NCERT Exemplar Class 12 Maths Solutions chapter 13 help students to interrelate such knowledge with real-life problems and display the application of such knowledge in different life scenarios and situations along with enhancing decision-making at an efficient level. It is a highly scoring chapter of NCERT Class 12 Maths Solutions that a student can utilize to gain higher scores in their exams.
Class 12 Maths Chapter 13 exemplar solutions Exercise: 13.3
Page number: 271-286
Total questions: 108 |
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) $1-\left(1-P_{1}\right)\left(1-P_{2}\right)$ \\$(\mathrm{iv}) \mathrm{P}_{1}+\mathrm{P}_{2}-2 \mathrm{P}_{1} \mathrm{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}$ ,\$=\frac{1}{3}$
and at X=4 if die shows a 3
$\mathrm{P}(\mathrm{X})=\mathrm{P}(3)$ ,\$\mathrm{P}(\mathrm{X})=\frac{1}{6}$$
$\begin{aligned} &\begin{array}{|l|c|l|l|} \hline \mathrm{X} & -1 & 1 & 4 \\ \hline \mathrm{P}(\mathrm{X}) & \frac{1}{2} & \frac{1}{3} & \frac{1}{6} \\ \hline \end{array}\\ &\therefore \text { Player's expected profit }=E(X)=X P(X)\\ &=-1 \times \frac{1}{2}+1 \times \frac{1}{3}+4 \times \frac{1}{6}\\ &=\frac{-3+2+4}{6}\\ &=\frac{3}{6}\\ &=\frac{1}{2}=\text { Rs } 0.50 \end{aligned}$
Question:14
Three dice are thrown at the same time. Find the probability of getting three twos’, if it is known that the sum of the numbers on the dice was six.
Answer:
Since three dice are thrown at the same time, the sample space is [n(S)] = 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.
$\\ \Rightarrow \quad E_{1}=\{(1,1,4),(1,2,3),(1,3,2),(1,4,1),(2,1,3),(2,2,2,),(2,3,1),(3,1,2),(3,2,1),(4,1,1)\} \\ \Rightarrow n\left(E_{1}\right)=10 \text { and } E_{2}\{2,2,2\} \\ \Rightarrow n\left(E_{2}\right)=1 \\ \text { And }\left(E_{1} \cap E_{2}\right)=1 \\ P\left(E_{1}\right)=\frac{10}{216} \\ P\left(E_{1} \cap E_{2}\right)=\frac{1}{216} \\ \therefore P\left(E_{2} \mid E_{1}\right)=\frac{P\left(E_{1} \cap E_{2}\right)}{P\left(E_{1}\right)} \\ \frac{\frac{1}{216}}{\frac{10}{216}}=\frac{1}{10}$
Question:15
Suppose 10,000 tickets are sold in a lottery each for Re 1. 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} \\$x1
$\begin{aligned} &\Rightarrow A=3\\ &\text { (ii) As we know that, }\\ &\operatorname{Var}(X)=E\left(X^{2}\right)-[E(X)]^{2}\\ &=\Sigma X^{2} P(X)-[\Sigma\{X P(X)\}]^{2}\\ &=\Sigma X^{2} P(X)-(2.94)^{2}\\ &\begin{array}{l} \text { We first find } \Sigma X^{2} P(X) \\ =1^{2} \times \frac{1}{2}+2^{2} \times \frac{1}{5}+4^{2} \times \frac{3}{25}+(2 A)^{2} \times \frac{1}{10}+(3 A)^{2} \times \frac{1}{25}+(5 A)^{2} \times \frac{1}{25} \end{array}\\ &=\frac{1}{2}+\frac{4}{5}+\frac{48}{25}+\frac{36}{10}+\frac{81}{25}+\frac{225}{25}\\ &=\frac{25+40+96+180+162+450}{50}\\ &=\frac{953}{50}\\ &=19.06\\ &\operatorname{Var}(X)=19.06-(2.94)^{2}\\ &=19.06-8.6436\\ &=10.4164 \end{aligned}$
Question:51
The probability distribution of a random variable x is given as under: $\mathrm{P}(\mathrm{X}=\mathrm{x})=\left\{\begin{array}{ll} \mathrm{kx}^{2} & \text { for } \mathrm{x}=1,2,3 \\ 2 \mathrm{kx} & \text { for } \mathrm{x}=4,5,6 \\ 0 & \text { otherwise } \end{array}\right.$ where k is a constant. Calculate (i) E(X) (ii) $E (3X^2)$ (iii) P(X ≥ 4)
Answer:
Given-
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, $\mathrm{p}=1 / 2$
The probability of getting tail,$\mathrm{q}=1 / 2$
The Required probability
$\\={ }^{8} \mathrm{C}_{3}\left(\frac{1}{2}\right)^{3}\left(\frac{1}{2}\right)^{8-3}\\$ =$\frac{8 \times 7 \times 6}{3 \times 2} \times \frac{1}{2^{8}}=\frac{7}{32}$
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 solutions for Class 12 Maths chapter 13 provided here for the NCERT books are very useful and detailed from the point of view of aiding practice, preparation and working for Board exams as well as the JEE Main exams.
NCERT Exemplar Class 12 Maths Solutions Chapter 13 PDF downloads are also available for students for extended learning. The topics covered are as follows:
