample Space, n(S) = 20
Let random variable be X which denotes the sum of the numbers on the cards drawn.
∴ X = 3, 4, 5, 6, 7, 8, 9
At X = 3
The cards whose sum is 3 are (1,2), (2,1)
$P(X)=\frac{2}{20}=\frac{1}{10}$$
At x=4
The cards whose sum is 4 are (1,3),(3,1)
$P(X)=\frac{2}{20}=\frac{1}{10}$$
At X=5
The cards whose sum is 5 are (1,4),(2,3),(3,2),(4,1)
$P(X)=\frac{4}{20}=\frac{1}{5}$$
At X=6
The cards whose sum is 6 are (1,5),(2,4),(4,2),(5,1)
$P(X)=\frac{4}{20}=\frac{1}{5}$$
At x=7
The cards whose sum is 7 are (2,5),(3,4),(4,3),(5,2)
$P(X)=\frac{4}{20}=\frac{1}{5}$$
At X = 8
The cards whose sum is 8 are (3,5), (5,3)
$P(X)=\frac{2}{20}=\frac{1}{10}$$
At X = 9
The cards whose sum is 9 are (4,5), (5,4)
$\begin{aligned} &\begin{array}{l} P(X)=\frac{2}{20}=\frac{1}{10} \\ \therefore \text { Mean, } E(X)=\Sigma \times P(X) \\ =3 \times \frac{1}{10}+4 \times \frac{1}{10}+5 \times \frac{1}{5}+6 \times \frac{1}{5}+7 \times \frac{1}{5}+8 \times \frac{1}{10}+9 \times \frac{1}{10} \\ =\frac{3}{10}+\frac{2}{5}+1+\frac{6}{5}+\frac{7}{5}+\frac{4}{5}+\frac{9}{10} \\ =\frac{3+4+10+12+14+8+9}{10} \\ =\frac{60}{10} \\ =6 \end{array}\\ &\text { And, }\\ &\Sigma X^{2} P(X)=3^{2} \times \frac{1}{10}+4^{2} \times \frac{1}{10}+5^{2} \times \frac{1}{5}+6^{2} \times \frac{1}{5}+7^{2} \times \frac{1}{5}+8^{2} \times \frac{1}{10}\\ &+9^{2} \times \frac{1}{10} \end{aligned}$
$\begin{aligned} &=\frac{9}{10}+\frac{16}{10}+5+\frac{36}{5}+\frac{49}{5}+\frac{64}{10}+\frac{81}{10}\\ &=\frac{9+16+50+72+98+64+81}{10}\\ &=\frac{390}{10}\\ &=39\\ &\text { Therefore, }\\ &\operatorname{Var} \mathrm{X}=\Sigma \mathrm{X}^{2} \mathrm{P}(\mathrm{X})-[\Sigma \mathrm{XP}(\mathrm{X})]^{2}\\ &=39-36\\ &=3 \end{aligned}$

Question:56

If $P(A)=\frac{4}{5}$ , and $P(A \cap B)=\frac{7}{10}$ , then P(B | A) is equal to
A. 1|10
B. 1|8
C. 7|8
D. 17|20

Answer:

Given- $P(A)=\frac{4}{5}$ , and $P(A \cap B)=\frac{7}{10}$
$\begin{aligned} &\text { As we know, }\\ &\mathrm{P}(\mathrm{B} \mid \mathrm{A}) \times \mathrm{P}(\mathrm{A})=\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \quad[\text { Property of Conditional Probability] }\\ &P(B \mid A) \times \frac{4}{5}=\frac{7}{10}\\ &P(B \mid A)=\frac{7}{10} \times \frac{5}{4}\\ &P(B \mid A)=\frac{7}{8} \end{aligned}$
Hence, Correct option is C

Question:57

If P(A ∩ B) = 7|10 and P(B) = 17|20, then P(A|B) equals
A. 14|17
B. 17|20
C. 7|8
D. 1|8

Answer:

Given-
$\begin{aligned} &P(B)=\frac{17}{20}\\ &P(A \cap B)=\frac{7}{10}\\ &\text { As we know, }\\ &\mathrm{P}(\mathrm{A} \mid \mathrm{B}) \times \mathrm{P}(\mathrm{B})=\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \text { [Property of Conditional Probability] }\\ &P(A \mid B) \times \frac{17}{20}=\frac{7}{10}\\ &P(B \mid A)=\frac{7}{10} \times \frac{20}{17}\\ &P(B \mid A)=\frac{14}{17} \end{aligned}$
Hence, Correct option is A

Question:58

If $\begin{aligned} &P(A)=\frac{3}{10}, P(B)=\frac{2}{5} \text { and } P(A \cup B)=\frac{3}{5}\\ \end{aligned}$ , then P(B|A) + P(A|B) equals
A. 1|4
B. 1|3
C. 5|12
C. 7|2

Answer:

Given-
$\begin{aligned} &P(A)=\frac{3}{10}, P(B)=\frac{2}{5} \text { and } P(A \cup B)=\frac{3}{5}\\ &\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \text { [Additive Law of Probability] }\\ &\therefore \frac{3}{5}=\frac{3}{10}+\frac{2}{5}-P(A \cap B)\\ &\Rightarrow P(A \cap B)=\frac{3}{10}-\frac{1}{5}\\ &\Rightarrow P(A \cap B)=\frac{3-2}{10}\\ &\Rightarrow P(A \cap B)=\frac{1}{10}\\ &\text { As we know the Property of Conditional Probability: }\\ &\mathrm{P}(\mathrm{A} \mid \mathrm{B}) \times \mathrm{P}(\mathrm{B})=\mathrm{P}(\mathrm{A} \cap \mathrm{B})\\ &\Rightarrow P(A \mid B)=\frac{P(A \cap B)}{P(B)} \end{aligned}$
$\begin{aligned} &\mathrm{P}(\mathrm{B} \mid \mathrm{A}) \times \mathrm{P}(\mathrm{A})=\mathrm{P}(\mathrm{B} \cap \mathrm{A})\\ &\Rightarrow P(B \mid A)=\frac{P(B \cap A)}{P(A)}\\ &\text { Multiplying eq. (i) and (ii), we get }\\ &\therefore P(B \mid A)+P(A \mid B)=\frac{P(B \cap A)}{P(A)}+\frac{P(A \cap B)}{P(B)}\\ &=\frac{\frac{1}{10}}{\frac{3}{10}}+\frac{\frac{1}{10}}{\frac{2}{5}}\\ &=\frac{1}{3}+\frac{1}{10} \times \frac{5}{2}\\ &=\frac{4+3}{12}\\ &=\frac{7}{12} \end{aligned}$
Hence, the Correct option is D

Question:59

If $P(A)=\frac{2}{5}, P(B)=\frac{3}{10} \text { and } P(A \cap B)=\frac{1}{5}$ , the P(A′|B′).P(B′|A′) is equal to
A. 5|6
B. 5|7
C. 25|42
D. 1

Answer:

Given-
$\begin{aligned} &P(A)=\frac{2}{5}, P(B)=\frac{3}{10} \text { and } P(A \cap B)=\frac{1}{5}\\ &\mathrm{P}\left(\mathrm{A}^{\prime} \mid \mathrm{B}^{\prime}\right) \times \mathrm{P}\left(\mathrm{B}^{\prime}\right)=\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)\\ &\Rightarrow P\left(A^{\prime} \mid B^{\prime}\right)=\frac{P\left(A^{\prime} \cap B^{\prime}\right)}{P\left(B^{\prime}\right)}\\ &\mathrm{P}\left(\mathrm{B}^{\prime} \mid \mathrm{A}^{\prime}\right) \times \mathrm{P}\left(\mathrm{A}^{\prime}\right)=\mathrm{P}\left(\mathrm{B}^{\prime} \cap \mathrm{A}^{\prime}\right)\\ &\Rightarrow P\left(B^{\prime} \mid A^{\prime}\right)=\frac{P\left(B^{\prime} \cap A^{\prime}\right)}{P\left(A^{\prime}\right)}\\ &\text { Multiplying eq. (i) and (ii), we get }\\ \end{aligned}$
$\begin{aligned} &P\left(A^{\prime} \mid B^{\prime}\right) \times P\left(B^{\prime} \mid A^{\prime}\right)=\frac{P\left(A^{\prime} \cap B^{\prime}\right)}{P\left(B^{\prime}\right)} \times \frac{P\left(A^{\prime} \cap B^{\prime}\right)}{P\left(A^{\prime}\right)}\\ &=\frac{1-P(A \cup B)}{P\left(B^{\prime}\right)} \times \frac{1-P(A \cup B)}{P\left(A^{\prime}\right)}\\ &\left[\because P\left(A^{\prime} \cap B^{\prime}\right)=P\left[(A \cup B)^{\prime}\right]=1-P(A \cup B)\right]\\ &=\frac{(1-P(A \cup B))^{2}}{(1-P(B)) \times(1-P(A))}\end{aligned}$

