NCERT Exemplar Class 12 Maths Solutions Chapter 13 Probability

NCERT Exemplar Class 12 Maths Solutions chapter 13 deals with the possibility of any event or explains how likely it is for an event to occur. The possibility of occurring or not occurring any event cannot be predicted but can be displayed in the form of probability with an absolute level of certainty. The probability of any event lies between 0 to 1 where 0 shows the absolute impossibility of an event, and 1 shows the maximum chances of happening of an event certainly.

NCERT Exemplar Class 12 Maths Solutions chapter 13 help students to interrelate such knowledge with real-life problems and display the application of such knowledge in different 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.

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

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

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

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

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

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

From equation (i) we get

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

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

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

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

$\begin{array}{rl}& \Rightarrow \frac{1}{18}\ne \frac{1}{6}\times \frac{1}{6}\\ & \Rightarrow \frac{1}{18}\ne \frac{1}{36}\end{array}$

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 $P(\overline{A})+P(\overline{B})$.

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- $P(\overline{A})+P(\overline{B})$
Therefore,
$$\begin{gathered} P(\overline{A})+P(\overline{B})=[1-P(A)+1-P(B)] \\ \Rightarrow P(\overline{A})+P(\overline{B})=2-[P(A)+P(B)] \\ \Rightarrow P(\overline{A})+P(\overline{B})=2-0.9 \\ P(\overline{A})+P(\overline{B})=1.1 \end{gathered}$$

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

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

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

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

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
$$\begin{gathered} P(E)=\frac{n(E)}{n(S)}=\frac{3}{36}=\frac{1}{12} \\ P(F)=\frac{n(F)}{n(S)}=\frac{10}{36}=\frac{5}{18} \\ P(G)=\frac{n(G)}{n(S)}=\frac{7}{36} \\ P(F\cap G)=\frac{3}{36}=\frac{1}{12} \\ P(F)\cdot P(G)=\frac{5}{18}\times \frac{7}{36}=\frac{35}{648} \end{gathered}$$
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 $r=0,1,2,\dots ..n$ \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 $q=\frac{1}{2}$
$\begin{array}{|lllll|}\hline \mathrm{X}& 0& 1& 2& 3\\ \mathrm{P}(\mathrm{X})& 3_{{\mathrm{c}}_0}{\mathrm{q}}^3& 3_{{\mathrm{c}}_1}{\mathrm{pq}}^2& 3_{{\mathrm{c}}_2}{\mathrm{p}}^2\mathrm{q}& 3_{{\mathrm{c}}_3}{\mathrm{p}}^3\\ \hline\end{array}$
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) $\begin{array}{rl}& \mathrm{P}({\mathrm{A}}^{\prime }\cap {\mathrm{B}}^{\prime })=1-\mathrm{P}(\mathrm{A}\cup \mathrm{B})\\ & =1-[\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A}\cap \mathrm{B})]\\ & =1-[\frac{1}{2}+\frac{1}{3}-\frac{1}{4}]\\ & =1-\left[\frac{6+4+3}{12}\right]\\ & =1-\frac{7}{12}\\ & =\frac{5}{12}\\ & \mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{B}}\right)=\frac{\mathrm{P}(\mathrm{A}\cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}\\ & ={}^{‘}(1/4)/(1/3)\\ & =\frac{3}{4}\end{array}$

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

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

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

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

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

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 $P(B\cap C)=\frac{1}{4}$, find the values of P (C | B) and P (A'∩ C').

Answer:

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

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 $\mathrm{P}\left({\mathrm{E}}_2\right)={\mathrm{P}}_2$

$\begin{array}{rl}& P_1{P}_2=P\left(E_1\right)\cdot P\left(E_2\right)\\ & =P(E_1\cap E_2)\end{array}$

So, $E_1$ and $E_2$ occur.

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

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

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 $\mathrm{P}\left({\mathrm{E}}_2\right)={\mathrm{P}}_2$

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

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 $\mathrm{P}\left({\mathrm{E}}_2\right)={\mathrm{P}}_2$

$\begin{array}{rl}& P_1+P_2-2P_1{P}_2=P\left(E_1\right)+P\left(E_2\right)-2P\left(E_1\right)\cdot P\left(E_2\right)\\ & =P\left(E_1\right)+P\left(E_2\right)-2P(E_1\cap E_2)\\ & =P(E_1\cup E_2)-2P(E_1\cap E_2)\end{array}$

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

Question:10

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

Answer:

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

But $\mathrm{k}\ge 0$

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

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

Mean of the distribution

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

Question:11

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

Answer:

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

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

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

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

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
$$\begin{gathered} \Rightarrow \mathrm{P}(\mathrm{X}=\mathrm{x}){=}^{\mathrm{n}}{C}_{\mathrm{x}}(\mathrm{p}{)}^{\mathrm{n}}{\mathrm{q}}^{\mathrm{n}-\mathrm{x}} \\ \text{ Where }\mathrm{n}=3,\mathrm{p}=\frac{1}{2},\mathrm{q}=\frac{1}{2}\text{ and }\mathrm{x}=0,1,2,3 \end{gathered}$$
$\begin{array}{|lccll|}\hline \mathrm{X}& 0& 1& 2& 3\\ \mathrm{P}(\mathrm{X})& \frac{1}{8}& \frac{3}{8}& \frac{3}{8}& \frac{1}{8}\\ \mathrm{XP}(\mathrm{X})& 0& \frac{3}{8}& \frac{3}{4}& \frac{3}{8}\\ {\mathrm{X}}^2\mathrm{P}(\mathrm{X})& 0& \frac{3}{8}& \frac{3}{2}& \frac{9}{8}\\ \hline\end{array}$
$\begin{array}{rl}& \text{ As we know, Var }(X)=E\left(X^2\right)-[E(X){]}^2\dots \dots \text{ (i) }\\ & \text{ Where, }E\left(X^2\right)=\sum _{i=1}^n{x}_i^{2}P(x)\text{ and }E(X)=\sum _{i=1}^n{x}_{i}P\left(x_i\right)\\ & \therefore \left[\mathrm{E}\left({\mathrm{X}}^2\right)\right]=\sum _{i=1}^{\mathrm{n}}{\mathrm{x}}_i^2\mathrm{P}\left({\mathrm{X}}_i\right)\\ & =0+\frac{3}{8}+\frac{3}{2}+\frac{9}{8}=\frac{24}{8}\\ & =3\\ & \text{ And }[\mathrm{E}(\mathrm{X}){]}^2={\left[\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_i\mathrm{P}\left({\mathrm{X}}_{\mathrm{i}}\right)\right]}^2\end{array}$
We know that $\mathrm{Var}(\mathrm{X})=\mathrm{E}\left({\mathrm{X}}^2\right)-[\mathrm{E}(\mathrm{X}){]}^2$

