Exercise 13.1 (Revised) - Chapter 13 - Probability - Ncert Solutions class 12 - Maths

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NCERT Class 12 Maths Solutions Chapter 13: Probability

Ex 13. 1 Question 1.

Given that $E$ and $F$ are events such that $P(E)=0.6, P(F)=0.3$ and $P(E \cap F)=0.02$, find $\mathbf{P}(E \mid F)$ and $\mathbf{P}(F \mid E)$

Answer.

Given: $\mathrm{P}(\mathrm{E})=0.6, \mathrm{P}(\mathrm{F})=0.3$ and $\mathrm{P}(\mathrm{E} \cap \mathrm{F})=0.2$
$
\begin{aligned}
& \therefore P(E \mid F)=\frac{P(E \cap F)}{P(F)}=\frac{0.2}{0.3}=\frac{2}{3} \\
& \text { And } P(F \mid E)=\frac{P(E \cap F)}{P(E)}=\frac{0.2}{0.6}=\frac{1}{3}
\end{aligned}
$

Ex 13. 1 Question 2.

Compute $P(A \mid B)$ if $P(B)=0.5$ and $P(A \cap B)=0.32$

Answer.

Given: $\mathrm{P}(\mathrm{B})=0.5, \mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.32$
$
\therefore P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{0.32}{0.50}=\frac{32}{50}=0.64
$

Ex 13. 1 Question 3.

If $P(A)=0.8, P(B)=0.5$ and $P(B \mid A)=0.4$, find:
(i) $\mathbf{P}(\mathbf{A} \cap \mathbf{B})$
(ii) $\mathbf{P}(\mathrm{A} \mid \mathrm{B})$
(iii) $\mathbf{P}(\mathrm{A} \cup \mathrm{B})$

Answer.

(i) $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{B} \mid \mathrm{A}) \cdot \mathrm{P}(\mathrm{A})=0.4 \times 0.8=0.32$

(ii) $\mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{0.32}{0.50}=\frac{32}{50}=0.64$
(iii) $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.8+0.5-0.32=0.98$

Ex 13. 1 Question 4.
Evaluate $P(A \cup B)$, if $2 P(A)=P(B)=\frac{5}{13}$ and $P(A \mid B)=\frac{2}{5}$
Answer:

It is given that,
$
2 P(A)=P(B)=\frac{5}{13}
$
$\therefore P(A)=\frac{5}{26}$
$
\begin{aligned}
& P(A \mid B)=\frac{2}{5} \\
& \Rightarrow \frac{P(A \cap B)}{P(B)}=\frac{2}{5} \\
& \Rightarrow P(A \cap B)=\frac{2}{5} \times P(B)=\frac{2}{5} \times \frac{5}{13}=\frac{2}{13}
\end{aligned}
$

It is known that, $P(A \cup B)=P(A)+P(B)-P(A \cap B)$
$
\begin{aligned}
& \Rightarrow P(A \cup B)=\frac{5}{26}+\frac{5}{13}-\frac{2}{13} \\
& \Rightarrow P(A \cup B)=\frac{5+10-4}{26} \\
& \Rightarrow P(A \cup B)=\frac{11}{26}
\end{aligned}
$

Ex 13. 1 Question 5:

If $P(A)=\frac{6}{11}, P(B)=\frac{5}{11}$ and $P(A \cup B)=\frac{7}{11}$, find
(i) $\mathrm{P}(\mathrm{A} \cap$B)
(ii) $\mathrm{P}(\mathrm{A} \mid \mathrm{B})$
(iii) $P(B \mid A)$

Answer:

It is given that
$
\mathrm{P}(\mathrm{A})=\frac{6}{11}, \mathrm{P}(\mathrm{B})=\frac{5}{11} \text {, and } \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{7}{11}
$
$
P(A \cup B)=\frac{7}{11}
$
$
\begin{aligned}
& \therefore P(A)+P(B)-P(A \cap B)=\frac{7}{11} \\
& \Rightarrow \frac{6}{11}+\frac{5}{11}-P(A \cap B)=\frac{7}{11} \\
& \Rightarrow P(A \cap B)=\frac{11}{11}-\frac{7}{11}=\frac{4}{11}
\end{aligned}
$

(ii) It is known that,
$
P(A \mid B)=\frac{P(A \cap B)}{P(B)}
$

Ex 13. 1 Question 6.

