NCERT Exemplar Class 12 Maths Solutions Chapter 13 Probability

NCERT Exemplar Class 12 Maths Solutions Chapter 13 Probability

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, the same number each time and a total score is 10 or more respectively. Determine whether or not A and B are independent.

Answer:

A loaded die is thrown such that $\mathrm{P}(1)=\mathrm{P}(2)=0.2, \mathrm{P}(3)=\mathrm{P}(5)=\mathrm{P}(6)=0.1$ and $\mathrm{P}(4)=0.3$ and die is thrown two times.

Also given that: $\mathrm{A}=$ Same number each time and

$\mathrm{B}=$ Total score is 10 or more.

So, $\mathrm{P}(\mathrm{A})=[\mathrm{P}(1,1)+\mathrm{P}(2,2)+\mathrm{P}(3,3)+\mathrm{P}(4,4)+\mathrm{P}(5,5)+\mathrm{P}(6,6)]$

$\begin{aligned} & =P(1) \cdot P(1)+P(2) \cdot P(2)+P(3) \cdot P(3)+P(4) \cdot P(4)+P(5) \cdot P(5)+P(6) \cdot P(6) \\ & 0.2 \times 0.2+0.2 \times 0.2+0.1 \times 0.1+0.3 \times 0.3+0.1 \times 0.1+0.1 \times 0.1 \\ & =0.04+0.04+0.01+0.09+0.01+0.01=0.20\end{aligned}$

Now $B=[(4,6),(6,4),(5,5),(5,6),(6,5),(6,6)]$

$\begin{aligned} & \mathrm{P}(\mathrm{B})=[\mathrm{P}(4) \cdot \mathrm{P}(6)+\mathrm{P}(6) \cdot \mathrm{P}(4)+\mathrm{P}(5) \cdot \mathrm{P}(5)+\mathrm{P}(5) \cdot \mathrm{P}(6)+\mathrm{P}(6) \cdot \mathrm{P}(5)+\mathrm{P}(6) \cdot \mathrm{P}(6) \\ & =0.3 \times 0.1+0.1 \times 0.3+0.1 \times 0.1+0.1 \times 0.1+0.1 \times 0.1+0.1 \times 0.1 \\ & =0.03+0.03+0.01+0.01+0.01+0.01=0.10\end{aligned}$

$A$ and $B$ both events will be independent if

$\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})$..............(i)

Here, $(\mathrm{A} \cap \mathrm{B})=\{(5,5),(6,6)\}$

$\begin{aligned} & \therefore \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(5,5)+\mathrm{P}(6,6)=\mathrm{P}(5) \cdot \mathrm{P}(5)+\mathrm{P}(6) \cdot \mathrm{P}(6) \\ & =0.1 \times 0.1+0.1 \times 0.1 \\ & =0.02\end{aligned}$

From equation (i) we get

$\begin{aligned} & 0.02=0.20 \times 0.10 \\ & 0.02=0.02\end{aligned}$

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:

We have $\mathrm{A}=\{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\}$
$\therefore \mathrm{n}(\mathrm{A})=6$ and $\mathrm{n}(\mathrm{S})=6 \times 6=36$

So, $\mathrm{P}(\mathrm{A})=\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{6}{36}=\frac{1}{6}$

And $B=\{(4,6),(6,4),(5,5),(5,6),(6,5),(6,6)\}$

$\begin{aligned} & n(B)=6 \text { and } n(S)=36 \\ & \therefore P(B)=\frac{n(B)}{n(S)}=\frac{6}{36}=\frac{1}{6} \\ & A \cap B=\{(5,5),(6,6)\} \\ & \therefore P(A \cap B)=\frac{2}{36}=\frac{1}{18}\end{aligned}$

Therefore, if A and B are independent Then $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})$

$\begin{aligned} & \Rightarrow \frac{1}{18} \neq \frac{1}{6} \times \frac{1}{6} \\ & \Rightarrow \frac{1}{18} \neq \frac{1}{36}\end{aligned}$

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:

Let red marble be represented with R and black marble with B.

The following three conditions are possible.

If at least one of the three marbles drawn is black and the first marble is red.

(i) $\mathbf{E}_{\mathbf{1}}$ : II ball is black and III is red

(ii) $\mathbf{E}_2:$ II ball is black and III is also black

(iii) $\mathbf{E}_3$ : II ball is red and III is black

$\begin{aligned} & \therefore \mathrm{P}\left(\mathrm{E}_1\right)=\mathrm{P}\left(\mathrm{R}_1\right) \cdot \mathrm{P}\left(\frac{\mathrm{B}_1}{\mathrm{R}_1}\right) \cdot \mathrm{P}\left(\frac{\mathrm{R}_2}{\mathrm{R}_1 \mathrm{~B}_1}\right) \\ & =\frac{5}{8} \cdot \frac{3}{7} \cdot \frac{4}{6} \\ & =\frac{60}{336} \\ & =\frac{5}{28} \\ & \mathrm{P}\left(\mathrm{E}_2\right)=\mathrm{P}\left(\mathrm{R}_1\right) \cdot \mathrm{P}\left(\frac{\mathrm{B}_1}{\mathrm{R}_1}\right) \cdot \mathrm{P}\left(\frac{\mathrm{B}_2}{\mathrm{R}_1 \mathrm{~B}_1}\right) \\ & =\frac{5}{8} \cdot \frac{3}{7} \cdot \frac{2}{6} \\ & =\frac{30}{336} \\ & =\frac{5}{36}\end{aligned}$

And $\mathrm{P}\left(\mathrm{E}_3\right)=\mathrm{P}\left(\mathrm{R}_1\right) \cdot \mathrm{P}\left(\frac{\mathrm{R}_2}{\mathrm{R}_1}\right) \cdot \mathrm{B}\left(\frac{\mathrm{B}_1}{\mathrm{R}_1 \mathrm{R}_2}\right)$

$\begin{aligned} & =\frac{5}{8} \cdot \frac{4}{7} \cdot \frac{3}{6} \\ & =\frac{60}{336} \\ & =\frac{5}{28} \\ & \therefore \mathrm{P}(\mathrm{E})=\mathrm{P}\left(\mathrm{E}_1\right)+\mathrm{P}\left(\mathrm{E}_2\right)+\mathrm{P}\left(\mathrm{E}_3\right)=\frac{5}{28}+\frac{5}{56}+\frac{5}{28}=\frac{25}{56}\end{aligned}$

Hence the required probability is $\frac{25}{56}$.

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 a total of 4
F= a total of 9 or more
G= a total divisible by 5
Therefore, for E,
E = a of a total of 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 independent pair

Question 6 Explain why the experiment of tossing a coin three times is said to have a 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 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:

i) $\begin{aligned} & \mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)=1-\mathrm{P}(\mathrm{A} \cup \mathrm{B}) \\ & =1-[\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})] \\ & =1-\left[\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right] \\ & =1-\left[\frac{6+4+3}{12}\right] \\ & =1-\frac{7}{12} \\ & =\frac{5}{12} \\ & \mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{B}}\right)=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})} \\ & ={ }^`(1 / 4) /(1 / 3) \\ & =\frac{3}{4}\end{aligned}$

ii) $\begin{aligned} & \mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)=1-\mathrm{P}(\mathrm{A} \cup \mathrm{B}) \\ & =1-[\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})] \\ & =1-\left[\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right] \\ & =1-\left[\frac{6+4+3}{12}\right] \\ & =1-\frac{7}{12} \\ & =\frac{5}{12} \\ & \mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{A}}\right)=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})} \\ & =\frac{\frac{1}{4}}{\frac{1}{2}} \\ & =\frac{1}{2}\end{aligned}$

iii) $\begin{aligned} & \mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)=1-\mathrm{P}(\mathrm{A} \cup \mathrm{B}) \\ & =1-[\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})] \\ & =1-\left[\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right] \\ & =1-\left[\frac{6+4+3}{12}\right] \\ & =1-\frac{7}{12} \\ & =\frac{5}{12} \\ & \mathrm{P}\left(\frac{A^{\prime}}{\mathrm{B}}\right)=\frac{\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}\right)}{\mathrm{P}(\mathrm{B})} \\ & =\frac{\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})} \\ & =1-\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}\end{aligned}$

$\begin{aligned} & =1-\frac{\frac{1}{4}}{\frac{1}{3}} \\ & =1-\frac{3}{4} \\ & =\frac{1}{4}\end{aligned}$

iv) $\begin{aligned} & \mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)=1-\mathrm{P}(\mathrm{A} \cup \mathrm{B}) \\ & =1-[\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})] \\ & =1-\left[\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right] \\ & =1-\left[\frac{6+4+3}{12}\right] \\ & =1-\frac{7}{12} \\ & =\frac{5}{12}\end{aligned}$

$\begin{aligned} & \mathrm{P}\left(\frac{\mathrm{A}^{\prime}}{\mathrm{B}^{\prime}}\right)=\frac{\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)}{\mathrm{P}\left(\mathrm{B}^{\prime}\right)} \\ & =\frac{\frac{5}{12}}{\frac{2}{3}} \\ & =\frac{5}{12} \times \frac{3}{12} \\ & =\frac{5}{8}\end{aligned}$

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 the events whose probabilities are
(i) $p_1 p_2$
(ii) $\left(1-p_1\right) p_2$
(iii) $1-\left(1-p_1\right)\left(1-p_2\right)$
(iv) $p_1+p_2-2 p_1 p_2$

Answer:

i) Here, $\mathrm{P}\left(\mathrm{E}_1\right)=\mathrm{P}_1$ and $\mathrm{P}\left(\mathrm{E}_2\right)=\mathrm{P}_2$

$\begin{aligned} & P_1 P_2=P\left(E_1\right) \cdot P\left(E_2\right) \\ & =P\left(E_1 \cap E_2\right)\end{aligned}$

So, $E_1$ and $E_2$ occur.

ii) Here, $\mathrm{P}\left(\mathrm{E}_1\right)=\mathrm{P}_1$ and $\mathrm{P}\left(\mathrm{E}_2\right)=\mathrm{P}_2$

$\begin{aligned} & \left(1-P_1\right) \cdot P_2=P\left(E_1\right)^{\prime} \cdot P\left(E_2\right) \\ & =P\left(E_1^{\prime} \cap E_2\right)\end{aligned}$

So, $\mathrm{E}_1$ does not occur but $\mathrm{E}_2$ occurs.