$\\ =\frac{\left(1-(P(A)+P(B)-P(A \cap B))^{2}\right.}{\left(1-\frac{3}{10}\right)\left(1-\frac{2}{5}\right)} \\ =\frac{\left[1-\left(\frac{2}{5}+\frac{3}{10}-\frac{1}{5}\right)\right]^{2}}{\left(\frac{10-3}{10}\right)\left(\frac{5-2}{5}\right)} \\ =\frac{\left[1-\left(\frac{4+3-2}{10}\right)\right]^{2}}{\frac{7}{10} \times \frac{3}{5}}$
$\\ =\frac{\left[1-\left(\frac{5}{10}\right)\right]^{2}}{\frac{7}{10} \times \frac{3}{5}} \\ =\frac{\left[1-\frac{1}{2}\right]^{2}}{\frac{7}{10} \times \frac{3}{5}} \\ =\frac{\left[\frac{1}{2}\right]^{2}}{\frac{7}{10} \times \frac{3}{5}} \\ =\frac{1}{4} \times \frac{50}{21} \\ =\frac{25}{42}$
Hence, the correct option is C

Question:60

If A and B are two events such that $\mathrm{P}(\mathrm{A})=\frac{1}{2}, \mathrm{P}(\mathrm{B})=\frac{1}{3}, \mathrm{P}(\mathrm{A} / \mathrm{B})=\frac{1}{4}$ , the P(A′ ∩ B′) equals
A. 1|12
B. 3|4
C. 1|4
D. 3|16

Answer:

Given-
$\mathrm{P}(\mathrm{A})=\frac{1}{2}, \mathrm{P}(\mathrm{B})=\frac{1}{3}, \mathrm{P}(\mathrm{A} / \mathrm{B})=\frac{1}{4}$
As we know,
P(A|B) × P(B) = P(A ∩ B) [Property of Conditional Probability]
$\\ \Rightarrow \frac{1}{4} \times \frac{1}{3}=P(A \cap B) \\ \Rightarrow P(A \cap B)=\frac{1}{12} \\ P\left(A^{\prime} \cap B^{\prime}\right)=1-P(A \cup B) \\ =1-[P(A)+P(B)-P(A \cap B)] \\ {[\because P(A \cup B)=P(A)+P(B)-P(A \cap B)]} \\ =1-\left[\frac{1}{2}+\frac{1}{3}-\frac{1}{12}\right] \\ =1-\left[\frac{6+4-1}{12}\right] \\ =1-\frac{9}{12} \\ =\frac{12-9}{12} \\ =\frac{3}{12} \\ =\frac{1}{4}$
Hence, the correct option is C

Question:61

If P(A) = 0.4, P(B) = 0.8 and P(B | A) = 0.6, then P(A ∪ B) is equal to
A. 0.24
B. 0.3
C. 0.48
D. 0.96

Answer:

Given-
P(A) = 0.4, P(B) = 0.8 and $P(B$ \vert $ A) = 0.6$
It is known that,
$\\P(B \vert A) \times P(A) = P(B \cap $ A) \\$ \Rightarrow $ 0.6 $ \times $ 0.4 = P(B $ \cap $ A) \\ \Rightarrow $ P(B $ \cap $ A) = 0.24 \\P(A \cup B) = P(A) + P(B) - P(A \cap B)$ [Additive Law of Probability] \\= 0.4 + 0.8 - 0.24 \\= 0.96$
Hence, $\textbf{the correct option is D}$

Question:62

If A and B are two events and A ≠ θ, B ≠ θ, then
A. P(A | B) = P(A).P(B)
B. $P(A \mid B)=\frac{P(A \cap B)}{P(B)}$

C. P(A | B).P(B | A)=1
D. P(A | B) = P(A) | P(B)

Answer:

Given-
$A \neq \phi, B \neq \phi$$
CASE 1 : If we take option $(\mathrm{A})$ i.e. $\mathrm{P}(\mathrm{A} \mid \mathrm{B})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})$$
$P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{P(A) \times P(B)}{P(B)}=P(A) \neq R H S$$
CASE 2 : If we take option (B) i.e.
$P(A \mid B)=\frac{P(A \cap B)}{P(B)}$$
this is true, knowing that this is conditional probability.
CASE 3: If we take option $\mathrm{C}$.i.e. $\mathrm{P}(\mathrm{A} \mid \mathrm{B}) \times \mathrm{P}(\mathrm{B} \mid \mathrm{A})=1$$
$\\\mathrm{LHS}=\mathrm{P}(\mathrm{A} \mid \mathrm{B}) \times \mathrm{P}(\mathrm{B} \mid \mathrm{A})$ \\$=\frac{P(A \cap B)}{P(B)} \times \frac{P(B \cap A)}{P(A)}$ \\$=\frac{[P(A \cap B)]^{2}}{P(B) P(A)}$ \\$\neq \mathrm{RHS}$$
Hence, the correct option is B

Question:63

A and B are events such that P(A) = 0.4, P(B) = 0.3 and P(A ∪ B) = 0.5. Then P (B′ ∩ A) equals
A. 2|3
B. 1|2
C. 3|10
D. 1|5

Answer:

Given-
$\\\mathrm{P}(\mathrm{A})=0.4, \mathrm{P}(\mathrm{B})=0.3$ and \\$\mathrm{P}(\mathrm{A} \cup \mathrm{B})=0.5$ \\$P(A \cup B)=P(A)+P(B)-P(A \cap B)[$ Additive Law of Probability $]$ \\$\Rightarrow 0.5=0.4+0.3-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$ \\$\Rightarrow P(A \cap B)=0.7-0.5$ \\$\Rightarrow P(A \cap B)=0.2$ \\$\therefore \mathrm{P}\left(\mathrm{B}^{\prime} \cap \\\mathrm{A}\right)=P(A)-P(A \cap B)$ \\$=0.4-0.2$ \\$=0.2$ \\$=\frac{1}{5}$$
Hence, the correct option is D

Question:64

You are given that A and B are two events such that P(B)= 3|5, P(A | B) = 1|2 and P(A ∪ B) = 4|5, then P(A) equals
A. 3|10
B. 1|5
C. 1/2
D. 3|5

Answer:

Given
$\begin{aligned} &P(B)=\frac{3}{5}, P(A \mid B)=\frac{1}{2} \text { and } P(A \cup B)=\frac{4}{5}\\ &\text { As we know, }\\ &\mathrm{P}(\mathrm{A} \mid \mathrm{B}) \times \mathrm{P}(\mathrm{B})=\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \text { [Property of conditional Probability] }\\ &\Rightarrow \frac{1}{2} \times \frac{3}{5}=P(A \cap B)\\ &\Rightarrow P(A \cap B)=\frac{3}{10}\\ &\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})[\text { Additive Law of Probability }]\\ &\therefore \frac{4}{5}=P(A)+\frac{3}{5}-\frac{3}{10}\\ &\Rightarrow P(A)=\frac{4}{5}-\frac{3}{5}+\frac{3}{10}\\ &\Rightarrow P(A)=\frac{1}{5}+\frac{3}{10}\\ &\Rightarrow P(A)=\frac{2+3}{10}\\ &\Rightarrow P(A)=\frac{5}{10}=\frac{1}{2} \end{aligned}$
Hence, the correct option is C

Question:65

In Exercise 64 above, P(B | A′) is equal to
A. 1|5
B. 3|10
C. 1|2
D. 3|5

Answer:

Referring to the above solution,
$\\ P\left(B \mid A^{\prime}\right)=\frac{P\left(B \cap A^{\prime}\right)}{P\left(A^{\prime}\right)} \\ =\frac{P(B)-P(B \cap A)}{1-P(A)} \\ =\frac{\frac{3}{5}-\frac{3}{10}}{1-\frac{1}{2}}$
$\\ =\frac{\frac{6-3}{10}}{\frac{1}{2}} \\ =\frac{\frac{3}{10}}{\frac{1}{2}} \\ =\frac{3}{5}$
Hence, the correct option is D

Question:66

If P(B) = 3|5, P(A|B) = 1|2 and P(A ∪ B) = 4|5, then P(A ∪ B)′ + P(A′ ∪ B) =
A. 1|5
B. 4|5
C. 1|2
D. 1

Answer:

Given-
$\begin{aligned} &P(B)=\frac{3}{5}, P(A \mid B)=\frac{1}{2} \text { and } P(A \cup B)=\frac{4}{5}\\ &\text { As we know, }\\ &\mathrm{P}(\mathrm{A} \mid \mathrm{B}) \times \mathrm{P}(\mathrm{B})=\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \text { [Property of Conditional Probability] }\\ &\Rightarrow \frac{1}{2} \times \frac{3}{5}=P(A \cap B)\\ &\Rightarrow P(A \cap B)=\frac{3}{10}\\ &\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \text { [Additive Law of Probability] }\\ &\therefore \frac{4}{5}=P(A)+\frac{3}{5}-\frac{3}{10}\\ &\Rightarrow P(A)=\frac{4}{5}-\frac{3}{5}+\frac{3}{10}\\ &\Rightarrow P(A)=\frac{1}{5}+\frac{3}{10}\\ &\Rightarrow P(A)=\frac{2+3}{10} \end{aligned}$
$\\ \Rightarrow P(A)=\frac{5}{10}=\frac{1}{2} \\ \therefore P(A \cup B)^{\prime}=P\left[A^{\prime} \cap B^{\prime}\right] \\ =1-P(A \cup B) \\ =1-\frac{4}{5} \\ =\frac{1}{5} \\ \text { and } P\left(A^{\prime} \cup B\right)=1-P\left(A^{\prime} \cap B\right) \\ =1-[P(A)-P(A \cap B)]$
$\\ =1-\left(\frac{1}{2}-\frac{3}{10}\right) \\ =1-\left(\frac{5-3}{10}\right) \\ =1-\frac{2}{10} \\ =\frac{10-2}{10} \\ =\frac{4}{5} \\ \Rightarrow P(A \cup B)^{\prime}+P\left(A^{\prime} \cup B\right)=\frac{1}{5}+\frac{4}{5} \\ =1$
Hence, the correct option is D

Question:67

Let P(A) = 7/13, P(B) = 9/13 and P(A ∩ B) = 4/13. Then P(A′|B) is equal to
A. 6/13
B. 4/13
C. 4/9
D. 5/9

Answer:

Given-
P(A) = 7/13, P(B) = 9/13 and P(A ∩ B) = 4/13

$\begin{aligned} &P\left(A^{\prime} \mid B\right)=\frac{P\left(A^{\prime} \cap B\right)}{P(B)}\\ &\text { From the above Venn diagram, }\\ &\frac{\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}\right)}{\mathrm{P}(\mathrm{B})}=\frac{\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}\\ &\Rightarrow \frac{\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{\frac{9}{13}-\frac{4}{13}}{\frac{9}{13}}\\ &\Rightarrow \frac{5}{13} \times \frac{13}{9}\\ &\Rightarrow \mathrm{P}\left(\mathrm{A}^{\prime} \mid \mathrm{B}\right)=\frac{5}{9}\\ &\text { Hence, } \mathrm{P}\left(\mathrm{A}^{\prime} \mid \mathrm{B}\right)=\frac{5}{9} \end{aligned}$

Question:68

If A and B such events that P(A) > 0 and P(B) ≠ 1, then P(A’|B’) equals
A. 1 – P(A|B)

B. 1 – P (A’|B)
C. $\frac{1-\mathrm{P}(\mathrm{A} \cup \mathrm{B})}{\mathrm{P}\left(\mathrm{B}^{\prime}\right)}$
D. P(A’) | P(B’)

Answer:

Given-
$\\P(A)>0$ and $P(B) \neq 1$ \\$\mathrm{P}\left(\mathrm{A}^{\prime} \mid \mathrm{B}^{\prime}\right)=\frac{P\left(A^{\prime} \cap \mathrm{B}^{\prime}\right)}{P\left(B^{\prime}\right)}$$
By de Morgan's Law:
$\\P\left(A^{\prime} \cap B^{\prime}\right)=P(A \cup B)^{\prime}$ \\$=\frac{1-P(A \cup B)}{P\left(B^{\prime}\right)}$$
Option c is correct answer.

Question:69

If A and B are two independent events with P(A) = 3/5 and P(B) = 4/9, then P (A′ ∩ B′) equals
A.4/15
B. 8/45
C. 1/3
D. 2/9

Answer:

$\begin{aligned} &\text { As } A \text { and } B \text { are independent } A^{\prime} \text { and } B^{\prime} \text { are also independent. }\\ &P\left(A^{\prime} \cap B^{\prime}\right)=P\left(A^{\prime}\right) \cdot P\left(B^{\prime}\right)\\ &\text { As we know, } \mathrm{P}\left(\mathrm{A}^{\prime}\right)=(1-\mathrm{P}(\mathrm{A})) \text { and } \mathrm{P}\left(\mathrm{B}^{\prime}\right)=(1-\mathrm{P}(\mathrm{B})\\ &\Rightarrow(1-\mathrm{P}(\mathrm{A})) \cdot(1-\mathrm{P}(\mathrm{B}))=\left(1-\frac{3}{5}\right)\left(1-\frac{4}{9}\right)\\ &=\left(\frac{2}{5}\right)\left(\frac{5}{9}\right)\\ &\Rightarrow\left(P\left(A^{\prime} \cap B^{\prime}\right)=\frac{2}{9}\right. \end{aligned}$

Question:70

If two events are independent, then
A. they must be mutually exclusive
B. the sum of their probabilities must be equal to 1
C. (A) and (B) both are correct
D. None of the above is correct

Answer:

Events which cannot happen at the same time are known as mutually exclusive events. For example: when tossing a coin, the result can either be heads or tails but cannot be both.
Events are independent if the occurrence of one event does not influence (and is not influenced by) the occurrence of the other(s).
Eg: Rolling a die and flipping a coin. The probability of getting any number on the die will not affect the probability of getting head or tail in the coin.
Therefore, if A and B events are independent, any information about A cannot tell anything about B while if they are mutually exclusive then we know if A occurs B does not occur.
Therefore, independent events cannot be mutually exclusive.
To test if probability of independent events is 1 or not:
Let A be the event of obtaining a head.
P(A) = 1/2
Let B be the event of obtaining 5 on a die.
P(B) = 1/6
Now A and B are independent events.
$\begin{aligned} &\text { Therefore, } \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})=\frac{1}{2}+\frac{1}{6}\\ &=\frac{3+1}{12}=\frac{4}{12}\\ &=\frac{1}{3}\\ &\text { Hence } P(A)+P(B) \neq 1\\ &\text { It is true in every case when two events are independent. } \end{aligned}$
Hence option D is correct.

Question:71

Let A and B be two events such that P(A) = 3/8, P(B) = 5/8 and P(A ∪ B) = 3/4. Then P(A | B).P(A′ | B) is equal to
A.2/5
B. 3/8
C. 3/20
D. 6/25

Answer:

$\begin{aligned} &\text { Given- }\\ &\mathrm{P}(\mathrm{A})=3 / 8, \mathrm{P}(\mathrm{B})=5 / 8 \text { and } \mathrm{P}(\mathrm{A} \cup \mathrm{B})=3 / 4\\ &\text { As we know, } \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})\\ &P(A \cap B)={\frac{3}{8}}+\frac{5}{8}-\frac{3}{4}\\ &\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{4}\\ &\text { Therefore, } \mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}, \text { then }\\ &\mathrm{P}(\mathrm{A} \mid \mathrm{B})=2 / 5\\ &\text { For, } \mathrm{P}\left(\mathrm{A}^{\prime} \mid \mathrm{B}\right) \end{aligned}$
$\\ P\left(A^{\prime} \mid B\right)=\frac{P\left(A^{\prime} \cap B\right)}{P(B)} \\ P\left(A^{\prime} \mid B\right)=\frac{P(B)-P(A \cap B)}{P(B)} \\ P\left(A^{\prime} \mid B\right)=\frac{\frac{5}{8}-\frac{1}{4}}{\frac{5}{8}} \\ P\left(A^{\prime} \mid B\right)=3 / 5 \\ \text { Therefore, } P(A \mid B) \cdot P\left(A^{\prime} \mid B\right)=\left(\frac{3}{5} \times \frac{2}{5}\right) \\ \text { Hence, } P(A \mid B) \cdot P\left(A^{\prime} \mid B\right)=6 / 25$
Option D is correct.

Question:72

If the events A and B are independent, then P(A ∩ B) is equal to
A. P (A) + P
B. (B) P(A) – P(B)
C. P (A) . P(B)
D. P(A) | P(B)

Answer:

We know that if events A and B are independent,
$\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})$$
From the definition of the independent Event,
$\\\mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}$ \\$P(A)=\frac{P(A \cap B)}{P(B)}$ \\$P(A \cap B)=P(A) \cdot P(B)$$
Option C is correct.

Question:73

Two events E and F are independent. If P(E) = 0.3, P(E ∪ F) = 0.5, then P(E | F)–P(F | E) equals
A. 2/7
B. 3/25
C. 1/70
D. 1/7

Answer:

Given-
P(E) = 0.3, P(E ∪ F) = 0.5
Also, E and F are independent, therefore,
P (E ∩ F)=P(E).P(F)
As we know , P(E ∪ F)=P(E)+P(F)- P(E ∩ F)
P(E ∪ F)=P(E)+P(F)- [P(E) P(F)]
$\\ 0.5=0.3+P(F)-0.3 P(F) \\ 0.5-0.3=(1-0.3) P(F) \\ P(F)=\frac{2}{7}$
$\\ \\ P(E \mid F)-P(F \mid E)=\frac{P(E \cap F)}{P(F)}-\frac{P(F \cap E)}{P(E)} \\ P(E \mid F)-P(F \mid E)=\frac{P(E \cap F) \cdot[P(E)-P(F)]}{P(E \cap F)} \\ P(E \mid F)-P(F \mid E)=P(E)-P(F) \\ P(F \mid F)-P(F \mid E)=\frac{3}{10} -\frac{2}{7}= \frac{21-20}{70} =\frac{1}{70}$

Question:74

A bag contains 5 red and 3 blue balls. If 3 balls are drawn at random without replacement the probability of getting exactly one red ball is
A. 45/196
B. 135/392
C. 15/56
D. 15/29

Answer:

The Probability of getting exactly one red ball is
P(R).P(B).P(B) + P(B).P(R).P(B) + P(B).P(B).P(R)
$\\ =\frac{5}{8} \cdot \frac{3}{7} \cdot \frac{2}{6}+\frac{3}{8} \cdot \frac{5}{7} \cdot \frac{2}{6}+\frac{3}{8} \cdot \frac{2}{7} \cdot \frac{5}{6} \\ =\frac{15}{56}$
Option C is correct.

Question:75

Refer to Question 74 above. The probability that exactly two of the three balls were red, the first ball being red, is
A. 1/3
B. 4/7
C. 15/28
D. 5/28

Answer:

Given-
A bag contains 5 red and 3 blue balls
Therefore, Total Balls in a Bag = 8
For exactly 1 red ball probability should be
3 Balls are drawn randomly then possibility for getting 1 red ball
P(E)=P(R).P(B)+P(B).P(R)
$\\ \mathrm{P}(\mathrm{E})=\frac{4}{7} \times \frac{3}{6}+\frac{3}{6} \times \frac{4}{7} \\ \text { Hence, } \mathrm{P}(\mathrm{E})=4 / 7$
Option B is correct.

Question:76

Three persons, A, B and C, fire at a target in turn, starting with A. Their probability of hitting the target are 0.4, 0.3 and 0.2 respectively. The probability of two hits is
A. 0.024
B. 0.188
C. 0.336
D. 0.452

Answer:

Given-
$\\ \mathrm{P}(\mathrm{A})=0.4 \mathrm{P}(\mathrm{B})=0.3 and \mathrm{P}(\mathrm{C})=0.2 \\ Therefore, \mathrm{P}\left(\mathrm{A}^{\prime}\right)=1-\mathrm{P}(\mathrm{A})=[1-0.4]=0.6 \\ \mathrm{P}\left(\mathrm{B}^{\prime}\right)=1-\mathrm{P}(\mathrm{B})=[1-0.3]=0.7 \\\mathrm{P}\left(\mathrm{C}^{\prime}\right)=1-\mathrm{P}(\mathrm{C})=[1-0.2]=0.8 \\\mathrm{P}(\mathrm{E})=\left[\mathrm{P}(\mathrm{A}) \times \mathrm{P}(\mathrm{B}) \times \mathrm{P}\left(\mathrm{C}^{\prime}\right)\right]+\left[\mathrm{P}(\mathrm{A}) \times \mathrm{P}\left(\mathrm{B}^{\prime}\right) \times \mathrm{P}(\mathrm{C})\right]+\left[\mathrm{P}\left(\mathrm{A}^{\prime}\right) \times \mathrm{P}(\mathrm{B}) \times \mathrm{P}(\mathrm{C})\right] \\ = (0.4 \times 0.3 \times 0.8)+(0.4 \times 0.7 \times 0.2)+(0.6 \times 0.3 \times 0.2) \\ =0.96+0.056+0.036 \\=0.188$ Hence, Probability of two hits is 0.188
Option B is correct.

Question:77

Assume that in a family, each child is equally likely to be a boy or a girl. A family with three children is chosen at random. The probability that the eldest child is a girl given that the family has at least one girl is
A. 1/2
B. 1/3
C. 2/3
D. 4/7

Answer:

The statement can be arranged in a set as S={(B,B,B),(G,G,G),(B,G,G),(G,B,G),(G,G,B),(G,B,B),(B,G,B),(B,B,G)}
Let A be Event that a family has at least one girl, therefore,
A={(G,B,B),(B,G,B),(B,B,G),(G,G,B),(B,G,G)(G,B,G),(G,G,G)
Let B be Event that eldest child is girl then, therefore,
B={(G,B,B)(G,G,B),(G,B,G),(G,G,G)
(A ∩ B)={(G,B,B),(G,G,B),(G,B,G,)(G,G,G)

since, $\mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}$$
$P(A \mid B)=\frac{\frac{4}{8}}{\frac{7}{8}}=\frac{4}{7}$$
Hence, $\mathrm{P}\left(\mathrm{E}_{2} \mid \mathrm{E}_{1}\right)=\frac{4}{7}$$
Option D is correct.

Question:78

A die is thrown and a card is selected at random from a deck of 52 playing cards. The probability of getting an even number on the die and a spade card is
A. 1/2
B. 1/4
C. 1/8
D. 3/4

Answer:

Let A be Event for getting number on dice and B be Event that a spade card is selected
Therefore,
A={2,4,6}
B={13}
$\begin{aligned} &\text { since, } P(A)=\frac{3}{6}=\frac{1}{2}\\ &\mathrm{P}(\mathrm{B})=\frac{13}{52}=\frac{1}{4}\\ &\text { As we know, If } \mathrm{E} \text { and } \mathrm{F} \text { are two independent events then, } \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})\\ &\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{2} \times \frac{1}{4}=\frac{1}{8}\\ &\text { Hence, } \mathrm{P}\left(\mathrm{E}_{1} \cap \mathrm{E}_{2}\right)=\frac{1}{8} \end{aligned}$

Question:79

A box contains 3 orange balls, 3 green balls and 2 blue balls. Three balls are drawn at random from the box without replacement. The probability of drawing 2 green balls and one blue ball is

$\\A. \frac{3}{28}$ \\B. $\frac{2}{21}$ \\C.$\frac{1}{28}$ \\D.$\frac{167}{168}$$

Answer:

Given-
There are total 8 balls in box.
Therefore, P(G) $=\frac{3}{8}$ , Probability of green ball
P(B) $=\frac{2}{8}$ , Probability of blue ball
The probability of drawing 2 green balls and one blue ball is
P(E)=P(G).P(G).P(B)+P(B).P(G).P(G)+P(G).P(B).P(G)
$\\ P(E)=\left(\frac{3}{8} \times \frac{2}{7} \times \frac{2}{6}\right)+\left(\frac{2}{8} \times \frac{3}{7} \times \frac{2}{6}\right)+\left(\frac{3}{8} \times \frac{2}{7} \times \frac{2}{6}\right) \\ P(E)=\frac{1}{28}+\frac{1}{28}+\frac{1}{28} \\ P(E)=\frac{3}{28} \\ $ Hence, $\quad P(E)=\frac{3}{28}$

Question:80

A flashlight has 8 batteries out of which 3 are dead. If two batteries are selected without replacement and tested, the probability that both are dead is

$\\ A.\frac{33}{56} \\\\ B.\frac{9}{64} \\\\ C.\frac{1}{14} \\\\ D.\frac{3}{28}$

Answer:

Given-
Total number of batteries: n= 8
Number of dead batteries are = 3
Therefore, Probability of dead batteries is $\frac{3}{8}$
If two batteries are selected without replacement and tested
Then, Probability of second battery without replacement is $\frac{2}{7}$
Required probability = $\frac{3}{8}\times \frac{2}{7}= \frac{3}{28}$

Question:81

Eight coins are tossed together. The probability of getting exactly 3 heads is
A. $\frac{1}{256}$
B. $\frac{7}{32}$
C. $\frac{5}{32}$
D. $\frac{3}{32}$

Answer:

Given-
probability distribution $\mathrm{P}(\mathrm{X}=\mathrm{r})={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}(\mathrm{p})^{r} \mathrm{ q}^{\mathrm{n}-\mathrm{r}}$$
Total number coin is tossed, n=8
The probability of getting head, $\mathrm{p}=1 / 2$$
The probability of getting tail, $\mathrm{q}=1 / 2$$
The Required probability
$\\={ }^{8} \mathrm{C}_{3}\left(\frac{1}{2}\right)^{3}\left(\frac{1}{2}\right)^{8-3}\\$ $\frac{8 \times 7 \times 6}{3 \times 2} \times \frac{1}{2^{8}}=\frac{7}{32}$$

Question:82

Two dice are thrown. If it is known that the sum of numbers on the dice was less than 6, the probability of getting a sum 3, is

A. $\frac{1}{18}$
B. $\frac{5}{18}$
C. $\frac{1}{5}$
D. $\frac{2}{5}$

Answer:

Let A be the event that the sum of numbers on the dice was less than 6
And B be the event that the sum of numbers on the dice is 3
Therefore,
A={(1,4)(4,1)(2,3)(3,2)(2,2)(1,3)(3,1)(1,2)(2,1)(1,1)
n(A)=10
B={(1,2)(2,1)
n(B)=2
Required probability = $\frac{nB}{nA}$
Required probability = $\frac{2}{10}$
Hence, the probability is $\frac{1}{5}$

Question:83

Which one is not a requirement of a binomial distribution?
A. There are 2 outcomes for each trial
B. There is a fixed number of trials
C. The outcomes must be dependent on each other
D. The probability of success must be the same for all the trials

Answer:

In the binomial distribution, there are 2 outcomes for each trial and there is a fixed number of trials and the probability of success must be the same for all trials.
Hence option c is correct.

Question:84

Two cards are drawn from a well shuffled deck of 52 playing cards with replacement. The probability, that both cards are queens, is

$\\ A.\frac{1}{13} \times \frac{1}{13} \\\\ B.\frac{1}{13}+\frac{1}{13} \\\\ C.\frac{1}{13} \times \frac{1}{17} \\\\ D.\frac{1}{13} \times \frac{4}{51}$

Answer:

We know that
Number of cards = 52
Number of queens = 4
Therefore, Probability of queen out of 52 cards = $\frac{4}{52}$
According to the question,
If a deck of card shuffled again with replacement, then
Probability of getting queen is , $\frac{4}{52}$
Therefore, The probability, that both cards are queen is , $\left [\frac{4}{52} \times\frac{4}{52} \right ]$
Hence, Probability is $\left [\frac{1}{13} \times \frac{1}{13} \right ]$

Question:85

The probability of guessing correctly at least 8 out of 10 answers on a true-false type examination is
A. $\frac{7}{64}$
B. $\frac{7}{128}$
C. $\frac{45}{1024}$
D. $\frac{7}{41}$

Answer:

We know that in the examination, we have only two option True and False.
Therefore, the probability of getting True is, p= 1/2
And, probability of getting False is, q=1/2
The total number of Answer in examination, n=10
The probability of guessing correctly at least 8 it means r=8,9,10
As we know, the probability distribution $\mathrm{P}(\mathrm{X}=\mathrm{r})={ }^{n} \mathrm{C}_{\mathrm{C}}(\mathrm{p})^{r} \mathrm{q}^{\mathrm{n}-\mathrm{r}}$$
$\mathrm{P}(\mathrm{X}=\mathrm{r})=\mathrm{P}(\mathrm{r}=8)+\mathrm{P}(\mathrm{r}=9)+\mathrm{P}(\mathrm{r}=10)$
$\\ =^{10} \mathrm{C}_{8}\left(\frac{1}{2}\right)^{8}\left(\frac{1}{2}\right)^{10-8}+^{10} \mathrm{C}_{9}\left(\frac{1}{2}\right)^{9}\left(\frac{1}{2}\right)^{10-9}+^{10} \mathrm{C}_{10}\left(\frac{1}{2}\right)^{10}\left(\frac{1}{2}\right)^{10-10}$ \\$=\left(\frac{10 !}{8 ! 2 !}+\frac{10 !}{9 !}+1\right)\left(\frac{1}{2}\right)^{10}$ \\$=[45+10+1]^{\left(\frac{1}{2}\right)^{10}}$ \\$=\frac{56}{1025}=\frac{7}{128}$$
Hence, Probability is $\frac{7}{128}$$

Question:86

The probability that a person is not a swimmer is 0.3. The probability that out of 5 persons 4 are swimmers is
$\\A. ${ }^{5} \mathrm{C}_{4}(0.7)^{4}(0.3)$ \\B. ${ }^{5} C_{1}(0.7)(0.3)^{4}$ \\C. ${ }^{5} \mathrm{C}_{4}(0.7)(0.3)^{4}$ $\\D.(0.7)^{4}(0.3)$$

Answer:

Given-
Total number of person n=5
Total number of swimmers among total person , r=4
Probability of not swimmer , Q=0.3
Therefore, The probability of swimmer , p=1-Q=0.7
As we know that the probability distribution $\mathrm{P}(\mathrm{X}=\mathrm{r})={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}(\mathrm{p})^{r} \mathrm{q}_{ }^{\mathrm{n-r}}$$
$\\P(X)={ }^{n} C_{r}(p)^{r}(q)^{n-r} \\ \\P(X)=\frac{n !}{(n-r) ! \times r !} \times(p)^{r} \times(q)^{n-r}$
= $\frac{5 !}{4 ! \times 1 !} \times(0.7)^{4} \times(0.3)$$
Hence, ${ }^{5} \mathrm{C}_{4}(0.7)^{4}(0.3)$$
Option A is correct.

Question:87

The probability distribution of a discrete random variable X is given below:

$\begin{array}{|c|c|c|c|c|} \hline X & 2 & 3 & 4 & 5 \\ \hline P(X) & \frac{5}{k} & \frac{7}{k} & \frac{9}{k} & \frac{11}{k} \\ \hline \end{array}$
The value of k is
A. 8
B. 16
C. 32
D. 48

Answer:

Given-
Probability distribution table
As we know $\sum_{i=1}^{n} P_{i}=1$$
$\\\Rightarrow \sum P_{i}=\left[\frac{5}{k}+\frac{7}{k}+\frac{9}{k}+\frac{11}{k}\right]=1$ \\$\Rightarrow\left[\frac{32}{\mathrm{k}}\right]=1$ \\$\mathrm{K}=32$$
Hence, the value of k is 32
Option C is correct.