Determine $P(E \mid F)$ : A coin is tossed three times.
(i) $\mathrm{E}$ : heads on third toss, $\mathrm{F}$ : heads on first two tosses.
(ii) $\mathrm{E}$ : at least two heads, $\mathrm{F}$ : at most two heads.
(iii) E : at most two tails, F : at least one tail.

Answer.

A coin tossed three times, i.e.,
S = (TTT, HTT, THT, TTH, HHT, HTH, THH, HHH)
$
\Rightarrow n(S)=8
$
(i) $\mathrm{E}$ : heads on third toss
$\mathrm{E}=(\mathrm{TTH}, \mathrm{HTH}, \mathrm{THH}, \mathrm{HHH})$
$
\begin{aligned}
& \Rightarrow n(\mathrm{E})=4 \\
& \mathrm{P}(\mathrm{E})=\frac{n(\mathrm{E})}{n(\mathrm{~S})}=\frac{4}{8}=\frac{1}{2}
\end{aligned}
$
$\mathrm{F}:$ heads on first two tosses
$\mathrm{F}=(\mathrm{HHT}, \mathrm{HHH})$
$
\begin{aligned}
& \Rightarrow n(\mathrm{~F})=2 \\
& \mathrm{P}(\mathrm{F})=\frac{n(\mathrm{~F})}{n(\mathrm{~S})}=\frac{2}{8}=\frac{1}{4} \\
& \therefore \mathrm{E} \cap \mathrm{F}=(\mathrm{HHH})
\end{aligned}
$

$
\begin{aligned}
& \Rightarrow n(\mathrm{E} \cap \mathrm{F})=1 \\
& \therefore \mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{n(\mathrm{E} \cap \mathrm{F})}{n(\mathrm{~S})}=\frac{1}{8}
\end{aligned}
$

And $P(E \mid F)=\frac{P(E \cap F)}{P(F)}=\frac{1 / 8}{2 / 8}=\frac{1}{2}$
(ii) E : at least two heads
$\mathrm{E}=(\mathrm{HHT}, \mathrm{HTH}, \mathrm{THH}, \mathrm{HHH})$
$
\begin{aligned}
& \Rightarrow n(\mathrm{E})=4 \\
& \mathrm{P}(\mathrm{E})=\frac{n(\mathrm{E})}{n(\mathrm{~S})}=\frac{4}{8}=\frac{1}{2}
\end{aligned}
$
$\mathrm{F}$ : at most two heads
F = (TTT, HTT, THT, TTH, HHT, HTH, THH)
$
\begin{aligned}
& \Rightarrow n(\mathrm{~F})=7 \\
& \mathrm{P}(\mathrm{F})=\frac{n(\mathrm{~F})}{n(\mathrm{~S})}=\frac{7}{8} \\
& \therefore \mathrm{E} \cap \mathrm{F}=(\mathrm{HHT}, \mathrm{HTH}, \mathrm{THH}) \\
& \Rightarrow n(\mathrm{E} \cap \mathrm{F})=3 \\
& \therefore \mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{n(\mathrm{E} \cap \mathrm{F})}{n(\mathrm{~S})}=\frac{3}{8}
\end{aligned}
$

$\text { And } P(E \mid F)=\frac{P(E \cap F)}{P(F)}=\frac{3 / 8}{7 / 8}=\frac{3}{7}$