iii) Here, $\mathrm{P}\left(\mathrm{E}_1\right)=\mathrm{P}_1$ and $\mathrm{P}\left(\mathrm{E}_2\right)=\mathrm{P}_2$

$\begin{aligned} & 1-\left(1-\mathrm{P}_1\right)\left(1-\mathrm{P}_2\right)=1-\mathrm{P}\left(\mathrm{E}_1\right)^{\prime} \mathrm{P}\left(\mathrm{E}_2\right)^{\prime} \\ & =1-\mathrm{P}\left(\mathrm{E}_1^{\prime} \cap \mathrm{E}_2^{\prime}\right) \\ & =1-\left[1-\mathrm{P}\left(\mathrm{E}_1 \cup \mathrm{E}_2\right)\right] \\ & =\mathrm{P}\left(\mathrm{E}_1 \cup \mathrm{E}_2\right)\end{aligned}$

So, either $E_1$ or $E_2$ or both $E_1$ and $E_2$ occur.

iv) Here, $\mathrm{P}\left(\mathrm{E}_1\right)=\mathrm{P}_1$ and $\mathrm{P}\left(\mathrm{E}_2\right)=\mathrm{P}_2$

$\begin{aligned} & P_1+P_2-2 P_1 P_2=P\left(E_1\right)+P\left(E_2\right)-2 P\left(E_1\right) \cdot P\left(E_2\right) \\ & =P\left(E_1\right)+P\left(E_2\right)-2 P\left(E_1 \cap E_2\right) \\ & =P\left(E_1 \cup E_2\right)-2 P\left(E_1 \cap E_2\right)\end{aligned}$

So, either $E_1$ or $E_2$ occurs but not both.

Question 10

A discrete random variable X has the probability distribution given 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:

i) $\begin{aligned} & \sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{P}_{\mathrm{i}}=1 \\ & \Rightarrow \mathrm{k}+\mathrm{k}^2+2 \mathrm{k}^2+\mathrm{k}=1 \\ & \Rightarrow 3 \mathrm{k} 2+2 \mathrm{k}-1=0 \\ & \Rightarrow 3 \mathrm{k}^2+3 \mathrm{k}-\mathrm{k}-1=0 \\ & \Rightarrow 3 \mathrm{k}(\mathrm{k}+1)-1(\mathrm{k}+1)=0 \\ & \Rightarrow(3 \mathrm{k}-1)(\mathrm{k}+1)=0 \\ & \therefore \mathrm{k}=\frac{1}{3} \text { and } \mathrm{k}=-1\end{aligned}$

But $\mathrm{k} \geq 0$

$\therefore \mathrm{k}=\frac{1}{3}$

ii) For a probability distribution, we know that if $\mathrm{P}_{\mathrm{i}} \geq 0$

Mean of the distribution

$\begin{aligned} & \mathrm{E}(\mathrm{X})=\sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{X}_{\mathrm{i}} \mathrm{P}_{\mathrm{i}} \\ & =0.5 \mathrm{k}+1 . \mathrm{k}^2+1.5\left(2 \mathrm{k}^2\right)+2 \mathrm{k} \\ & =\frac{\mathrm{k}}{2}+\mathrm{k}^2+3 \mathrm{k}^2+2 \mathrm{k} \\ & =4 \mathrm{k}^2+\frac{5}{2} \mathrm{k} \\ & =4\left(\frac{1}{3}\right)^2+\frac{5}{2}\left(\frac{1}{3}\right) \\ & =\frac{4}{9}+\frac{5}{6} \\ & =\frac{23}{18}\end{aligned}$

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:

i) To prove, $\mathrm{P}(\mathrm{A})=\mathrm{P}(\mathrm{A} \cap \mathrm{B})+\mathrm{P}(\mathrm{A} \cap \overline{\mathrm{B}})$

$\begin{aligned} & \text { R.H.S. }=P(A \cap B)+P(A \cap \bar{B}) \\ & =P(A) \cdot P(B)+P(A) \cdot P(\bar{B}) \\ & =P(A)[P(B)+P(\bar{B})] \\ & =P(A) \cdot 1 \\ & =P(A) \\ & =\text { L.H.S. }\end{aligned}$

ii) To prove, $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A} \cap \mathrm{B})+\mathrm{P}(\mathrm{A} \cap \overline{\mathrm{B}})+\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})$

$\begin{aligned} & \text { R.H.S. }=P(A) \cdot P(B)+P(A) \cdot P(\overline{\mathrm{~B}})+\mathrm{P}(\overline{\mathrm{A}}) \cdot \mathrm{P}(\mathrm{B}) \\ & =\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{A})[1-\mathrm{P}(\mathrm{B})]+[1-\mathrm{P}(\mathrm{A})] \cdot \mathrm{P}(\mathrm{B}) \\ & =\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B}) \\ & =\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\ & =\mathrm{P}(\mathrm{A} \cup \mathrm{B}) \\ & =\text { L.H.S. }\end{aligned}$

Question 12

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

Answer:

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

$\begin{aligned} & =3-\left(\frac{3}{2}\right)^2 \\ & =3-\frac{9}{4} \\ & =\frac{3}{4}\end{aligned}$

$\therefore$ Standard deviation $=\sqrt{\operatorname{Var}(\mathrm{X})}$

$\begin{aligned} & =\sqrt{\frac{3}{4}} \\ & =\frac{\sqrt{3}}{2} .\end{aligned}$

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
The 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 the ice shows other 1 or 6.
$\mathrm{P}(\mathrm{X})=\mathrm{P}(1)+\mathrm{P}(6)$
$P(X)=\frac{1}{6}+\frac{1}{6}$
$P(X)=\frac{1}{3}$
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}\\
&\text { ∴ } \text { Player's expected profit }=E(X)=X P(X)\\
&\begin{aligned}
& =-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}
\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)] = 6 3 = 216.
Let E 1 be the event when the sum of numbers on the dice was six and
E 2 be the event when three twos occur.
$ \Rightarrow 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. The first prize is Rs 3000 and the second prize is 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 the third prize is $\frac{3}{10000}$.
1 first prize of 3000, so the probability of winning the third prize is $\frac{1}{10000}$.
1 second prize of 2000, so the probability of winning the 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, and Band ag 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 E 1 be the event that bag I is selected
E 2 be the event that bag II is selected
E 3 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 the second ball being blue?

Answer:

Given-
The box has 5 blue and 4 red balls.
Let E 1 be the event that the first ball drawn is blue
E 2 be the event that the first ball drawn is red and
E be the event that the 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 $E_1, E_2, E_3$ and $E_4$ be the events tthe hat first, second, third and fourth card is King respectively.

$\begin{aligned} & \therefore P\left(E_1 \cap E_2 \cap E_3 \cap E_4\right) \\ & =P\left(E_1\right) \cdot P\left(\frac{E_2}{E_1}\right) \cdot P\left[\frac{E_3}{\left(\mathrm{E}_1 \cap \mathrm{E}_2\right)}\right] \cdot \mathrm{P}\left[\frac{\mathrm{E}_4}{\left(\mathrm{E}_1 \cap \mathrm{E}_2 \cap \mathrm{E}_3 \cap \mathrm{E}_4\right)}\right] \\ & =\frac{4}{52} \times \frac{3}{51} \times \frac{2}{50} \times \frac{1}{49} \\ & =\frac{24}{52 \cdot 51 \cdot 50 \cdot 49} \\ & =\frac{1}{13 \cdot 17 \cdot 25 \cdot 49} \\ & =\frac{1}{27075}\end{aligned}$

Hence, the required probability is $\frac{1}{27075}$.

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
Here, $p=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{1}{2}$

$\mathrm{q}=1-\frac{1}{2}=\frac{1}{2}$

$\begin{aligned} & \therefore \mathrm{P}(\mathrm{x}=\mathrm{r})={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{p}^{\mathrm{r}} \mathrm{q}^{\mathrm{n}-\mathrm{r}} \\ & ={ }^5 \mathrm{C}_3\left(\frac{1}{3}\right)^3\left(\frac{1}{2}\right)^{5-3} \\ & =\frac{5!}{3!2!} \cdot\left(\frac{1}{2}\right)^3 \cdot\left(\frac{1}{2}\right)^2 \\ & =10 \cdot \frac{1}{8} \cdot \frac{1}{4} \\ & =\frac{5}{16}\end{aligned}$

Hence, the required probability is $\frac{5}{16}$.

Question 21

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

Answer:

Let X = the random variable for getting ahead.
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:

Here $\mathrm{n}=7$

$\mathrm{p}=0.25=\frac{25}{100}=\frac{1}{4}$

And $\mathrm{q}=1-\frac{1}{4}=\frac{3}{4}$

$\begin{aligned} & \mathrm{P}(\mathrm{X} \geq 2)=1-[\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)] \\ & =1-\left[{ }^7 \mathrm{C}_0\left(\frac{1}{4}\right)^0\left(\frac{3}{4}\right)^7+{ }^7 \mathrm{C}_1\left(\frac{1}{4}\right)^1\left(\frac{3}{4}\right)^6\right] \\ & =1-\left[\left(\frac{3}{4}\right)^7+\frac{7}{4}\left(\frac{3}{4}\right)^6\right] \\ & =1-\left(\frac{3}{4}\right)^6\left(\frac{3}{4}+\frac{7}{4}\right) \\ & =1-\left(\frac{3}{4}\right)^6\left(\frac{10}{4}\right) \\ & =1-\frac{729}{4096} \times \frac{10}{4} \\ & =1-\frac{7290}{16384} \\ & =\frac{16384-7290}{16384} \\ & =\frac{9094}{16384}\end{aligned}$

$=\frac{4547}{8192}$

Hence, the required probability is $\frac{4547}{8192}$.

Question 23

A lot of 100 watches are 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:

Probability of defective watch out of 100 watches $=\frac{10}{100}=\frac{1}{10}$. Here, $\mathrm{n}=8$

$\begin{aligned} & p=\frac{1}{10} \\ & q=1-\frac{1}{10}=\frac{9}{10}\end{aligned}$

And $r \geq 1$

$\begin{aligned} & \mathrm{P}(\mathrm{X} \geq 1)=1-\mathrm{P}(\mathrm{x}=0) \\ & =1-{ }^8 \mathrm{C}_0\left(\frac{1}{10}\right)^0\left(\frac{9}{10}\right)^{8-0} \\ & =1-\left(\frac{9}{10}\right)^8\end{aligned}$

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) We know that $\mathrm{P}(0)+\mathrm{P}(1)+\mathrm{P}(2)+\mathrm{P}(3)=1$

$\begin{aligned} & \Rightarrow \mathrm{k}+\frac{\mathrm{k}}{2}+\frac{\mathrm{k}}{4}+\frac{\mathrm{k}}{8}=1 \\ & \Rightarrow \frac{8 \mathrm{k}+4 \mathrm{k}+2 \mathrm{k}+\mathrm{k}}{8}=1 \\ & \Rightarrow 15 \mathrm{k}=8 \\ & \therefore \mathrm{k}=\frac{8}{15}\end{aligned}$

ii) $\begin{aligned} & \mathrm{P}(\mathrm{X} \leq 2)=\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2) \\ & =\mathrm{k}+\frac{\mathrm{k}}{2}+\frac{\mathrm{k}}{4} \\ & =\frac{7 \mathrm{k}}{4} \\ & =\frac{7}{4} \times \frac{8}{15} \\ & =\frac{14}{15}\end{aligned}$

And $\mathrm{P}(\mathrm{X}>2)=\mathrm{P}(\mathrm{X}=3)$

$\begin{aligned} & =\frac{\mathrm{k}}{8} \\ & =\frac{1}{8} \times \frac{8}{15} \\ & =\frac{1}{15}\end{aligned}$

iii) $\begin{aligned} & P(X \leq 2)+P(X>2)=\frac{14}{15}+\frac{1}{15} \\ & =\frac{14+1}{15} \\ & =\frac{15}{15} \\ & =1\end{aligned}$
Question 26

For the following probability distribution determine the 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 are 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 are 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 that: for the first die, $\mathrm{P}(6)=\frac{1}{2}$ And $P(\overline{6})=1-\frac{1}{2}=\frac{1}{2}$

$\Rightarrow \mathrm{P}(1)+\mathrm{P}(2)+\mathrm{P}(3)+\mathrm{P}(4)+\mathrm{P}(5)=\frac{1}{2}$

But $\mathrm{P}(1)=\mathrm{P}(2)=\mathrm{P}(3)=\mathrm{P}(4)=\mathrm{P}(5)$

$\begin{aligned} & \therefore 5 . P(1)=\frac{1}{2} \\ & \Rightarrow P(1)=\frac{1}{10} \text { and } P(\overline{1})=1-\frac{1}{10}=\frac{9}{10}\end{aligned}$

For the second die, $\mathrm{P}(1)=\frac{2}{5}$ and $\mathrm{P}(\overline{1})=1-\frac{2}{5}=\frac{3}{5}$
Let X be the number of one's seen

$\begin{aligned} & \therefore \mathrm{X}=0,1,2 \\ & \Rightarrow \mathrm{P}(\mathrm{X}=0)=\mathrm{P}(\overline{1}) \cdot \mathrm{P}(\overline{1}) \\ & =\frac{9}{10} \cdot \frac{3}{5} \\ & =\frac{27}{50} \\ & =0.54\end{aligned}$

$\begin{aligned} & \mathrm{P}(\mathrm{X}=1)=\mathrm{P}(\overline{1}) \cdot \mathrm{P}(1)+\mathrm{P}(1) \cdot \mathrm{P}(\overline{1}) \\ & =\frac{9}{10} \cdot \frac{2}{5}+\frac{1}{10} \cdot \frac{3}{5} \\ & =\frac{18+3}{50} \\ & =\frac{21}{50} \\ & =0.42 \\ & \mathrm{P}(\mathrm{X}=2)=\mathrm{P}(1)_{\mathrm{I}} \cdot \mathrm{P}(1)_{\mathrm{II}} \\ & =\frac{1}{10} \cdot \frac{2}{5} \\ & =\frac{2}{50} \\ & =0.04\end{aligned}$

Hence, the required probability distribution is

$
\begin{array}{|c|c|c|c|}
\hline \mathrm{X} & 0 & 1 & 2 \\
\hline \mathrm{P}(\mathrm{X}) & 0.54 & 0.42 & 0.04 \\
\hline
\end{array}
$

Question 30

Two probability distributions of the discrete random variables 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:

We know that, $\mathrm{E}(\mathrm{X})=\sum_{\mathrm{i}=11}^{\mathrm{n}} \mathrm{P}_{\mathrm{i}} \mathrm{X}_{\mathrm{i}}$

$\begin{aligned} & \Rightarrow \mathrm{E}(\mathrm{X})=0 \cdot \frac{1}{5}+1 \cdot \frac{2}{5}+2 \cdot \frac{1}{5}+3 \cdot \frac{1}{5} \\ & =0+\frac{2}{5}+\frac{2}{5}+\frac{3}{5} \\ & =\frac{7}{5}\end{aligned}$

$\begin{aligned} & \mathrm{E}\left(\mathrm{Y}^2\right)=0 \cdot \frac{1}{5}+1 \cdot \frac{3}{10}+4 \cdot \frac{2}{5}+9 \cdot \frac{1}{10} \\ & =0+\frac{3}{10}+\frac{8}{5}+\frac{9}{10} \\ & =\frac{28}{10} \\ & =\frac{14}{5}\end{aligned}$

Now $E\left(Y^2\right)=\frac{14}{5}$ and $2 E(X)=2 \cdot \frac{7}{5}=\frac{14}{5}$ Hence, $\mathrm{E}\left(\mathrm{Y}^2\right)=2 \mathrm{E}(\mathrm{X})$.

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

Question 33

Suppose that 6% of the people with blood group O are left-handed 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 E 1 be the event that the person selected is of group O
E 2 be the event that the person selected is of other than blood group O
And E 3 be the event that the person selected is left-handed
∴P(E 1 ) =0.30, P(E 2 ) =0.70
P(E 3 |E 1 ) = 0.060 And P(E 3 |E 2 ) =0.10
Using Bayes theorem, we have:
$\begin{aligned} & \mathrm{P}\left(\frac{\mathrm{E}_1}{\mathrm{H}}\right)=\frac{\mathrm{P}\left(\mathrm{E}_1\right) \cdot \mathrm{P}\left(\frac{\mathrm{H}}{\mathrm{E}_1}\right)}{\mathrm{P}\left(\mathrm{E}_1\right) \cdot \mathrm{P}\left(\frac{H}{\mathrm{E}_1}\right)+\mathrm{P}\left(\mathrm{E}_2\right) \cdot \mathrm{P}\left(\frac{H}{\mathrm{E}_2}\right)} \\ & =\frac{0.30 \times 0.06}{0.30 \times 0.06+0.70 \times 0.10} \\ & =\frac{0.018}{0.018+0.070} \\ & =\frac{0.018}{0.088} \\ & =\frac{9}{44}\end{aligned}$

Hence, the required probability is $\frac{9}{44}$.

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:

Given that: $\mathrm{S}=\{1,2,3, \ldots, \mathrm{n}\}$

$\begin{aligned} & \therefore P(r \leq p \mid s \leq p)=\frac{P(P \cap S)}{P(S)} \\ & =\frac{p-1}{n} \times \frac{n}{n-1} \\ & =\frac{p-1}{n-1}\end{aligned}$

Hence, the required probability is $\frac{\mathrm{p}-1}{\mathrm{n}-1}$.

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 when a die is thrown twice.

X=1,2,3,4,5,6

And $S=\{(1,1),(1,2),(2,1),(2,2),(1,3),(2,3),(3,1),(3,2),(3,3),(3,4),(3,5), \ldots,(6,6)\}$

So, $\mathrm{P}(\mathrm{X}=1)=\frac{1}{6} \cdot \frac{1}{6}=\frac{1}{36}$

$\begin{aligned} & \mathrm{P}(\mathrm{X}=2)=\frac{1}{6} \cdot \frac{1}{6}+\frac{1}{6} \cdot \frac{1}{6}+\frac{1}{6} \cdot \frac{1}{6}=\frac{3}{36} \\ & \mathrm{P}(\mathrm{X}=3)=\frac{1}{6} \cdot \frac{1}{6}+\frac{1}{6} \cdot \frac{1}{6}+\frac{1}{6} \cdot \frac{1}{6}+\frac{1}{6} \cdot \frac{1}{6}+\frac{1}{6} \cdot \frac{1}{6}=\frac{5}{36}\end{aligned}$

Similarly $\mathrm{P}(\mathrm{X}=4)=\frac{7}{36}$

$\begin{aligned} & \mathrm{P}(\mathrm{X}=5)=\frac{9}{36} \\ & \text { And } \mathrm{P}(\mathrm{X}=6)=\frac{11}{36}\end{aligned}$

Now, the mean $\mathrm{E}(\mathrm{X})=\sum_{\mathrm{i}=1}^{\mathrm{n}} x_{\mathrm{i}} \mathrm{p}_{\mathrm{i}}$

$\begin{aligned} & =1 \times \frac{1}{36}+2 \times \frac{3}{36}+3 \times \frac{5}{36}+4 \times \frac{7}{36}+5 \times \frac{9}{36}+6 \times \frac{11}{36} \\ & =\frac{1}{36}+\frac{6}{36}+\frac{15}{36}+\frac{28}{36}+\frac{45}{36}+\frac{66}{36} \\ & =\frac{161}{36}\end{aligned}$

Hence, the required mean $=\frac{161}{36}$.

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 that: $\mathrm{X}=0,1,2$

And $P(X)$ at $X=0$ and 1 is $p$.

Let $\mathrm{P}(\mathrm{X})$ at $\mathrm{X}=2$ is x

$\begin{aligned} & \Rightarrow \mathrm{p}+\mathrm{p}+\mathrm{x}=1 \\ & \Rightarrow \mathrm{x}=1-2 \mathrm{p}\end{aligned}$

$\begin{aligned} & \therefore \mathrm{E}(\mathrm{X})=0 \cdot \mathrm{p}+1 \cdot \mathrm{p}+2(1-2 \mathrm{p}) \\ & =\mathrm{p}+2-4 \mathrm{p} \\ & =2-3 \mathrm{p}\end{aligned}$

And $E\left(X^2\right)=0 \cdot p+1 \cdot p+4(1-2 p)$

$\begin{aligned} & =p+4-8 p \\ & =4-7 p\end{aligned}$

Given that: $\mathrm{E}\left(\mathrm{X}^2\right)=\mathrm{E}(\mathrm{X})$

$\begin{aligned} & \therefore 4-7 p=2-3 p \\ & \Rightarrow 4 p=2 \\ & \Rightarrow p=\frac{1}{2}\end{aligned}$

Hence, the required value of p is $\frac{1}{2}$.