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

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

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

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)
$P(X)=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}$ $=\frac{3}{6}$\mathrm{\$}=\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}$$
$\begin{array}{rl}& \begin{array}{|lcll|}\hline \mathrm{X}& -1& 1& 4\\ \mathrm{P}(\mathrm{X})& \frac{1}{2}& \frac{1}{3}& \frac{1}{6}\\ \hline\end{array}\\ & \therefore \text{ Player's expected profit }=E(X)=XP(X)\\ & =-1\times \frac{1}{2}+1\times \frac{1}{3}+4\times \frac{1}{6}\\ & =\frac{-3+2+4}{6}\\ & =\frac{3}{6}\\ & =\frac{1}{2}=\text{ Rs }0.50\end{array}$

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.
$$\begin{gathered} \Rightarrow E_1=\{(1,1,4),(1,2,3),(1,3,2),(1,4,1),(2,1,3),(2,2,2,),(2,3,1),(3,1,2),(3,2,1),(4,1,1)\} \\ \Rightarrow n\left(E_1\right)=10\text{ and }{E}_2\{2,2,2\} \\ \Rightarrow n\left(E_2\right)=1 \\ \text{ And }(E_1\cap E_2)=1 \\ P\left(E_1\right)=\frac{10}{216} \\ P(E_1\cap E_2)=\frac{1}{216} \\ \therefore P(E_2\mid E_1)=\frac{P(E_1\cap E_2)}{P\left(E_1\right)} \\ \frac{\frac{1}{216}}{\frac{10}{216}}=\frac{1}{10} \end{gathered}$$

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 $\frac{1}{10000}$.
$\begin{array}{rl}& \begin{array}{|lllll|}\hline \mathrm{X}& 0& 500& 2000& 3000\\ \mathrm{P}(\mathrm{X})& \frac{9995}{10000}& \frac{3}{10000}& \frac{1}{10000}& \frac{1}{10000}\\ \hline\end{array}\\ & \text{ since, }E(X)=X(PX)\\ & \text{ Therefore, }\\ & \mathrm{E}(\mathrm{X})=0\times \frac{9995}{10000}+\frac{1500}{10000}+\frac{2000}{10000}+\frac{3000}{10000}\\ & =\frac{1500+2000+3000}{10000}\\ & =\frac{6500}{10000}\\ & =\frac{13}{20}=\mathrm{Rs}0.65\end{array}$

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.
$\begin{array}{rl}& \therefore \mathrm{P}(\mathrm{E}\mid {\mathrm{E}}_1)=\frac{0}{17},\mathrm{P}(\mathrm{E}\mid {\mathrm{E}}_2)=\frac{9}{17}\\ & \text{ And }\\ & \mathrm{P}\left({\mathrm{E}}_1\right)=\frac{4}{9}\text{ and }\mathrm{P}\left({\mathrm{E}}_2\right)=\frac{5}{9}\\ & \therefore P(E)=P\left(E_1\right)\cdot P(E\mid E_1)+P\left(E_2\right)\cdot P(E\mid E_2)\\ & =\frac{4}{9}\times \frac{10}{17}+\frac{5}{9}\times \frac{9}{17}\\ & =\frac{40+45}{153}\\ & =\frac{85}{153}\\ & =\frac{5}{9}\end{array}$

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,
$$\begin{gathered} \Rightarrow P(E1)=P(E2)=\frac{1}{2} \\ P(E\mid E_2)=\frac{3}{5} \\ P(E\mid E_2)=\frac{2}{6}=\frac{1}{3} \\ \therefore P(E)=P\left(E_1\right)\cdot P(E\mid E_1)+P\left(E_2\right)\cdot P(E\mid E_2) \\ =\frac{1}{2}\times \frac{3}{5}+\frac{1}{2}\times \frac{2}{6} \\ =\frac{3}{10}+\frac{2}{12} \\ =\frac{18+10}{60} \\ =\frac{28}{60} \\ =\frac{7}{15} \end{gathered}$$

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.
$$\begin{gathered} P\left(E_1\right)=\frac{5}{9} \\ P\left(E_2\right)=\frac{4}{9} \\ P(E\mid E_1)=\frac{4}{8} \\ P(E\mid E_2)=\frac{5}{8} \\ \therefore P(E)=P\left(E_1\right).P(E\mid E_1)+P\left(E_2\right)\cdot P(E\mid E_2) \\ =\frac{5}{9}\times \frac{4}{8}+\frac{4}{9}\times \frac{5}{8} \\ =2\left(\frac{20}{72}\right) \\ =\frac{40}{72} \\ =\frac{5}{9} \end{gathered}$$