(iii) $\mathrm{E}$ : at most two tails
$\mathrm{E}=(\mathrm{HTT}$, THT, TTH, HHT, HTH, THH, HHH $)$
$
\begin{aligned}
& \Rightarrow n(\mathrm{E})=7 \\
& \mathrm{P}(\mathrm{E})=\frac{n(\mathrm{E})}{n(\mathrm{~S})}=\frac{7}{8}
\end{aligned}
$
$\mathrm{F}:$ at least one tail
F = (TTT, HTT, THT, TTH, HHT, HTH, THH)
$
\begin{aligned}
& \Rightarrow n(\mathrm{~F})=7 \\
& \mathrm{P}(\mathrm{F})=\frac{n(\mathrm{~F})}{n(\mathrm{~S})}=\frac{7}{8}
\end{aligned}
$
$\therefore \mathrm{E} \cap \mathrm{F}=(\mathrm{HTT}, \mathrm{THT}$, TTH, HHT, HTH, THH)
$
\begin{aligned}
& \Rightarrow n(\mathrm{E} \cap \mathrm{F})=6 \\
& \therefore \mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{n(\mathrm{E} \cap \mathrm{F})}{n(\mathrm{~S})}=\frac{6}{8}
\end{aligned}
$

And $P(E \mid F)=\frac{P(E \cap F)}{P(F)}=\frac{6 / 8}{7 / 8}=\frac{6}{7}$

Ex 13. 1 Question 7.

Determine $\mathbf{P}(E \mid F)$ : Two coins are tossed once.
(i) E : tail appears on one coin, $F$ : one coin shows head.
(ii) E : no tail appears, $F:$ no head appears.

Answer.

$\mathrm{S}=(\mathrm{HH}, \mathrm{TH}, \mathrm{HT}, \mathrm{TT}) \Rightarrow n(\mathrm{~S})=4$

(i) E : tail appears on one coin
$
\begin{aligned}
& \mathrm{E}=(\mathrm{TH}, \mathrm{HT}) \Rightarrow n(\mathrm{E})=2 \\
& \mathrm{P}(\mathrm{E})=\frac{n(\mathrm{E})}{n(\mathrm{~S})}=\frac{2}{4}=\frac{1}{2}
\end{aligned}
$

F : one coin shows head
$
\mathrm{F}=(\mathrm{TH}, \mathrm{HT}) \Rightarrow n(\mathrm{~F})=2
$
$
\begin{aligned}
& \mathrm{P}(\mathrm{F})=\frac{n(\mathrm{~F})}{n(\mathrm{~S})}=\frac{2}{4}=\frac{1}{2} \\
& \therefore \mathrm{E} \cap \mathrm{F}=(\mathrm{TH}, \mathrm{HT}) \Rightarrow n(\mathrm{E} \cap \mathrm{F})=2 \\
& \therefore \mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{n(\mathrm{E} \cap \mathrm{F})}{n(\mathrm{~S})}=\frac{2}{4}=\frac{1}{2}
\end{aligned}
$

And $P(E \mid F)=\frac{P(E \cap F)}{P(F)}=\frac{1 / 2}{1 / 2}=1$

(ii) $\mathrm{E}$ : no tail appears
$
\begin{aligned}
& \mathrm{E}=(\mathrm{HH}) \Rightarrow n(\mathrm{E})=1 \\
& \mathrm{P}(\mathrm{E})=\frac{n(\mathrm{E})}{n(\mathrm{~S})}=\frac{1}{4}
\end{aligned}
$

F: no head appears
$
\begin{aligned}
& \mathrm{F}=(\mathrm{TT}) \Rightarrow n(\mathrm{~F})=1 \\
& \mathrm{P}(\mathrm{F})=\frac{n(\mathrm{~F})}{n(\mathrm{~S})}=\frac{1}{4}
\end{aligned}
$

$\begin{aligned}
& \therefore \mathrm{E} \cap \mathrm{F}=\phi \Rightarrow n(\mathrm{E} \cap \mathrm{F})=0 \\
& \therefore \mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{n(\mathrm{E} \cap \mathrm{F})}{n(\mathrm{~S})}=\frac{0}{4}=0 \\
& \text { And } \mathrm{P}(\mathrm{E} \mid \mathrm{F})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}=\frac{0}{1 / 4}=0
\end{aligned}$

Ex 13. 1 Question 8.

Determine $P(E \mid F)$ : A dice is thrown three times.
$E: 4$ appears on the third toss, $F: 6$ and 5 appears respectively on first two tosses.
Answer.

Since a dice has six faces. Therefore $n(S)=6 \times 6 \times 6=216$
$
\begin{aligned}
& \mathrm{E}=(1,2,3,4,5,6) \times(1,2,3,4,5,6) \times(4) \\
& \mathrm{F}=(6) \times(5) \times(1,2,3,4,5,6) \\
& \Rightarrow n(\mathrm{~F})=1 \times 1 \times 6=6 \\
& \mathrm{P}(\mathrm{F})=\frac{n(\mathrm{~F})}{n(\mathrm{~S})}=\frac{6}{216} \\
& \therefore \mathrm{E} \cap \mathrm{F}=(6,5,4) \\
& \Rightarrow n(\mathrm{E} \cap \mathrm{F})=1 \\
& \therefore \mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{n(\mathrm{E} \cap \mathrm{F})}{n(\mathrm{~S})}=\frac{1}{216}
\end{aligned}
$

And $\mathrm{P}(\mathrm{E} \mid \mathrm{F})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}=\frac{1 / 216}{6 / 216}=\frac{1}{6}$
Ex 13. 1 Question 9.

Determine $(E \mid F)$ : Mother, father and son line up at random for a family picture.
$E$ : Son on one end, $F$ : Father in middle.

Answer.

$\mathrm{S}=(\mathrm{MFS}, \mathrm{MSF}, \mathrm{SFM}, \mathrm{SMF}, \mathrm{FMS}, \mathrm{FSM}) \Rightarrow n(\mathrm{~S})=6$
$\mathrm{E}$ : Son on one end
$\mathrm{E}=(\mathrm{MFS}, \mathrm{SFM}, \mathrm{SMF}, \mathrm{FMS}) \Rightarrow n(\mathrm{E})=4$
$\mathrm{F}:$ Father in middle
$\mathrm{F}=(\mathrm{MFS}, \mathrm{SFM}) \quad \Rightarrow n(\mathrm{~F})=2$
$\mathrm{P}(\mathrm{F})=\frac{n(\mathrm{~F})}{n(\mathrm{~S})}=\frac{2}{6}=\frac{1}{3}$
$\therefore \mathrm{E} \cap \mathrm{F}=(\mathrm{MFS}, \mathrm{SFM}) \Rightarrow n(\mathrm{E} \cap \mathrm{F})=2$
$\therefore \mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{n(\mathrm{E} \cap \mathrm{F})}{n(\mathrm{~S})}=\frac{2}{6}=\frac{1}{3}$
And $P(E \mid F)=\frac{P(E \cap F)}{P(F)}=\frac{1 / 3}{1 / 3}=1$

Ex 13. 1 Question 10.

A black and a red die are rolled.
(a) Find the conditional probability of obtaining a sum greater than 9 , given that the black die resulted in a 5.
(b) Find the conditional probability of obtaining the sum 8 , given that the red die resulted in a number less than 4.

Answer.

(a) $n(S)=6 \times 6=36$
Let A represents obtaining a sum greater than 9 and B represents black die resulted in a 5 .
$
\mathrm{A}=(46,64,55,36,63,45,54,65,56,66) \Rightarrow n(\mathrm{~A})=10
$

$
\begin{aligned}
& \mathrm{P}(\mathrm{A})=\frac{n(\mathrm{~A})}{n(\mathrm{~S})}=\frac{10}{36} \\
& \mathrm{~B}=(51,52,53,54,55,56) \Rightarrow n(\mathrm{~B})=6 \\
& \mathrm{P}(\mathrm{B})=\frac{n(\mathrm{~B})}{n(\mathrm{~S})}=\frac{6}{36} \\
& \mathrm{~A} \cap \mathrm{B}=(55,56) \Rightarrow n(\mathrm{~A} \cap \mathrm{B})=2 \\
& \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{2}{36} \\
& \mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{2 / 36}{6 / 36}=\frac{2}{6}=\frac{1}{3}
\end{aligned}
$
(b) Let A denoted the sum is 8
$
\therefore A=\{(2,6) .(3,5),(4,4),(5,3),(6,2)\}
$
$B=$ Red die results in a number less than 4, either first or second die is red
$
\begin{aligned}
& \therefore \mathrm{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)\} \\
& \mathrm{P}(\mathrm{B})=\frac{n(\mathrm{~B})}{n(\mathrm{~S})}=\frac{18}{36}=\frac{1}{2} \\
& \mathrm{~A} \cap \mathrm{B}=\{(2,6),(3,5) \Rightarrow n(\mathrm{~A} \cap \mathrm{B})=2 \\
& \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{2}{36}=\frac{1}{18}
\end{aligned}
$

$\mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{1 / 18}{1 / 2}=\frac{2}{18}=\frac{1}{9}$

Ex 13. 1 Question 11.