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:

We know that, Variance $(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2$

$\begin{aligned} & \mathrm{E}(\mathrm{X})=\sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{p}_{\mathrm{i}} x_{\mathrm{i}} \\ & =0 \times \frac{1}{6}+1 \times \frac{5}{18}+2 \times \frac{2}{9}+3 \times \frac{1}{6}+4 \times \frac{1}{9}+5 \times \frac{1}{18} \\ & =0+\frac{5}{18}+\frac{4}{9}+\frac{3}{6}+\frac{4}{9}+\frac{5}{118} \\ & =\frac{5+8+9+8+5}{18} \\ & =\frac{35}{18} \\ & \mathrm{E}(\mathrm{X})^2=0 \times \frac{1}{6}+1 \times \frac{5}{18}+4 \times \frac{2}{9}+9 \times \frac{1}{6}+16 \times \frac{1}{9}+25 \times \frac{1}{18} \\ & =\frac{5}{18}+\frac{8}{9}+\frac{9}{6}+\frac{16}{9}+\frac{25}{18} \\ & =\frac{5+16+27+32+25}{18} \\ & =\frac{105}{18}\end{aligned}$

$\begin{aligned} & \therefore \operatorname{Var}(\mathrm{x})=\frac{105}{18}-\frac{35}{18} \times \frac{35}{18} \\ & =\frac{1890-1225}{324} \\ & =\frac{665}{324}\end{aligned}$

Hence, the required variance is $\frac{665}{324}$.

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 the 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 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:

Let A be the event having m white and n black balls

$\mathrm{E}_1=\{$ first ball drawn of white colour $\}$

$\mathrm{E}_2=\{$ first ball drawn of black colour $\}$

$E_3=\{$ second ball drawn of white colour $\}$

$\begin{aligned} & \therefore P\left(E_1\right)=\frac{m}{m+n} \text { and } P\left(E_2\right)=\frac{n}{m+n} \\ & P\left(\frac{E_3}{E_1}\right)=\frac{m+k}{m+n+k} \text { and } P\left(\frac{E_3}{E_2}\right)=\frac{m}{m+n+k}\end{aligned}$

$\operatorname{Now} P\left(E_3\right)=P\left(E_1\right) \cdot P\left(\frac{E_3}{E_1}\right)+P\left(E_2\right)\left(\frac{E_3}{E_2}\right)$

$\begin{aligned} & =\frac{\mathrm{m}}{\mathrm{m}+\mathrm{n}} \times \frac{\mathrm{m}+\mathrm{k}}{\mathrm{m}+\mathrm{n}+\mathrm{k}}+\frac{\mathrm{n}}{\mathrm{m}+\mathrm{n}} \times \frac{\mathrm{m}}{\mathrm{m}+\mathrm{n}+\mathrm{k}} \\ & =\frac{\mathrm{m}}{\mathrm{m}+\mathrm{n}+\mathrm{k}}\left[\frac{\mathrm{m}+\mathrm{k}}{\mathrm{m}+\mathrm{n}}+\frac{\mathrm{n}}{\mathrm{m}+\mathrm{n}}\right] \\ & =\frac{\mathrm{m}}{\mathrm{m}+\mathrm{n}+\mathrm{k}}\left[\frac{\mathrm{m}+\mathrm{n}+\mathrm{k}}{\mathrm{m}+\mathrm{n}}\right] \\ & =\frac{\mathrm{m}}{\mathrm{m}+\mathrm{n}}\end{aligned}$

Hence, the probability of drawing a white ball does not depend upon k.

Question 41

Three bags contain several 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:

Given that,

Bag I: 3 red balls and no white ball

Bag II: 2 red balls and 1 white ball

Bag III: no red ball and 3 white balls

Let $E_1, E_2$ and $E_3$ be the events of choosing Bag I, Bag II and Bag III respectively and a ball is drawn from it.

$\begin{aligned} & \therefore \mathrm{P}\left(\mathrm{E}_1\right)=\frac{1}{6} \\ & \mathrm{P}\left(\mathrm{E}_2\right)=\frac{2}{6}\end{aligned}$

And $P\left(E_3\right)=\frac{3}{6}$

i) $\begin{aligned} & \therefore \mathrm{P}(\mathrm{E})=\mathrm{P}\left(\mathrm{E}_1\right) \cdot \mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{E}_1}\right)+\mathrm{P}\left(\mathrm{E}_2\right) \cdot \mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{E}_2}\right)+\mathrm{P}\left(\mathrm{E}_3\right) \cdot \mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{E}_3}\right) \\ & =\frac{1}{6} \cdot \frac{3}{3}+\frac{2}{6} \cdot \frac{2}{3}+\frac{3}{6} \cdot 0 \\ & =\frac{3}{18}+\frac{4}{18} \\ & =\frac{7}{18}\end{aligned}$

ii) Let F be the event that a white ball is selected

$\begin{aligned} & \therefore \mathrm{P}(\mathrm{F})=1-\mathrm{P}(\mathrm{E}) \quad \ldots \ldots[\mathrm{P}(\mathrm{E})+\mathrm{P}(\mathrm{F})=1] \\ & =1-\frac{7}{18} \\ & =\frac{11}{18}\end{aligned}$

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:

i) $\begin{aligned} & \mathrm{P}\left(\frac{\mathrm{E}_2}{\mathrm{~F}}\right)=\frac{\mathrm{P}\left(\mathrm{E}_2\right) \cdot \mathrm{P}\left(\frac{\mathrm{F}}{\mathrm{E}_2}\right)}{\mathrm{P}\left(\mathrm{E}_1\right) \cdot \mathrm{P}\left(\frac{\mathrm{F}}{\mathrm{E}_1}\right)+\mathrm{P}\left(\mathrm{E}_2\right) \cdot \mathrm{P}\left(\frac{\mathrm{F}}{\mathrm{E}_2}\right)+\mathrm{P}\left(\mathrm{E}_3\right) \cdot \mathrm{p}\left(\frac{\mathrm{F}}{\mathrm{E}_3}\right)} \\ & =\frac{\frac{2}{6} \cdot \frac{1}{3}}{\frac{1}{6} \cdot 0+\frac{2}{6} \cdot \frac{1}{3}+\frac{3}{6} \cdot 1} \\ & =\frac{\frac{2}{18}}{\frac{2}{18}+\frac{3}{6}} \\ & =\frac{2}{11}\end{aligned}$

ii) $\begin{aligned} & \mathrm{P}\left(\frac{\mathrm{E}_3}{\mathrm{~F}}\right)=\frac{\mathrm{P}\left(\mathrm{E}_3\right) \cdot \mathrm{P}\left(\frac{\mathrm{F}}{\mathrm{E}_3}\right)}{\mathrm{P}\left(\mathrm{E}_1\right) \cdot \mathrm{P}\left(\frac{\mathrm{F}}{\mathrm{E}_1}\right)+\mathrm{P}\left(\frac{\mathrm{F}}{\mathrm{E}_2}\right)+\mathrm{P}\left(\mathrm{E}_3\right) \cdot \mathrm{P}\left(\frac{\mathrm{F}}{\mathrm{E}_3}\right)} \\ & =\frac{\frac{3}{6} \cdot 1}{\frac{1}{6} \cdot 0+\frac{2}{6} \cdot \frac{1}{3}+\frac{3}{6} \cdot 1} \\ & =\frac{\frac{3}{6}}{\frac{2}{18}+\frac{3}{6}} \\ & =\frac{3}{6} \times \frac{18}{11} \\ & =\frac{9}{11}\end{aligned}$

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:

i) $\begin{aligned} & \mathrm{P}(\mathrm{E})=\mathrm{P}\left(\mathrm{A}_1\right) \cdot \mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{A}_1}\right)+\mathrm{P}\left(\mathrm{A}_2\right) \cdot \mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{A}_2}\right)+\mathrm{P}\left(\mathrm{A}_3\right) \cdot \mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{A}_3}\right) \\ & =\frac{4}{10} \cdot \frac{45}{100}+\frac{4}{10} \cdot \frac{60}{100}+\frac{2}{10} \cdot \frac{35}{100} \\ & =\frac{180}{1000}+\frac{240}{1000}+\frac{70}{1000} \\ & =\frac{490}{1000} \\ & =0.49\end{aligned}$

ii) $\begin{aligned} & P\left(\frac{\overline{\mathrm{E}}}{\mathrm{A}_3}\right)=1-\mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{A}_3}\right) \\ & =1-\frac{35}{1000} \\ & =\frac{65}{100} \\ & =0.65\end{aligned}$

iii) Given that $\mathrm{A}_1: \mathrm{A}_2: \mathrm{A}_3=4: 4: 2$

$\begin{aligned} & \therefore \mathrm{P}\left(\mathrm{A}_1\right)=\frac{4}{10} \\ & \mathrm{P}\left(\mathrm{A}_2\right)=\frac{4}{10}\end{aligned}$

And $\mathrm{P}\left(\mathrm{A}_3\right)=\frac{2}{10}$

Where $A_1, A_2$ and $A_3$ are the three types of seeds.

Let E be the event that seed germinates and $\overline{\mathrm{E}}$ be the event that a seed does not germinate

$\therefore \mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{A}_1}\right)=\frac{45}{100} \mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{A}_2}\right)=\frac{60}{100}$ and $\mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{A}_3}\right)=\frac{35}{100}$

And $\mathrm{P}\left(\frac{\overline{\mathrm{E}}}{\mathrm{A}_1}\right)=\frac{55}{100}, \mathrm{P}\left(\frac{\overline{\mathrm{E}}}{\mathrm{A}_2}\right)=\frac{40}{100}$ and $\mathrm{P}\left(\frac{\overline{\mathrm{E}}}{\mathrm{A}_3}\right)=\frac{65}{100}$

Using Bayes' Theorem, we get

$\begin{aligned} & \mathrm{P}\left(\frac{\mathrm{A}_2}{\overline{\mathrm{E}}}\right)=\frac{\mathrm{P}\left(\mathrm{A}_2\right) \cdot \mathrm{P}\left(\frac{\overline{\mathrm{E}}}{\mathrm{A}_2}\right)}{\mathrm{P}\left(\mathrm{A}_1\right) \cdot \mathrm{P}\left(\frac{\overline{\mathrm{E}}}{\mathrm{A}_1}\right)+\mathrm{P}\left(\mathrm{A}_2\right) \cdot \mathrm{P}\left(\frac{\overline{\mathrm{E}}}{\mathrm{A}_2}\right)+\mathrm{P}\left(\mathrm{A}_3\right) \cdot \mathrm{P}\left(\frac{\overline{\mathrm{E}}}{\mathrm{A}_3}\right)} \\ & =\frac{\frac{4}{10} \cdot \frac{40}{100}}{\frac{4}{10} \cdot \frac{55}{100}+\frac{4}{10} \cdot \frac{40}{100}+\frac{2}{10} \cdot \frac{65}{100}} \\ & =\frac{\frac{160}{1000}}{\frac{220}{1000}+\frac{160}{1000}+\frac{130}{1000}} \\ & =\frac{160}{220+160+130} \\ & =\frac{160}{510} \\ & =\frac{16}{51} \\ & =0.314\end{aligned}$

Hence, the required probability is $\frac{16}{51}$ or 0.314

Question 44

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

Answer:

Let events E1, and E2 be the following events-
E1 be the event that the letter is from TATA NAGAR and E2 be the event that the 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 the letter comes from TATA NAGAR
P(E|E2) is the probability that two consecutive letters are visible when the letter comes 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 as 1 or 3, a ball is taken from the first bag; but if it shows up as any other number, a ball is chosen from the second bag. Find the probability of choosing a black ball.