Question:88

For the following probability distribution:
$\begin{array}{|l|l|l|l|l|l|} \hline \mathrm{X} & -4 & -3 & -2 & -1 & 0 \\ \hline \mathrm{P}(\mathrm{X}) & 0.1 & 0.2 & 0.3 & 0.2 & 0.2 \\ \hline \end{array}$
E(X) is equal to:
A. 0
B. –1
C. –2
D. –1.8

Answer:

Given-
Probability distribution table
$\\ \mathrm{E}(\mathrm{X})=\sum \mathrm{X.P}(\mathrm{X}) \\ \mathrm{E}(\mathrm{X})=[(-4) \times(0.1)+(-3) \times(0.2)+(-2) \times(0.3)+(-1) \times(0.2)+(0 \times 0.2)] \\ \mathrm{E}(\mathrm{X})=[-0.4-0.6-0.6-0.2+0] \\ \mathrm{E}(\mathrm{X})=[-1.8] \\ \text { Hence, } \mathrm{E}(\mathrm{X})=-1.8$
Option D is correct.

Question:89

For the following probability distribution
$\begin{array}{|l|l|l|l|l|} \hline \mathrm{X} & 1 & 2 & 3 & 4 \\ \hline \mathrm{P}(\mathrm{X}) & 1 / 10 & 1 / 5 & 3 / 10 & 2 / 5 \\ \hline \end{array}$
$E(X^2)$ is equal to
A. 3
B. 5
C. 7
D. 10

Answer:

Given-
Probability distribution table
$\\ \mathrm{E}\left(\mathrm{X}^{2}\right)=\sum \mathrm{X}^{2} \cdot \mathrm{P}(\mathrm{X}) \\ \mathrm{E}\left(\mathrm{X}^{2}\right)=\left[1^{2} \times \frac{1}{10}+2^{2} \times \frac{1}{5}+3^{2} \times \frac{3}{10}+4^{2} \times \frac{2}{5}\right] \\ \mathrm{E}\left(\mathrm{X}^{2}\right)=\left[\frac{1}{10}+\frac{4}{5}+\frac{27}{10}+\frac{32}{5}\right] \\ \mathrm{E}\left(\mathrm{X}^{2}\right)=\left[\frac{1+8+27+64}{10}\right] \\ \mathrm{E}\left(\mathrm{X}^{2}\right)=\left[\frac{100}{10}\right] \\ \text { Hence, } \mathrm{E}\left(\mathrm{X}^{2}\right)=10$
Option D is correct.

Question:90

Suppose a random variable X follows the binomial distribution with parameters n and p, where 0 < p < 1. If P(x = r) / P(x = n–r) is independent of n and r, then p equals
A. 1/2
B. 1/3
C. 1/5
D. 1/7

Answer:

$\begin{aligned} &\text { As we know that in binomial distribution } \mathrm{P}(\mathrm{X}=\mathrm{r})={ }^{n} \mathrm{C}_{r}(\mathrm{p})^{\mathrm{r}}(\mathrm{q})^{\mathrm{n}-\mathrm{r}}\\ &\mathrm{P}(\mathrm{x}=\mathrm{r})=\frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}) ! \mathrm{r} !}(\mathrm{p})^{\mathrm{r}}(1-\mathrm{p})^{\mathrm{n}-\mathrm{r}} \text { where } \mathrm{q}=1-\mathrm{p}\\ &\text { Therefore, }\\ &\frac{\mathrm{P}(\mathrm{x}=\mathrm{r})}{\mathrm{P}(\mathrm{x}=\mathrm{n}-\mathrm{r})}=\frac{(\mathrm{p})^{\mathrm{r}}(1-\mathrm{p})^{\mathrm{n}-\mathrm{r}}}{(\mathrm{p})^{\mathrm{n}-\mathrm{r}}(1-\mathrm{p})^{\mathrm{r}}}{}\\ &\text { since, }{ }^{n} \mathrm{C}_{r}={ }^{n} \mathrm{C}_{n-r}\\ &\frac{P(x=r)}{P(x=n-r)}=\left(\frac{1-p}{p}\right)^{n-2 r} \end{aligned}$
According to the question, this expression is independent of n and r if
$\frac{1-p}{p}=1 \Rightarrow p=\frac{1}{2}$
Hence $p=\frac{1}{2}$
Option A is correct.

Question:91

In a college, 30% students fail in physics, 25% fail in mathematics and 10% fail in both. One student is chosen at random. The probability that she fails in physics if she has failed in mathematics is
A. $\frac{1}{10}$
B. $\frac{2}{5}$
C. $\frac{9}{20}$
D. $\frac{1}{3}$

Answer:

Let A be the event that students failed in physics.
As per the question, 30% students failed in physics.
∴ P(A) = 0.30
Similarly, if we denote the event of failing in maths with B.
We get P(B) = 0.25
And probability of failing in both subjects can be represented using intersection as
P (A ∩ B) = 0.1
To find- a conditional probability of failing of student in physics given that she has failed in mathematics.
The situation can be represented mathematically as-
P(A|B) =?
Using the fundamental idea of conditional probability, we know that:
$\mathrm{P}(\mathrm{E} \mid \mathrm{F})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}$ where $\mathrm{E}$ \& F denotes 2 random events.$
$\therefore P(A \mid B)=\frac{P(A \cap B)}{P(B)}$
$\Rightarrow P(A \mid B)=\frac{0.1}{0.25}=\frac{10}{25}=\frac{2}{5}$$
Our answers clearly match with option B
$\therefore$ Option $(\mathrm{B})$$ is the only correct choice.

Question:92

A and B are two students. Their chances of solving a problem correctly are 1/3 and 1/4, respectively. If the probability of their making a common error is, 1/20 and they obtain the same answer, then the probability of their answer to be correct is

A. $\frac{1}{12}$
B. $\frac{1}{40}$
C. $\frac{13}{120}$
D. $\frac{10}{13}$

Answer:

Let E be the event that student ‘A’ solves the problem correctly.
∴ P(E) = 1/3
In the same way if we denote the event of ’B’ solving the problem correctly with F
We get P(F) = 1/4
Since both the events are independent.
∴ Probability that both the students solve the question correctly can be represented as-
$P(E \cap F)=\frac{1}{3} \times \frac{1}{4}=\frac{1}{12}=P\left(E_{1}\right)\{$ say $\}$ \{we can multiply because events are independent $\}$$
$\therefore$$ Probability that both the students could not solve the question correctly can be represented as-
$\mathrm{P}\left(\mathrm{E}^{\prime} \cap \mathrm{F}^{\prime}\right)=\frac{2}{3} \times \frac{3}{4}=\frac{1}{2}=\mathrm{P}\left(\mathrm{E}_{2}\right)\{\mathrm{say}\}$$
Given: probability of making a common error and both getting same answer.
If they are making an error, we can be sure that answer coming out is wrong.
Let S be the event of getting same answer.
$\therefore$ above situation can be represented using conditional probability.
$\mathrm{P}\left(\mathrm{S} \mid \mathrm{E}_{2}\right)=1 / 20$$
And if their answer is correct obviously, they will get same answer.
$\therefore P\left(S \mid E_{1}\right)=1$$
To find- the probability of getting a correct answer if they committed a common error and got the same answer.
Mathematically, i.e, $\mathrm{P}\left(\mathrm{E}_{1} \mid \mathrm{S}\right)=?$$
By observing our requirement and availability of equations, we can use Bayes theorem to solve this.
$\therefore$$ Using Bayes theorem, we get-
$P\left(E_{1} \mid S\right)=\frac{P\left(E_{1}\right) P\left(S \mid E_{1}\right)}{P\left(E_{1}\right) P\left(S \mid E_{1}\right)+P\left(E_{2}\right) P\left(S \mid E_{2}\right)}$
Substituting the values from above -
$P\left(E_{1} \mid S\right)=\frac{\frac{1}{12} \times 1}{\frac{1}{12} \times 1+\frac{1}{2} \times \frac{1}{20}}=\frac{\frac{1}{12}}{\frac{1}{12}+\frac{1}{40}}=\frac{10}{13}$
Our answers clearly match with option D.
∴ Option (D) is the only correct choice.

Question:93

A box has 100 pens of which 10 are defective. What is the probability that out of a sample of 5 pens drawn one by one with replacement at most one is defective?