A fair die is rolled. Consider events $E=\{1,3,5\}, F=\{2,3\}$ and $G=\{2,3,4,5\}$. Find:
(i) $\mathbf{P}(E \mid F)$ and $\mathbf{P}(F \mid E)$
(ii) $\mathbf{P}(E \mid G)$ and $\mathbf{P}(G \mid E)$
(iii) $\mathbf{P}[(E \cup F) \mid G]$ and $\mathbf{P}[(E \cap F) \mid G]$

Answer.

$S=(1,2,3,4,5,6) \quad \Rightarrow n(S)=6$
$
\begin{aligned}
& \mathrm{E}=(1,3,5) \mathrm{F}=(2,3) \mathrm{G}=(2,3,4,5) \\
& \Rightarrow n(\mathrm{E})=3 n(\mathrm{~F})=2 n(\mathrm{G})=4
\end{aligned}
$
(i) $\mathrm{P}(\mathrm{E})=\frac{n(\mathrm{E})}{n(\mathrm{~S})}=\frac{3}{6} \quad \mathrm{P}(\mathrm{F})=\frac{n(\mathrm{~F})}{n(\mathrm{~S})}=\frac{2}{6}$
$\mathrm{E} \cap \mathrm{F}=(3) \Rightarrow n(\mathrm{E} \cap \mathrm{F})=1$
$
\mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{n(\mathrm{E} \cap \mathrm{F})}{n(\mathrm{~S})}=\frac{1}{6}
$

$
P(E \mid F)=\frac{P(E \cap F)}{P(F)}=\frac{1 / 6}{2 / 6}=\frac{1}{2} \text { and } P(F \mid E)=\frac{P(E \cap F)}{P(E)}=\frac{1 / 6}{3 / 6}=\frac{1}{3}
$
(ii) P (E) $=\frac{n(\mathrm{E})}{n(\mathrm{~S})}=\frac{3}{6} \quad \mathrm{P}(\mathrm{G})=\frac{n(\mathrm{G})}{n(\mathrm{~S})}=\frac{4}{6}$
$
\begin{aligned}
& \mathrm{E} \cap \mathrm{G}=(3,5) \Rightarrow n(\mathrm{E} \cap \mathrm{G})=2 \\
& \mathrm{P}(\mathrm{E} \cap \mathrm{G})=\frac{n(\mathrm{E} \cap \mathrm{G})}{n(\mathrm{~S})}=\frac{2}{6}
\end{aligned}
$

$
P(E \mid G)=\frac{P(E \cap G)}{P(G)}=\frac{2 / 6}{4 / 6}=\frac{2}{4}=\frac{1}{2} \text { and } P(G \mid E)=\frac{P(E \cap G)}{P(E)}=\frac{2 / 6}{3 / 6}=\frac{2}{3}
$
(iii) P (G) $=\frac{n(\mathrm{G})}{n(\mathrm{~S})}=\frac{4}{6}$
$
\begin{aligned}
& E \cup F=(1,2,3,5) \text { and } G(2,3,4,5) \\
& (E \cup F) \cap G=(2,3,5) \Rightarrow n[(E \cup F) \cap G]=3 \\
& P[(E \cup F) \cap G]=\frac{3}{6} \\
& P(E \cup F \mid G)=\frac{P[(E \cup F) \cap G]}{P(G)}=\frac{3 / 6}{4 / 6}=\frac{3}{4}
\end{aligned}
$

Again $\mathrm{E} \cap \mathrm{F}=(3)$
$(\mathrm{E} \cap \mathrm{F}) \cap \mathrm{G}=(3) \Rightarrow n[(\mathrm{E} \cap \mathrm{F}) \cap \mathrm{G}]=1$
$
\begin{aligned}
& P[(E \cap F) \cap G]=\frac{1}{6} \\
& P(E \cap F \mid G)=\frac{P[(E \cap F) \cap G]}{P(G)}=\frac{1 / 6}{4 / 6}=\frac{1}{4}
\end{aligned}
$

Ex 13. 1 Question 12.