Answer:

Let $\mathrm{E}_1$ be the event of selecting Bag I
And $E_2$ be the event of selecting Bag II

Let $E_3$ be the event that the black ball is selected

$\begin{aligned} & \therefore \mathrm{P}\left(\mathrm{E}_1\right)=\frac{2}{6}=\frac{1}{3} \text { and } \mathrm{P}\left(\mathrm{E}_2\right)=1-\frac{1}{3}=\frac{2}{3} \\ & \mathrm{P}\left(\frac{\mathrm{E}_3}{\mathrm{E}_1}\right)=\frac{3}{7} \text { and } \mathrm{P}\left(\frac{\mathrm{E}_3}{\mathrm{E}_2}\right)=\frac{4}{7} \\ & \therefore \mathrm{P}\left(\mathrm{E}_3\right)=\mathrm{P}\left(\mathrm{E}_1\right) \cdot \mathrm{P}\left(\frac{\mathrm{E}_3}{\mathrm{E}_1}\right)+\mathrm{P}\left(\mathrm{E}_2\right) \cdot \mathrm{P}\left(\frac{\mathrm{E}_3}{\mathrm{E}_2}\right) \\ & =\frac{1}{3} \cdot \frac{3}{7}+\frac{2}{3} \cdot \frac{4}{7} \\ & =\frac{3+8}{21} \\ & =\frac{11}{21}\end{aligned}$

Hence, the required probability is $\frac{11}{21}$.

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:

We have 3 urns:

$\therefore$ Probabilities of choosing either of the urns are

$\mathrm{P}\left(\mathrm{U}_1\right)=\mathrm{P}\left(\mathrm{U}_2\right)=\mathrm{P}\left(\mathrm{U}_3\right)=\frac{1}{3}$

Let H be the event of drawing a white ball from the chosen urn.

$\begin{aligned} & \therefore \mathrm{P}\left(\frac{\mathrm{H}}{\mathrm{U}_1}\right)=\frac{2}{5} \\ & \mathrm{P}\left(\frac{\mathrm{H}}{\mathrm{U}_2}\right)=\frac{3}{5}\end{aligned}$

And $\mathrm{P}\left(\frac{\mathrm{H}}{\mathrm{U}_3}\right)=\frac{4}{5}$

$\begin{aligned} & \therefore \mathrm{P}\left(\frac{\mathrm{U}_2}{\mathrm{H}}\right)=\frac{\mathrm{P}\left(\mathrm{U}_2\right) \cdot \mathrm{P}\left(\frac{\mathrm{H}}{\mathrm{U}_2}\right)}{\mathrm{P}\left(\mathrm{U}_1\right) \cdot \mathrm{P}\left(\frac{\mathrm{H}}{\mathrm{U}_1}\right)+\mathrm{P}\left(\mathrm{U}_2\right) \cdot \mathrm{P}\left(\frac{\mathrm{H}}{\mathrm{U}_2}\right)+\mathrm{P}\left(\mathrm{U}_3\right) \cdot \mathrm{P}\left(\frac{\mathrm{H}}{\mathrm{U}_3}\right)} \\ & =\frac{\frac{1}{3} \cdot \frac{3}{5}}{\frac{1}{3} \cdot \frac{2}{5}+\frac{1}{3} \cdot \frac{3}{5}+\frac{1}{3} \cdot \frac{4}{5}} \\ & =\frac{\frac{3}{5}}{\frac{2}{5}+\frac{3}{5}+\frac{4}{5}} \\ & =\frac{3}{9} \\ & =\frac{1}{3}\end{aligned}$

Hence, the required probability is $\frac{1}{3}$.

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 a healthy person being diagnosed with 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 $\mathbf{E}_{\mathbf{1}}$ : Event that a person has TB

$\mathbf{E}_{\mathbf{2}}$ : Event that a person does not have TB

And $\mathbf{H}$: Event that the person is diagnosed to have TB.

$\begin{aligned} & \therefore \mathrm{P}\left(\mathrm{E}_1\right)=\frac{1}{1000}=0.001 \\ & \mathrm{P}\left(\mathrm{E}_2\right)=1-\frac{1}{1000}=\frac{999}{1000}=0.999 \\ & \mathrm{P}\left(\frac{\mathrm{H}}{\mathrm{E}_1}\right)=0.99 \\ & \mathrm{P}\left(\frac{\mathrm{H}}{\mathrm{E}_2}\right)=0.001\end{aligned}$

$\begin{aligned} & \therefore \mathrm{P}\left(\frac{\mathrm{E}_1}{\mathrm{H}}\right)=\frac{\mathrm{P}\left(\mathrm{E}_1\right) \cdot \mathrm{P}\left(\frac{\mathrm{H}}{\mathrm{E}_1}\right)}{\mathrm{P}\left(\mathrm{E}_1\right) \cdot \mathrm{P}\left(\frac{\mathrm{H}}{\mathrm{E}_1}\right)+\mathrm{P}\left(\mathrm{E}_2\right) \cdot \mathrm{P}\left(\frac{\mathrm{H}}{\mathrm{E}_2}\right)} \\ & =\frac{0.001 \times 0.99}{0.001 \times 0.99+0.999 \times 0.001} \\ & =\frac{0.99}{00.99+0.999} \\ & =\frac{0.990}{0.990+0.999} \\ & =\frac{990}{1989} \\ & =\frac{110}{221}\end{aligned}$

Hence, the required probability is $\frac{110}{221}$.

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 $\mathbf{E}_{\mathbf{1}}$ : The event that the item is manufactured on machine A

$\mathbf{E}_{\mathbf{2}}$ : The event that the item is manufactured on machine B

$\mathbf{E}_{\mathbf{3}}$ : The event that the item is manufactured on machine C

Let H be the event that the selected item is defective.

$\therefore$ Using Bayes' Theorem,

$\begin{aligned} & \mathrm{P}\left(\mathrm{E}_1\right)=\frac{50}{100} \\ & \mathrm{P}\left(\mathrm{E}_2\right)=\frac{30}{100} \\ & \mathrm{P}\left(\mathrm{E}_3\right)=\frac{20}{100}\end{aligned}$

$\mathrm{P}\left(\frac{\mathrm{H}}{\mathrm{E}_1}\right)=\frac{2}{100}$

$\mathrm{P}\left(\frac{\mathrm{H}}{\mathrm{E}_2}\right)=\frac{30}{100}$

And $P\left(\frac{H}{E_3}\right)=\frac{3}{100}$

$\begin{aligned} & \therefore \mathrm{P}\left(\frac{\mathrm{E}_1}{\mathrm{H}}\right)=\frac{\mathrm{P}\left(\mathrm{E}_1\right) \cdot \mathrm{P}\left(\frac{\mathrm{H}}{\mathrm{E}_1}\right)}{\mathrm{P}\left(\mathrm{E}_1\right) \cdot \mathrm{P}\left(\frac{\mathrm{H}}{\mathrm{E}_1}\right)+\mathrm{P}\left(\mathrm{E}_2\right) \cdot \mathrm{P}\left(\frac{\mathrm{H}}{\mathrm{E}_2}\right)+\mathrm{P}\left(\mathrm{E}_3\right) \cdot \mathrm{P}\left(\frac{\mathrm{H}}{\mathrm{E}_3}\right)} \\ & =\frac{\frac{50}{100} \times \frac{2}{100}}{\frac{50}{100} \times \frac{2}{100}+\frac{30}{100} \times \frac{2}{100}+\frac{20}{100} \times \frac{3}{100}} \\ & =\frac{100}{100+60+60} \\ & =\frac{100}{220} \\ & =\frac{10}{22} \\ & =\frac{5}{11}\end{aligned}$

Hence, the required probability is $\frac{5}{11}$.

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) The standard deviation of X.