$\\A. \left(\frac{9}{10}\right)^{5}$ \\\\B.$\frac{1}{2}\left(\frac{9}{10}\right)^{4}$ \\\\C.. $\frac{1}{2}\left(\frac{9}{10}\right)^{5}$ \\\\D. $\left(\frac{9}{10}\right)^{5}+\frac{1}{2}\left(\frac{9}{10}\right)^{4}$$

Answer:

We can solve this using Bernoulli trials.
Here n = 5 (as we are drawing 5 pens only)
Success is defined when we get a defective pen.
Let p be the probability of success and q probability of failure.
∴ p = 10/100 = 0.1
And q = 1 – 0.1 = 0.9
To find- the probability of getting at most 1 defective pen.
Let X be a random variable denoting the probability of getting r number of defective pens.
∴ P (drawing atmost 1 defective pen) = P(X = 0) + P(X = 1)
The binomial distribution formula is:
$\mathrm{P}(\mathrm{x})=^{n} \mathrm{C}_{x} \mathrm{p}^{\mathrm{x}}(1-\mathrm{P})^{\mathrm{n}-\mathrm{x}}$$
Where:
x = total number of “successes.”
P = probability of success on an individual trial
n = number of trials
$\Rightarrow P(X=0)+P(X=1)=5_{C_{0}} p^{0} q^{5}+{ }^{5} C_{1} p^{1} q^{4}$$
$\mathrm{P}($ drawing at most 1 defective pen $)=\left(\frac{9}{10}\right)^{5}+5\left(\frac{1}{10}\right)\left(\frac{9}{10}\right)^{4}$$
$\Rightarrow \mathrm{P}($ drawing at most 1 defective pen $) = \left ( \frac{9}{10} \right )^5+\frac{1}{2}\left ( \frac{9}{10} \right )^4$
Our answer matches with option D.
∴ Option (D) is the only correct choice.

Question:94

State True or False for the statements in the Exercise.
Let P(A) > 0 and P(B) > 0. Then A and B can be both mutually exclusive and independent.

Answer:

FALSE
Events are mutually exclusive when–
P(A∪B) = P(A) + P(B)
But as per the conditions in question, it is not necessary that they will meet the condition because it might be possible that
P(A ∩ B) ≠ 0
Events are independent when–
P(A ∩ B) = P(A)P(B)
Again P(A) > 0 and P(B)> 0 are not sufficient conditions to validate them.

Question:95

State True or False for the statements in the Exercise.
If A and B are independent events, then A′ and B′ are also independent.

Answer:

TRUE
As A and B are independent
$\begin{aligned} &\Rightarrow P({A} \cap {B})=P(A) P(B)\\ &\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)=\mathrm{P}({\mathrm{A}} \underline{\cup} \underline{\mathrm{B}})^{\prime}\{\text { using De morgan's law }\}\\ &P(A \cup B)^{\prime}=1-P(A \cup B)\\ &\text { We know } P(A \cup B)=P(A)+P(B)-P(A \cap B)\\ &\Rightarrow P(A \cup B)^{\prime}=1-[P(A)+P(B)-P(A \cap B)]\\ &\Rightarrow \mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)=1-\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B}) \text { as } \mathrm{A} \& \mathrm{~B} \text { are independent }\}\\ &=[1-\mathrm{P}(\mathrm{A})]-\mathrm{P}(\mathrm{B})(1-\mathrm{P}(\mathrm{A})]\\ &\Rightarrow P\left(A^{\prime} \cap B^{\prime}\right)=(1-P(A))(1-P(B))\\ &=\mathrm{P}\left(\mathrm{A}^{\prime}\right) \mathrm{P}\left(\mathrm{B}^{\prime}\right) \end{aligned}$
hence proved

Question:96

State True or False for the statements in the Exercise.
If A and B are mutually exclusive events, then they will be independent also.

Answer:

False
If A and B are mutually exclusive, that means
P(A∪B) = P(A) + P(B)
From this equation it cannot be proved that
P(A ∩ B)= P(A)P(B).
Hence, it is a false statement.

Question:97

State True or False for the statements in the Exercise.
Two independent events are always mutually exclusive.

Answer:

False
If A and B are independent events, it means that
P(A ∩ B) = P(A)P(B)
From the equation it cannot be proved that
P(A∪B) = P(A) + P(B)
It is only possible if either P(A) or P(B) = 0, which is not given in question.
Hence, it is a false statement.

Question:98

State True or False for the statements in the Exercise.
If A and B are two independent events then P(A and B) = P(A).P(B)

Answer:

TRUE
If A and B are independent events it means that
P(A ∩ B) = P(A)P(B)
Thus, from the definition of independent event we say that statement is true.

Question:99

State True or False for the statements in the Exercise.

Another name for the mean of a probability distribution is expected value.

Answer:

TRUE
Mean gives the average of values and if it is related with probability or random variable it is often called expected value.

Question:100

State True or False for the statements in the Exercise.
If A and B are independent events, then P(A′ ∪ B) = 1 – P (A) P(B′)

Answer:

TRUE
If A and B are independent events, it means that
P(A ∩ B) = P(A)P(B)
P(A′ ∪ B) = P(A’) + P(B) – P(A’ ∩ B)
and P(A′ ∪ B) represents the probability of event ‘only B’ excluding common points.
$\\ \therefore P\left(A^{\prime} \cap B\right)=P(B)-P(A \cap B) \\ \Rightarrow P\left(A^{\prime} \cup B\right)=P\left(A^{\prime}\right)+P(B)-P(B)+P(A \cap B) \\ \Rightarrow P\left(A^{\prime} \cup B\right)=1-P(A)+P(A) P(B)\{\text { independent events }\} \\ \Rightarrow P\left(A^{\prime} \cup B\right)=1-P(A)\{1-P(B)\} \\ \Rightarrow P\left(A^{\prime} \cup B\right)=1-P(A) P\left(B^{\prime}\right)$
Hence Proved

Question:101

State True or False for the statements in the Exercise.
If A and B are independent, then
P (exactly one of A, B occurs) = P(A)P(B′)+P(B) P(A′)

Answer:

TRUE
If A and B are independent events, that means
$\\ P(A \cap B)=P(A) P(B) \\ P\left(A^{\prime} \cap B\right)=P\left(A^{\prime}\right) P(B) \\ \text { And } P\left(A \cap B^{\prime}\right)=P(A) P\left(B^{\prime}\right) \\ P(\text { exactly one of } A, B \text { occurs })=P\left(A^{\prime} \cap B\right)+P\left(A \cap B^{\prime}\right) \\ \Rightarrow P(\text { exactly one of } A, B \text { occurs })=P\left(A^{\prime}\right) P(B)+P(A) P\left(B^{\prime}\right)$
∴ Statement is true.

Question:102

State True or False for the statements in the Exercise.
If A and B are two events such that P(A) > 0 and P(A) + P(B) >1, then
$\mathrm{P}(\mathrm{B} \mid \mathrm{A}) \geq 1-\frac{\mathrm{P}\left(\mathrm{B}^{\prime}\right)}{\mathrm{P}(\mathrm{A})}$

Answer:

True
$\operatorname{As}, \mathrm{P}(\mathrm{B} \mid \mathrm{A})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}=\frac{\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{AUB})}{\mathrm{P}(\mathrm{A})}$ \\$\Rightarrow P(B \mid A)=\frac{P(A)+1-P\left(B^{\prime}\right)-P(A \cup B)}{P(A)}$ \\$\Rightarrow \mathrm{P}(\mathrm{B} \mid \mathrm{A})=1-\frac{\mathrm{P}\left(\mathrm{B}^{\prime}\right)}{\mathrm{P}(\mathrm{A})}+\frac{1-\mathrm{P}(\mathrm{AUB})}{\mathrm{P}(\mathrm{A})}$$
The above equation means that:
$\\P(B \mid A) \geq 1-\frac{P\left(B^{\prime}\right)}{P(A)}$ \\$\frac{1-\mathrm{P}(\mathrm{AUB})}{\mathrm{P}(\mathrm{A})}$$
we need to add $\mathrm{P}(\mathrm{A}) \quad$$ to get the equal term
$\Rightarrow$ LHS is greater.