Assume that each child born is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that (i) the youngest is a girl (ii) at least one is a girl?

Answer.

Let first and second girl are denoted by $\mathrm{G}_1$ and $\mathrm{G}_2$ and boys $\mathrm{B}_1$ and $\mathrm{B}_2$.
$
\therefore S=\left\{\left(G_1 G_2\right),\left(G_1 B_2\right),\left(G_2 B_1\right),\left(B_1 B_2\right)\right\}
$

Let $A=$ Both the children are girls $=\left(G_1 G_2\right)$
$B=$ Youngest child is girl $=\left\{\left(G_1 G_2\right),\left(B_1 G_2\right)\right\}$
$\mathrm{C}=$ at least one is a girl $=\left\{\left(\mathrm{G}_1 \mathrm{~B}_2\right),\left(\mathrm{G}_1 \mathrm{G}_2\right),\left(\mathrm{B}_1 \mathrm{G}_2\right)\right\}$
$A \cap B=\left(G_1 G_2\right) \Rightarrow P(A \cap B)=\frac{1}{4}$
$A \cap C=\left(G_1 G_2\right) \Rightarrow P(A \cap C)=\frac{1}{4}$
$P(B)=\frac{2}{4}$ and $P(C)=\frac{3}{4}$
(i) $P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{1 / 4}{2 / 4}=\frac{1}{2}$
(ii) $\mathrm{P}(\mathrm{A} \mid \mathrm{C})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{C})}{\mathrm{P}(\mathrm{C})}=\frac{1 / 4}{3 / 4}=\frac{1}{3}$

Ex 13. 1 Question 13.

An instructor has a test bank consisting of 300 easy True/False questions, 200 difficult True/False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the test bank, what is the probability that it will be an easy question given that it is a multiple choice question?

Answer.

Number of easy True/False questions $=300$
Number of difficult True/False questions $=200$
Number of easy multiple choice questions $=500$
Number of difficult multiple choice questions $=400$
Total number of all such questions $=n(\mathrm{~S})=1400$

Let $\mathrm{E}$ represents an easy question and $\mathrm{F}$ represents a multiple choice question.
$
\begin{aligned}
& \therefore n(\mathrm{E})=300+500=800 \text { and } n(\mathrm{~F})=500+400=900 \\
& \mathrm{P}(\mathrm{F})=\frac{n(\mathrm{~F})}{n(\mathrm{~S})}=\frac{900}{1400} \\
& n(\mathrm{E} \cap \mathrm{F})=500 \Rightarrow \mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{n(\mathrm{E} \cap \mathrm{F})}{n(\mathrm{~S})}=\frac{500}{1400} \\
& \mathrm{P}(\mathrm{E} \mid \mathrm{F})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}=\frac{500 / 1400}{900 / 1400}=\frac{500}{900}=\frac{5}{9}
\end{aligned}
$

Ex 13. 1 Question 14.

Given that the two numbers appearing on throwing two dice are different. Find the probability of the event 'the sum of numbers on the dice is 4 '.
$
\begin{aligned}
& \text { Answer: } S=(1,1),(2,1),(3,1),(4,1),(5,1),(6,1) \\
& (1,2),(2,2),(3,2),(4,2),(5,2),(6,2) \\
& (1,3),(2,3),(3,3),(4,3),(5,3),(6,3) \\
& (1,4),(2,4),(3,4),(4,4),(5,4),(6,4) \\
& (1,5),(2,5),(3,5),(4,5),(5,5),(6,5) \\
& (1,6),(2,6),(3,6),(4,6),(5,6),(6,6) \\
& \therefore n(S)=36
\end{aligned}
$