Answer:

i) Here, $\mathrm{P}(\mathrm{X}=\mathrm{x})=\mathrm{k}(\mathrm{x}+1)$ for $\mathrm{x}=1,2,3,4$

So, $\mathrm{P}(\mathrm{X}=1)=\mathrm{k}(1+1)=2 \mathrm{k}$

$\begin{aligned} & \mathrm{P}(\mathrm{X}=2)=\mathrm{k}(2+1)=3 \mathrm{k} \\ & \mathrm{P}(\mathrm{X}=3)=\mathrm{k}(3+1)=4 \mathrm{k} \\ & \mathrm{P}(\mathrm{X}=4)=\mathrm{k}(4+1)=5 \mathrm{k}\end{aligned}$

Also, $\mathrm{P}(\mathrm{X}=\mathrm{x})=2 \mathrm{kx}$ for $\mathrm{x}=5,6,7$

$\begin{aligned} & \mathrm{P}(\mathrm{X}=5)=2(5) \mathrm{k}=10 \mathrm{k} \\ & \mathrm{P}(\mathrm{X}=6)=2(6) \mathrm{k}=12 \mathrm{k} \\ & \mathrm{P}(\mathrm{X}=7)=2(7) \mathrm{k}=14 \mathrm{k}\end{aligned}$

We know that $\sum_{i=1}^n P\left(X_i\right)=1$

So, $2 \mathrm{k}+3 \mathrm{k}+4 \mathrm{k}+5 \mathrm{k}+10 \mathrm{k}+12 \mathrm{k}+14 \mathrm{k}=1$

$\begin{aligned} & \Rightarrow 50 \mathrm{k}=1 \\ & \Rightarrow \mathrm{k}=\frac{1}{50}\end{aligned}$

Hence, the value of k is $\frac{1}{50}$

ii) $\begin{aligned} & \mathrm{E}(\mathrm{X})=1 \times \frac{2}{50}+2 \times \frac{3}{50}+3 \times \frac{4}{50}+4 \times \frac{5}{50}+5 \times \frac{10}{50}+6 \times \frac{12}{50}+7 \times \frac{14}{50} \\ & =\frac{2}{50}+\frac{6}{50}+\frac{12}{50}+\frac{50}{50}+\frac{72}{50}+\frac{98}{50} \\ & =\frac{260}{50} \\ & =\frac{26}{5} \\ & ={ }^` 5.2\end{aligned}$

iii) $\begin{aligned} & \text { Variance }=\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2 \\ & \mathrm{E}\left(\mathrm{X}^2\right)=1 \times \frac{2}{50}+4 \times \frac{3}{50}+9 \times \frac{4}{50}+16 \times \frac{5}{50}+25 \times \frac{10}{50}+36 \times \frac{12}{50}+49 \times \frac{14}{50} \\ & =\frac{2}{50}+\frac{12}{50}+\frac{36}{50}+\frac{80}{50}+\frac{250}{50}+\frac{432}{50}+\frac{686}{50} \\ & =\frac{1498}{50} \\ & \therefore \text { Variance }(\mathrm{X})=\frac{1498}{50}-\left(\frac{26}{5}\right)^2 \\ & =\frac{1498}{50}-\frac{676}{25} \\ & =\frac{1498-1352}{50} \\ & =\frac{146}{50} \\ & =2.92\end{aligned}$

$\begin{aligned} & \therefore \mathrm{S} . \mathrm{D}=\sqrt{2.92} \\ & =1.7 \quad \ldots . .(\text { Approx })\end{aligned}$

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} ×1$
$\begin{aligned} &\Rightarrow A=3\\ &\text { (ii) As we know that, }\\ &\operatorname{Var}(X)=E\left(X^{2}\right)-[E(X)]^{2}\\ &=\Sigma X^{2} P(X)-[\Sigma\{X P(X)\}]^{2}\\ &=\Sigma X^{2} P(X)-(2.94)^{2}\\ &\begin{array}{l} \text { We first find } \Sigma X^{2} P(X) \\ =1^{2} \times \frac{1}{2}+2^{2} \times \frac{1}{5}+4^{2} \times \frac{3}{25}+(2 A)^{2} \times \frac{1}{10}+(3 A)^{2} \times \frac{1}{25}+(5 A)^{2} \times \frac{1}{25} \end{array}\\ &=\frac{1}{2}+\frac{4}{5}+\frac{48}{25}+\frac{36}{10}+\frac{81}{25}+\frac{225}{25}\\ &=\frac{25+40+96+180+162+450}{50}\\ &=\frac{953}{50}\\ &=19.06\\ &\operatorname{Var}(X)=19.06-(2.94)^{2}\\ &=19.06-8.6436\\ &=10.4164 \end{aligned}$

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-

i) Given that: $\mathrm{P}(\mathrm{X}=\mathrm{x})= \begin{cases}\mathrm{k} x^2 & \text { for } x=1,2,3 \\ 2 \mathrm{k} x & \text { for } x=4,5,6 \\ 0 & \text { otherwise }\end{cases}$

We know that $\sum_{i=1}^{\mathrm{n}} \mathrm{P}\left(\mathrm{X}_{\mathrm{i}}\right)=1$

$\begin{aligned} & \therefore \mathrm{k}+4 \mathrm{k}+9 \mathrm{k}+8 \mathrm{k}+10 \mathrm{k}+12 \mathrm{k}=1 \\ & \Rightarrow 44 \mathrm{k}=1 \\ & \Rightarrow \mathrm{k}=\frac{1}{44} \\ & \mathrm{E}(\mathrm{X})=\sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{P}_{\mathrm{i}} \mathrm{X}_{\mathrm{i}} \\ & =1 \times \mathrm{k}+2 \times 4 \mathrm{k}+3 \times 9 \mathrm{k}+4 \times 8 \mathrm{k}+5 \times 10 \mathrm{k}+6 \times 12 \mathrm{k} \\ & =\mathrm{k}+8 \mathrm{k}+27 \mathrm{k}+32 \mathrm{k}+50 \mathrm{k}+72 \mathrm{k} \\ & =190 \mathrm{k} \\ & =190 \times \frac{1}{44} \\ & =\frac{95}{22} \\ & =4.32 \ldots \ldots .(\text { Approx })\end{aligned}$

ii) Given that: $\mathrm{P}(\mathrm{X}=\mathrm{x})= \begin{cases}\mathrm{k} x^2 & \text { for } x=1,2,3 \\ 2 \mathrm{k} x & \text { for } x=4,5,6 \\ 0 & \text { otherwise }\end{cases}$

We know that $\sum_{i=1}^n P\left(X_i\right)=1$

$\begin{aligned} & \therefore \mathrm{k}+4 \mathrm{k}+9 \mathrm{k}+8 \mathrm{k}+10 \mathrm{k}+12 \mathrm{k}=1 \\ & \Rightarrow 44 \mathrm{k}=1 \\ & \Rightarrow \mathrm{k}=\frac{1}{44} \\ & \mathrm{E}\left(3 \mathrm{X}^2\right)=3[\mathrm{k}+4 \times 4 \mathrm{k}+9 \times 9 \mathrm{k}+16 \times 8 \mathrm{k}+25 \times 10 \mathrm{k}+36 \times 12 \mathrm{k}] \\ & =3[\mathrm{k}+16 \mathrm{k}+81 \mathrm{k}+128 \mathrm{k}+250 \mathrm{k}+432 \mathrm{k}] \\ & =3[908 \mathrm{k}] \\ & =3 \times 908 \times \frac{1}{44} \\ & =\frac{2724}{44} \\ & =61.9 \ldots . . .(\text { Approx })\end{aligned}$
iii) Given that: $\mathrm{P}(\mathrm{X}=\mathrm{x})= \begin{cases}\mathrm{k} x^2 & \text { for } x=1,2,3 \\ 2 \mathrm{k} x & \text { for } x=4,5,6 \\ 0 & \text { otherwise }\end{cases}$

We know that $\sum_{i=1}^n P\left(X_i\right)=1$

$\begin{aligned} & \therefore \mathrm{k}+4 \mathrm{k}+9 \mathrm{k}+8 \mathrm{k}+10 \mathrm{k}+12 \mathrm{k}=1 \\ & \Rightarrow 44 \mathrm{k}=1 \\ & \Rightarrow \mathrm{k}=\frac{1}{44} \\ & \mathrm{P}(\mathrm{X} \geq 4)=\mathrm{P}(\mathrm{X}=4)+\mathrm{P}(\mathrm{X}=5)+\mathrm{P}(\mathrm{X}=6) \\ & =8 \mathrm{k}+10 \mathrm{k}+12 \mathrm{k}=30 \mathrm{k} \\ & =30 \times \frac{1}{44} \\ & =\frac{15}{22}\end{aligned}$

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 a head on both sides and (n + 1) coins are fair coins
Therefore, Total coins = 2n + 1
Let E1, and 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…….Enaree 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 an unfair coin is tossed
P(E|E2) is the probability of getting a head when a 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 the random variable such that $\mathrm{X}=0,1,2$
And $\mathrm{E}=$ The event of drawing an ace
And $\mathrm{F}=$ The event of drawing non-ace.

$\therefore \mathrm{P}(\mathrm{E})=\frac{4}{52}$ and $\mathrm{P}(\overline{\mathrm{E}})=\frac{48}{52}$

Now $P(X=0)=P(\overline{\mathrm{E}}) \cdot P(\overline{\mathrm{E}})$

$\begin{aligned} & =\frac{48}{52} \cdot \frac{47}{51} \\ & =\frac{188}{221} \\ & \mathrm{P}(\mathrm{X}=1)=\mathrm{P}(\mathrm{E}) \cdot \mathrm{P}(\overline{\mathrm{E}})+\mathrm{P}(\overline{\mathrm{E}}) \cdot \mathrm{P}(\mathrm{E}) \\ & =\frac{4}{52} \times \frac{48}{51} \times \frac{48}{52} \times \frac{4}{51} \\ & =\frac{32}{221} \\ & \mathrm{P}(\mathrm{X}=2)=\mathrm{P}(\mathrm{E}) \cdot \mathrm{P}(\mathrm{E}) \\ & =\frac{4}{52} \cdot \frac{3}{51} \\ & =\frac{1}{221}\end{aligned}$

Now, Mean $\mathrm{E}(\mathrm{X})=0 \times \frac{188}{221}+1 \times \frac{32}{221}+2 \times \frac{1}{221}$

$\begin{aligned} & =\frac{32}{221}+\frac{2}{221} \\ & =\frac{34}{221} \\ & =\frac{2}{13} \\ & \mathrm{E}\left(\mathrm{X}^2\right)=0 \times \frac{188}{221}+1 \times \frac{32}{221}+4 \times \frac{1}{221} \\ & =\frac{32}{221}+\frac{4}{221} \\ & =\frac{36}{221}\end{aligned}$

$\begin{aligned} & \therefore \text { Variance }=\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2 \\ & =\frac{36}{221}-\left(\frac{2}{13}\right)^2 \\ & =\frac{36}{221}-\frac{4}{169} \\ & =\frac{468-68}{13 \times 221} \\ & =\frac{400}{2873} \\ & \text { Standard deviation }=\sqrt{\frac{400}{2873}}\end{aligned}$

= 0.377...(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 the 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. $\frac{1}{10}$ B. $\frac{1}{8}$ C. $\frac{7}{8}$ D. $\frac{17}{20}$

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, the answer is option (C).