Question:103

State True or False for the statements in the Exercise.
If A, B and C are three independent events such that P(A) = P(B) = P(C) = p, then P (At least two of A, B, C occur) = $3p^2 - 2p^3$

Answer:

True
Let A, B,C be the occurrence of events A,B and C and A’,B’ and C’ not occurrence.
P(A) = P(B) = P(C) = p and P(A’) = P(B’) = P(C’) = 1-p
P (At least two of A, B, C occur) = P(A ∩ B ∩ C’) + P(A ∩ B’ ∩ C) + P(A’ ∩ B ∩ C) + P(A ∩ B ∩ C)
events are independent:
$\mathrm{P}$ (At least two of $\mathrm{A}, \mathrm{B},$ C occur $)=\mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B}) \mathrm{P}\left(\mathrm{C}^{\prime}\right)+\mathrm{P}(\mathrm{A}) \mathrm{P}\left(\mathrm{B}^{\prime}\right) \mathrm{P}(\mathrm{C})+\mathrm{P}\left(\mathrm{A}^{\prime}\right) \mathrm{P}(\mathrm{B}) \mathrm{P}(\mathrm{C})+\mathrm{P}\left(\mathrm{A}\right) \mathrm{P}(\mathrm{B}) \mathrm{P}(\mathrm{C})=3 \mathrm{p}^{2}(1-\mathrm{p})+p^3$$
$=3 p^{2}-2p^{3}$$
Hence, statement is true.

Question:104

Fill in the blanks in the following question:
If A and B are two events such that
$\mathrm{P}(\mathrm{A} \mid \mathrm{B})=\mathrm{p}, \mathrm{P}(\mathrm{A})=\mathrm{p}, \mathrm{P}(\mathrm{B})=\frac{1}{3}$ and $P(A \cup B)=\frac{5}{9}$ , then p =

Answer:

$\begin{aligned} &p=1 / 3\\ &\text { We know }\\ &P(A \mid B)=\frac{P(A \cap B)}{P(B)}\\ &=\frac{\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cup \mathrm{B})}{\mathrm{P}(\mathrm{B})}\ \end{aligned}$
$\\ p=\frac{p+\frac{1}{3}-\frac{5}{9}}{\frac{1}{3}}\\ p=\frac{9 p+3-5}{3}\\ 3 p=9 p-2\\ \Rightarrow 6 p=2\\ \therefore p=1 / 3$

Question:105

Fill in the blanks in the following question:
If A and B are such that
$\mathrm{P}\left(\mathrm{A}^{\prime} \cup \mathrm{B}^{\prime}\right)=\frac{2}{3} \text { and } \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{5}{9}$ ,
then P(A') + P(B') = ..................

Answer:

$\begin{aligned} &10 / 9\\ &\mathrm{P}\left(\mathrm{A}^{\prime} \cup \mathrm{B}^{\prime}\right)=\mathrm{P}\left(\mathrm{A}^{\prime}\right)+\mathrm{P}\left(\mathrm{B}^{\prime}\right)-\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)\{\mathrm{using} \text { union of two sets }\}\\ &\Rightarrow P\left(A^{\prime}\right)+P\left(B^{\prime}\right)=P\left(A^{\prime} \cup B^{\prime}\right)+P\left(A^{\prime} \cap B^{\prime}\right)\\ &\because \mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)=\mathrm{P}(\mathrm{A} \cup \mathrm{B})^{\prime}\{\text { using De Morgan's law }\}\\ &\Rightarrow \mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)=1-\mathrm{P}(\mathrm{A} \cup \mathrm{B})=1-5 / 9=4 / 9\\ &\therefore P\left(A^{\prime}\right)+P\left(B^{\prime}\right)=2 / 3+4 / 9=10 / 9 \end{aligned}$

Question:106

Fill in the blanks in the following question:
If X follows binomial distribution with parameters n = 5, p and P (X = 2) = 9.P (X = 3), then p = ___________

Answer:

p = 1/10
As n = 5 {representing no. of trials}
p = probability of success
As it is a binomial distribution.
∴ probability of failure = q = 1 – p
Given-
P(X = 2) = 9.P(X = 3)
The binomial distribution formula is:
$\mathrm{P}(\mathrm{x})=^{n} \mathrm{C}_{x} \mathrm{P}^{\mathrm{x}}(1-\mathrm{P})^{\mathrm{n}-\mathrm{x}}$$

Where:
x = total number of “successes.”
P = probability of success on an individual trial
n = number of trials
using binomial distribution,
$\\\Rightarrow{ }^{5} \mathrm{C}_{2} \mathrm{p}^{2} \mathrm{q}^{5-2}=9^{5} \mathrm{C}_{3} \mathrm{p}^{3} \mathrm{q}^{5-3}$ \\$\Rightarrow 10 p^{2} q^{3}=9 \times 10 p^{3} q^{2}$ \\$\Rightarrow 10 q=90 p\{A s, p \neq 0$ and $q \neq 0\}$ \\$\Rightarrow q=9 p$ \\$\Rightarrow 1-p=9 p \Rightarrow 10 p=1$ \\$\therefore p=1 / 10$$

Question:107

Fill in the blanks in the following question:
Let X be a random variable taking values x₁, x₂,..., x_n with probabilities p₁, p₂, ..., p_n, respectively. Then var (X) =

Answer:

$\\\operatorname{var}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^{2}\right)-[\mathrm{E}(\mathrm{X})]^{2},$ where $\mathrm{E}(\mathrm{X})$ represents mean or expected value for random variable $\mathrm{X}$$
Variance is the mean of deviation of Random variable from its expected value.
$\operatorname{Var}(X)=\sum_{D i}\left(X_{i}-X\right)^{2}$
On expanding we get the formula: $\operatorname{var}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^{2}\right)-[\mathrm{E}(\mathrm{X})]^{2}$$

Question:108

Fill in the blanks in the following question:
Let A and B be two events. If P(A | B) = P(A), then A is ___________ of B.

Answer:

we know that
$P(A \mid B)=\frac{P(A \cap B)}{P(B)}$
Given-
$\\\frac{P(A \cap B)}{P(B)}=P(A)$ \\$\Rightarrow P(A \cap B)=P(A) P(B)$$
This implies that A and B are independent of each other.
$\therefore \mathrm{A}$ is independent of $\mathrm{B}$.$

NCERT exemplar solutions for Class 12 Maths chapter 13 provided here for the NCERT books are very useful and detailed from the point of view of aiding practice, preparation and working for Board exams as well as the JEE Main exams.

NCERT exemplar Class 12 Maths solutions chapter 13 PDF download are also available for students for extended learning. The topics covered are as follows:

Main Subtopics

Introduction
Conditional Probability
Properties of conditional probability
Multiplication theorem on probability
Independent events
Bayes’ Theorem
Partition of a sample space
Theorem of total probability
Random Variables and its Probability Distributions
Probability distribution of a random variable
Mean of a random variable
Variance of a random variable
Bernoulli Trials and Binomial Distribution
Bernoulli trials
Binomial distribution

What will the students learn from NCERT Exemplar Class 12 Math Solutions Chapter 13?

Today’s society has adopted a very practical approach to life and has understood the principle of chance in nature. Similarly, the importance of studying probability could be inferred from the basic lives of people which displays a great example at how there is a constant need to understand the possibility and extract information out of it for one’s benefit. NCERT exemplar Class 12 Math solutions chapter 13 will help students to understand each topic in a better way. Such instances of life display the necessity and importance of learning the concept of probability or possibility of any event to make important decisions in their life and how they are drawn based on chances. Such as in life you come across a lot of data which need to be interpreted, and decisions are to be extracted out of such data on the basis of probable thinking which necessitates the learning of probabilistic thinking.

NCERT exemplar Class 12 Math solutions chapter 13 are easy to grasp and can be helpful in scoring well in the exams. It is not only theoretical/academic oriented but provides a great exposure regarding the nature of chance and variation in life when you encounter such situations in real life. It also develops and exhibits real-life situations teaching the art of risk analysis and management, providing a better approach towards understanding risks in real life and displaying relative chances of events in one’s life. This topic also helps in scientific reasoning and also has its imprint in various fields such as engineering, mathematician, research analyst, etc. Probability also supports statistics and helps improve and analyze data-driven decision-making skills for real life.