Let A represents the event "the sum of numbers on the dice is 4 " and B represents the event "the two numbers appearing on throwing two dice are different".
$
\text { Therefore, } \mathrm{A}=\{(1,3),(2,2),(3,1)\} \Rightarrow n(\mathrm{~A})=3
$

$
\mathrm{P}(\mathrm{A})=\frac{n(\mathrm{~A})}{n(\mathrm{~S})}=\frac{3}{36}
$

Also $B=\{(2,1),(3,1),(4,1),(5,1),(6,1),(1,2),(3,2),(4,2),(5,2),(6,2),(1,3),(2,3),(4,3),(5,3)$, $(6,3),(1,4),(2,4),(3,4),(5,4),(6,4),(1,5),(2,5),(3,5),(4,5),(6,5),(1,6),(2,6),(3,6),(4,6),(5$, 6)\}
$
\begin{aligned}
& n(\mathrm{~B})=30 \\
& \mathrm{P}(\mathrm{B})=\frac{n(\mathrm{~B})}{n(\mathrm{~S})}=\frac{30}{36}
\end{aligned}
$

Now $\mathrm{A} \cap \mathrm{B}=\{(1,3),(3,1)\}$
$
\begin{aligned}
& \Rightarrow n(\mathrm{~A} \cap \mathrm{B})=2 \\
& \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{n(\mathrm{~A} \cap \mathrm{B})}{n(\mathrm{~S})}=\frac{2}{36}
\end{aligned}
$

Hence, $P(A \mid B)=\frac{P(A \cap B)}{P(B)}$
$
=\frac{2 / 36}{30 / 36}=\frac{2}{30}=\frac{1}{15}
$

Ex 13. 1 Question 15.

Consider the experiment of throwing a die, if a multiple of 3 comes up throw the die again and if any other number comes toss a coin. Find the conditional probability of the event "the coin shows a tail", given that "at least one die shows a 3".
$
\begin{aligned}
& \text { Ans. } S=\{(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \\
& \quad(1, H),(2, H),(3, H),(4, H),(5, H),(1, T),(2, T),(3, T),(4, T),(5, T)\} \\
& \therefore n(S)=20
\end{aligned}
$

$P($ first die shows a multiple of 3$)=\frac{12}{36}=\frac{1}{3}$
$P$ (first die shows a number which is not a multiple of 3 ) $=\frac{4}{6} \times \frac{1}{2}+\frac{4}{6} \times \frac{1}{2}=\frac{8}{12}=\frac{2}{3}$
Let $\mathrm{A}=$ the coin shows a tail $=\{(1, \mathrm{~T}),(2, \mathrm{~T}),(4, \mathrm{~T}),(5, \mathrm{~T})\}$
$B=$ at least one die shows a $3=\{(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)\}$
$\mathrm{A} \cap \mathrm{B}=\varnothing$
$n(\mathrm{~A})=4, n(\mathrm{~B})=6, n(\mathrm{~A} \cap \mathrm{B})=0$
$\mathrm{P}(\mathrm{B})=\frac{6}{36}=\frac{1}{6}$ and $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=0$
$P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{0}{6 / 36}=0$

In each of the following choose the correct answer:
Ex 13. 1 Question 16.

If $P(A)=\frac{1}{2}, P(B)=0$, then $P(A \mid B)$ is:
(A) 0
(B) $\frac{1}{2}$
(C) not defined
(D) 1

Answer.

$\mathrm{P}(\mathrm{A})=\frac{1}{2}, \mathrm{P}(\mathrm{B})=0$
$
\therefore P(A \cap B)=0
$

$
\therefore P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{0}{0}=\text { not defined }
$

Therefore, option (C) is correct.
Ex 13. 1 Question 17.

If $A$ and $B$ are events such that $P(A \mid B)=\mathbf{P}(B \mid A)$, then:
(A) $\mathbf{A} \subset \mathbf{B}$
(B) $A=B$
(C) $\mathbf{A} \cap \mathbf{B}=\varnothing$
(D) $\mathbf{P}(\mathrm{A})=\mathbf{P}(\mathbf{B})$

Answer.

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

Therefore, option (D) is correct.