Question 57

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

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, the answer is option (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. $\frac{1}{4}$ B. $\frac{1}{3}$ C. $\frac{5}{12}$ D. $\frac{7}{2}$

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 answer is option (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. $\frac{5}{6}$ B. $\frac{5}{7}$ C. $\frac{25}{42}$ 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 answer is the option (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. $\frac{1}{12}$ B. $\frac{3}{4}$ C. $\frac{1}{4}$ D. $\frac{3}{16}$

Answer:

Given that: $\mathrm{P}(\mathrm{A})=\frac{1}{2}, \mathrm{P}(\mathrm{B})=\frac{1}{3}$ and $\mathrm{P}(\mathrm{A} / \mathrm{B})=\frac{1}{4}$

$\begin{aligned} & \mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{B}}\right)=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})} \\ & \frac{1}{4}=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\frac{1}{3}} \\ & \Rightarrow \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{4} \times \frac{1}{3}=\frac{1}{12}\end{aligned}$

Now $\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)=1-\mathrm{P}(\mathrm{A} \cup \mathrm{B})$

$\begin{aligned} & =1-[\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})] \\ & =1-\left[\frac{1}{2}+\frac{1}{3}-\frac{1}{12}\right] \\ & =1-\left[\frac{5}{6}-\frac{1}{12}\right] \\ & =1-\frac{9}{12} \\ & =\frac{3}{12} \\ & =\frac{1}{4}\end{aligned}$

Hence, the answer is the option (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 that: $\mathrm{P}(\mathrm{A})=0.4$

$\mathrm{P}(\mathrm{B})=0.8$

And $P\left(\frac{B}{A}\right)=0.6$

$\begin{aligned} & \mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{A}}\right)=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})} \\ & \Rightarrow 0.6=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{0.4} \\ & \therefore \mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.6 \times 0.4=0.24 \\ & \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\ & =0.4+0.8-0.24 \\ & =1.20-0.24 \\ & =0.96\end{aligned}$

Hence, the answer is the option (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 that: $\mathrm{A} \neq \Phi$ and $\mathrm{B} \neq \Phi$

Then $\mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}$
Hence, the answer is the option (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 that: $\mathrm{P}(\mathrm{A})=0.4, \mathrm{P}(\mathrm{B})=0.3$ and $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=0.5$

$\begin{aligned} & \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\ & 0.5=0.4+0.3-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\ & \mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.4+0.3-0.5=0.2 \\ & \therefore \mathrm{P}\left(\mathrm{B}^{\prime} \cap \mathrm{A}\right)=\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\ & =0.4-0.2 \\ & =0.2 \\ & =\frac{1}{5}\end{aligned}$
Hence, the answer is option (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. $\frac{3}{10}$ B. $\frac{1}{5}$ C. $\frac{1}{2}$ D. $\frac{3}{5}$

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 answer is option (C).

Question 65

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

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 answer is option (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. $\frac{1}{5}$ B. $\frac{4}{5}$ C. $\frac{1}{2}$ 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 answer is option (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

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

Hence, the answer is option (D).

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 that: $\mathrm{P}(\mathrm{A})>0$ and $\mathrm{P}(\mathrm{B}) \neq 1$

$\begin{aligned} & \therefore \mathrm{P}\left(\mathrm{A}^{\prime} \mid \mathrm{B}^{\prime}\right)=\frac{\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)}{\mathrm{P}\left(\mathrm{B}^{\prime}\right)} \\ & =\frac{1-\mathrm{P}(\mathrm{A} \cup \mathrm{B})}{\mathrm{P}\left(\mathrm{B}^{\prime}\right)}\end{aligned}$

Hence, the answer is the option (C).

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:

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

Hence, the answer is option (D).

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 that 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 a 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 the 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, the answer is option (D).

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:

Given that: $\mathrm{P}(\mathrm{A})=\frac{3}{8}, \mathrm{P}(\mathrm{B})=\frac{5}{8}$ and $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{3}{4}$

$\begin{aligned} & \therefore \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\ & \frac{3}{4}=\frac{3}{8}+\frac{5}{8}-\mathrm{P}(\mathrm{A} \cap \mathrm{B})\end{aligned}$

$\Rightarrow \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{3}{8}+\frac{5}{8}-\frac{3}{4}=\frac{1}{4}$

$\begin{aligned} & \text { Now } P\left(\frac{A}{B}\right) \cdot P\left(\frac{A^{\prime}}{B}\right)=\frac{P(A \cap B)}{P(B)} \cdot \frac{P\left(A^{\prime} \cap B\right)}{P(B)} \\ & =\frac{P(A \cap B)}{P(B)} \cdot \frac{P(B)-P(A \cap B)}{P(B)}\end{aligned}$

$\begin{aligned} & =\frac{\frac{1}{4}}{\frac{5}{8}} \cdot \frac{\left(\frac{5}{8}-\frac{1}{4}\right)}{\frac{5}{8}} \\ & =\frac{2}{5} \cdot \frac{3}{5} \\ & =\frac{6}{25}\end{aligned}$
Hence, the answer is option (D).

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:

If the events A and B are independent, then $\mathrm{P}(\mathrm{A} \cap \mathrm{B})$ is equal to $\underline{\mathbf{P}}(\underline{\mathbf{A}}) \cdot \mathbf{P}(\underline{\mathbf{B}})$.

Since $A$ and $B$ are two independent events

$\therefore \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})$

Hence, the answer is option (C).

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}$

Hence, the answer is option (C).

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)
$\\ =\frac{5}{8} \cdot \frac{3}{7} \cdot \frac{2}{6}+\frac{3}{8} \cdot \frac{5}{7} \cdot \frac{2}{6}+\frac{3}{8} \cdot \frac{2}{7} \cdot \frac{5}{6} \\ =\frac{15}{56}$
Hence, the answer is option (C).

Question 75

Refer to Question 74 above. The probability that exactly two of the three balls were red, the first ball is 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 the possibility of getting 1 red ball
P(E)=P(R).P(B)+P(B).P(R)
$\\ \mathrm{P}(\mathrm{E})=\frac{4}{7} \times \frac{3}{6}+\frac{3}{6} \times \frac{4}{7} \\ \text { Hence, } \mathrm{P}(\mathrm{E})=4 / 7$
Hence, the answer is option (B).

Question 76

Three persons, A, B and C, fire at a target in turn, starting with A. Their probability of hitting the target ais0.4, 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-
$\begin{aligned} & \mathrm{P}(\mathrm{A})=0.4 \mathrm{P}(\mathrm{B})=0.3 \text { and } \mathrm{P}(\mathrm{C})=0.2 \\ & \text { Therefore }, \mathrm{P}\left(\mathrm{A}^{\prime}\right)=1-\mathrm{P}(\mathrm{A})=[1-0.4]=0.6 \\ & \mathrm{P}\left(\mathrm{B}^{\prime}\right)=1-\mathrm{P}(\mathrm{B})=[1-0.3]=0.7 \\ & \mathrm{P}\left(\mathrm{C}^{\prime}\right)=1-\mathrm{P}(\mathrm{C})=[1-0.2]=0.8 \\ & \mathrm{P}(\mathrm{E})=\left[\mathrm{P}(\mathrm{A}) \times \mathrm{P}(\mathrm{B}) \times \mathrm{P}\left(\mathrm{C}^{\prime}\right)\right]+\left[\mathrm{P}(\mathrm{A}) \times \mathrm{P}\left(\mathrm{B}^{\prime}\right) \times \mathrm{P}(\mathrm{C})\right]+\left[\mathrm{P}\left(\mathrm{A}^{\prime}\right) \times \mathrm{P}(\mathrm{B}) \times \mathrm{P}(\mathrm{C})\right] \\ & =(0.4 \times 0.3 \times 0.8)+(0.4 \times 0.7 \times 0.2)+(0.6 \times 0.3 \times 0.2) \\ & =0.96+0.056+0.036 \\ & =0.188\end{aligned}$

Hence, the answer is option (B).

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 the eldest child is a 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, $\mathrm{P}\left(\mathrm{E}_{2} \mid \mathrm{E}_{1}\right)=\frac{4}{7}$
Hence, the answer is option (D).

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}
$\begin{aligned} &\text { since, } P(A)=\frac{3}{6}=\frac{1}{2}\\ &\mathrm{P}(\mathrm{B})=\frac{13}{52}=\frac{1}{4}\\ &\text { As we know, If } \mathrm{E} \text { and } \mathrm{F} \text { are two independent events then, } \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})\\ &\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{2} \times \frac{1}{4}=\frac{1}{8}\\ &\text { Hence, } \mathrm{P}\left(\mathrm{E}_{1} \cap \mathrm{E}_{2}\right)=\frac{1}{8} \end{aligned}$

Hence, the answer is option (C).

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 a total of 8 balls in the 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, $\quad P(E)=\frac{3}{28}$

Hence, the answer is the option (A).

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
The number of dead batteries is = 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 = $\frac{3}{8}\times \frac{2}{7}= \frac{3}{28}$

Hence, the answer is option (D).

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 $\mathrm{P}(\mathrm{X}=\mathrm{r})={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}(\mathrm{p})^{r} \mathrm{ q}^{\mathrm{n}-\mathrm{r}}$
The total number of coin tossed, n=8
The probability of getting head, $\mathrm{p}=\frac 12$
The probability of getting tail,$\mathrm{q}=\frac12$
The Required probability
$\\={ }^{8} \mathrm{C}_{3}\left(\frac{1}{2}\right)^{3}\left(\frac{1}{2}\right)^{8-3}\\$ =$\frac{8 \times 7 \times 6}{3 \times 2} \times \frac{1}{2^{8}}=\frac{7}{32}$

Hence, the answer is option (B).

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 $\frac{1}{5}$

Hence, the answer is option (C).

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, the answer is option (C).

Question 84

Two cards are drawn from a well-shuffled deck of 52 playing cards with replacements. 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 cards is shuffled again with replacement, then
Probability of getting queen is , $\frac{4}{52}$
Therefore, the probability, that both cards are queens, $\left [\frac{4}{52} \times\frac{4}{52} \right ]$
Hence, Probability is $\left [\frac{1}{13} \times \frac{1}{13} \right ]$

Hence, the answer is option (A).

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:

Here, $\mathrm{n}=10$

$\mathrm{p}=\frac{1}{2}$ and $\mathrm{q}=\frac{1}{2} \quad \ldots$. (For true/false questions)

And $r \geq 8$ i.e. $8,9,10$

$\begin{aligned} & \therefore \mathrm{P}(\mathrm{X} \geq 8)=\mathrm{P}(\mathrm{x}=8)+\mathrm{P}(\mathrm{x}=9)+\mathrm{P}(\mathrm{x}=10) \\ & ={ }^{10} \mathrm{C}_8\left(\frac{1}{2}\right)^8\left(\frac{1}{2}\right)^2+{ }^{10} \mathrm{C}_9\left(\frac{1}{2}\right)^9\left(\frac{1}{2}\right)+{ }^{10} \mathrm{C}_{10}\left(\frac{1}{2}\right)^{10}\left(\frac{1}{2}\right)^0 \\ & =45 \cdot\left(\frac{1}{2}\right)^{10}+10 \cdot\left(\frac{1}{2}\right)^{10}+\left(\frac{1}{2}\right)^{10}+\left(\frac{1}{2}\right)^{10} \\ & =\left(\frac{1}{2}\right)^{10}(45+10+1) \\ & =56 \times \frac{1}{1024} \\ & =\frac{7}{128}\end{aligned}$

Hence, the answer is option (B).

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 that: $\overline{\mathrm{P}}=0.3$

$\begin{aligned} & \therefore \mathrm{p}=0.7 \text { and } \mathrm{q}=1-0.7=0.3 \\ & \mathrm{n}=5 \text { and } \mathrm{r}=4\end{aligned}$

We know that

$\begin{aligned} & \mathrm{P}(\mathrm{X}=\mathrm{r})={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}(\mathrm{p})^{\mathrm{r}} \cdot(\mathrm{q})^{\mathrm{n}-\mathrm{r}} \\ & \therefore \mathrm{P}(\mathrm{x}=4)={ }^5 \mathrm{C}_4(0.7)^4(0.3)^{5-4} \\ & ={ }^5 \mathrm{C}_4(0.7)^4(0.3)\end{aligned}$

Hence, the answer is option (A).

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
Hence, the answer is option (C).

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$
Hence, the answer is option (D).

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$
Hence, the answer is option (D).

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 $p=\frac{1}{2}$
Hence, the answer is option (A).

Question 91

In a college, 30% of 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 $\mathrm{E}_1$ be the event that the student fails in Physics and $\mathrm{E}_2$ be the event that she fails in Mathematics.

$\begin{aligned} & \therefore \mathrm{P}\left(\mathrm{E}_1\right)=\frac{30}{100} \\ & \mathrm{P}\left(\mathrm{E}_2\right)=\frac{25}{100}\end{aligned}$

And $P\left(E_1 \cap E_2\right)=\frac{10}{100}$

$\begin{aligned} & \therefore \mathrm{P}\left(\frac{\mathrm{E}_1}{\mathrm{E}_2}\right)=\frac{\mathrm{P}\left(\mathrm{E}_1 \cap \mathrm{E}_2\right)}{\mathrm{P}\left(\mathrm{E}_2\right)} \\ & =\frac{\frac{10}{100}}{\frac{25}{100}} \\ & =\frac{2}{5}\end{aligned}$

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 $\mathrm{E}_1$ be the event that both of them solve the problem.

$\therefore \mathrm{P}\left(\mathrm{E}_1\right)=\frac{1}{3} \times \frac{1}{4}=\frac{1}{12}$

And $E_2$ be the event that both of them incorrectly the problem.

$\begin{aligned} & \therefore \mathrm{P}\left(\mathrm{E}_2\right)=\left(1-\frac{1}{3}\right) \times\left(1-\frac{1}{4}\right) \\ & =\frac{2}{3} \times \frac{3}{4}=\frac{1}{2}\end{aligned}$

Let H be the event that both of them get the same answer. Here, $\mathrm{P}\left(\frac{\mathrm{H}}{\mathrm{E}_1}\right)=1$

$\begin{aligned} & \mathrm{P}\left(\frac{\mathrm{H}}{\mathrm{E}_2}\right)=\frac{1}{20} \\ & \therefore \mathrm{P}\left(\frac{\mathrm{E}_1}{\mathrm{H}}\right)=\frac{\mathrm{P}\left(\mathrm{E}_1\right) \cdot \mathrm{P}\left(\frac{\mathrm{H}}{\mathrm{E}_1}\right)}{\mathrm{P}\left(\mathrm{E}_1\right) \cdot \mathrm{P}\left(\frac{\mathrm{H}}{\mathrm{E}_1}\right)+\mathrm{P}\left(\mathrm{E}_2\right) \cdot \mathrm{P}\left(\frac{\mathrm{H}}{\mathrm{E}_2}\right)} \\ & =\frac{\frac{1}{12} \times 1}{\frac{1}{12} \times 1+\frac{1}{2} \times \frac{1}{20}} \\ & =\frac{\frac{1}{12}}{\frac{1}{12}+\frac{1}{40}} \\ & =\frac{\frac{1}{12}}{\frac{10+3}{120}} \\ & =\frac{\frac{1}{12}}{\frac{13}{120}} \\ & =\frac{10}{13}\end{aligned}$
Hence, the answer is option (D).

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 as 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 defective pens.
∴ P (drawing almost 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.
Hence, the answer is option (D).

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:
Events are mutually exclusive when–
P(A∪B) = P(A) + P(B)
But as per the conditions in question, they don't need to meet the condition because it might be possible.
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.

Hence, the statement is false.

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:
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, the statement is true.

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:
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, the statement is false.

Question 97

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

Answer:
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, the statement is false.

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:
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 the statement is true.

Hence, the 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 the expected value.

Answer:
Mean gives the average of values and if it is related with probability or variable it is called the led expected value.

Hence, the statement is true.

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:
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, the statement is true.

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:

Exactly one of A and B occurs.

This means if occurs B does not occur and if B occurs A does not occur.

$\therefore$ Required probability $=\mathrm{P}\left(\mathrm{A} \cap \mathrm{B}^{\prime}\right)+\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}\right)$

$=\mathrm{P}(\mathrm{A}) \mathrm{P}\left(\mathrm{B}^{\prime}\right)+\mathrm{P}\left(\mathrm{A}^{\prime}\right) \mathrm{P}(\mathrm{B})$

Since $A$ and $B$ are independent the $n A^{\prime}$ and $B^{\prime}, A$ and $B^{\prime}$ are also independent

Hence, the 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:

$\begin{aligned} & \because \mathrm{P}(\mathrm{B} \mid \mathrm{A})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})} \\ & =\frac{\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cup \mathrm{B})}{\mathrm{P}(\mathrm{A})}>\frac{1-\mathrm{P}(\mathrm{A} \cup B)}{\mathrm{P}(\mathrm{A})}\end{aligned}$

Hence, the statement is false.

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:

Since $P$ (at least two of A, B and C occur)

$\begin{aligned} & =p \times p \times(1-p)+(1-p) \cdot p \cdot p+p(1-p) \cdot p+p \cdot p \cdot p \\ & =3 p^2(1-p)+p^3 \\ & =3 p^2-3 p^3+p^3 \\ & =3 p^2-2 p^3\end{aligned}$
Hence, the 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:

Given that, $\mathrm{P}(\mathrm{A})=\mathrm{p}$

$P(B)=\frac{1}{3}$

And $P(A \cup B)=\frac{5}{9}$

$\mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\mathrm{p}$

$\Rightarrow \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{p}$

$\begin{aligned} & \mathrm{P}(\mathrm{B})=\mathrm{p} \cdot \frac{1}{3} \text { and } \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\ & \frac{5}{9}=\mathrm{p}+\frac{1}{3}-\frac{\mathrm{p}}{3} \\ & \Rightarrow \frac{5}{9}-\frac{1}{3}=\frac{2 \mathrm{p}}{3} \\ & \Rightarrow \frac{2}{9}=\frac{2 \mathrm{p}}{3} \\ & \Rightarrow \mathrm{p}=\frac{1}{3}\end{aligned}$

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:

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

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:

Given that: $\mathrm{P}(\mathrm{X}=2)=9 \mathrm{P}(\mathrm{X}=3)$

$\begin{aligned} & \Rightarrow{ }^5 \mathrm{C}_2 \mathrm{p}^2 \mathrm{q}^3=9 .{ }^5 \mathrm{C}_3 \mathrm{p}^3 \mathrm{q}^2 \\ & \Rightarrow \frac{1}{9}=\frac{{ }^5 \mathrm{C}_3 \mathrm{p}^2 \mathrm{q}^2}{{ }^5 \mathrm{C}_2 \mathrm{p}^2 \mathrm{q}^3} \\ & \Rightarrow \frac{1}{9}=\frac{\mathrm{p}}{\mathrm{q}} \ldots \ldots . .\left[\because{ }^5 \mathrm{C}_3={ }^5 \mathrm{C}_2\right] \\ & \Rightarrow 9 \mathrm{p}=\mathrm{q} \\ & \Rightarrow 9 \mathrm{p}=1-\mathrm{p} \\ & \Rightarrow 9 \mathrm{p}+\mathrm{p}=1 \\ & \Rightarrow 10 \mathrm{p}=1 \\ & \therefore \mathrm{p}=\frac{1}{10}\end{aligned}$

Question 107

Fill in the blanks in the following Question
Let X be a random variable taking values x 1 , x 2 ,..., x n with probabilities p 1 , p 2 , ..., p n , respectively. Then var (X) =

Answer:

$\begin{aligned} & \operatorname{Var}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2 \\ & =\sum \mathrm{X}^2 \mathrm{P}(\mathrm{X})-\left[\sum \mathrm{X} \cdot \mathrm{P}(\mathrm{X})\right]^2 \\ & =\sum \mathrm{p}_{\mathrm{i}} x_{\mathrm{i}}^2-\left(\sum \mathrm{p}_{\mathrm{i}} x_{\mathrm{i}}\right)^2\end{aligned}$

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:

$\begin{aligned} & \because \mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})} \\ & \Rightarrow \mathrm{P}(\mathrm{A})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})} \\ & \Rightarrow \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})\end{aligned}$

$\therefore \mathrm{A}$ is independent of $\mathrm{B}$.

NCERT Exemplar Class 12 Maths Solutions Chapter 13 Probability: Topics

• 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
• The mean of a random variable
• Variance of a random variable
• Bernoulli Trials and Binomial Distribution
• Bernoulli trials
• Binomial distribution

Importance of Solving NCERT Exemplar Class 12 Maths Solutions Chapter 13

• The exemplar problems go beyond the basics, helping students grasp more advanced concepts with greater clarity.
• Various types of questions, like MCQs, fill-in-the-blanks, true-false, short-answer type, and long-answer type, will enhance the logical and analytical skills of the students.
• These NCERT exemplar exercises cover all the important topics and concepts so that students can be well-prepared for various exams.
• By solving these NCERT exemplar problems, students will get to know about all the real-life applications of probability.

NCERT Exemplar Class 12 Maths Solutions Chapterwise

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