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Class 12 NCERT Solutions

Chapter 13: Probability

Master the conditional dependencies, the power of Bayes’ Theorem, and the logic of random variables with our step-by-step logic.

Exercise 13.1
1. Given that $\mathbf{E}$ and $\mathbf{F}$ are events such that $\mathbf{P}(\mathbf{E})=0.6, \mathbf{P}(\mathbf{F})=0.3$ and $P(E \cap F)=0.2$, find $P(E / F)$ and $P(F / E)$.
Solution

We are given that $\mathrm{P}(\mathrm{E})=0.6, \mathrm{P}(\mathrm{F})=0.3, \mathrm{P}(\mathrm{E} \cap \mathrm{F})=0.2$

$\therefore \quad \mathrm{P}(\mathrm{E} / \mathrm{F})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}=\frac{0.2}{0.3}=\frac{2}{3}$

and $\mathrm{P}(\mathrm{F} / \mathrm{E})=\frac{\mathrm{P}(\mathrm{F} \cap \mathrm{E})}{\mathrm{P}(\mathrm{E})}=\frac{0.2}{0.6}=\frac{1}{3}$.

2. Compute $\mathbf{P}(\mathbf{A} / \mathbf{B})$, if $\mathbf{P}(\mathbf{B})=0.5$ and $\mathbf{P}(\mathbf{A} \cap \mathbf{B})=0.32$.
Solution

$\mathrm{P}(\mathrm{A} / \mathrm{B})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{0.32}{0.5}=\frac{32}{50}=\frac{16}{25}$.

3. If $P(A)=0.8, P(B)=0.5$ and $P(B / A)=0.4$, find $\begin{array}{lll}\text { (i) } \mathbf{P}(\mathbf{A} \cap \mathbf{B}) & \text { (ii) } \mathbf{P}(\mathbf{A} / \mathbf{B}) & \text { (iii) } \mathbf{P}(\mathbf{A} \cup \mathbf{B}) \text {. }\end{array}$
Solution

(i) We are given that $\mathrm{P}(\mathrm{A})=0.8, \mathrm{P}(\mathrm{B})=0.5$ and $\mathrm{P}(\mathrm{B} / \mathrm{A})=0.4$


Next, $\mathrm{P}(\mathrm{B} / \mathrm{A})=0.4$ (given)

$$ \begin{aligned} & \Rightarrow \quad \frac{\mathrm{P}(\mathrm{~B} \cap \mathrm{~A})}{\mathrm{P}(\mathrm{~A})}=0.4 \Rightarrow \frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{0.8}=0.4 \\ & \Rightarrow \quad \mathrm{P}(\mathrm{~A} \cap \mathrm{~B})=0.8 \times 0.4=0.32 \end{aligned} $$

(ii) $\quad \mathrm{P}(\mathrm{A} / \mathrm{B})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{0.32}{0.5}=\frac{32}{100} \times \frac{10}{5}=\frac{64}{100}=0.64$

(iii) $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$

$$ =0.8+0.5-0.32=1.3-0.32=0.98 $$
4. Evaluate $\mathbf{P}(\mathbf{A} \cup \mathbf{B})$, if $\mathbf{2 P}(\mathbf{A})=\mathbf{P}(\mathbf{B})=\frac{\mathbf{5}}{\mathbf{1 3}}$ and $\mathbf{P}(\mathbf{A} / \mathbf{B})=\frac{\mathbf{2}}{\mathbf{5}}$.
Solution

We are given that $2 \mathrm{P}(\mathrm{A})=\mathrm{P}(\mathrm{B})=\frac{5}{13} \quad \Rightarrow \quad \mathrm{P}(\mathrm{A})=\frac{5}{26}, \mathrm{P}(\mathrm{B})=\frac{5}{13}$

Next, $\mathrm{P}(\mathrm{A} / \mathrm{B})=\frac{2}{5} \quad$ (given) $\Rightarrow \frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{2}{5}$

$\Rightarrow \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{2}{5} \mathrm{P}(\mathrm{B})=\frac{2}{5} \times \frac{5}{13}=\frac{2}{13}$

$\therefore \quad \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$

$$ =\frac{5}{26}+\frac{5}{13}-\frac{2}{13}=\frac{5+10-4}{26}=\frac{11}{26} . $$
5. If $P(A)=\frac{6}{11}, P(B)=\frac{5}{11}$ and $P(A \cup B)=\frac{7}{11}$, find (i) $\mathbf{P}(\mathbf{A} \cap \mathbf{B})$ (ii) $\mathbf{P}(\mathbf{A} / \mathbf{B})$ (iii) $\mathbf{P}(\mathbf{B} / \mathbf{A})$.
Solution

(i) We are given that $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{7}{11}$

$$ \begin{aligned} & \Rightarrow \quad \mathrm{P}(\mathrm{~A})+\mathrm{P}(\mathrm{~B})-\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})=\frac{7}{11} \\ & \Rightarrow \quad \frac{6}{11}+\frac{5}{11}-\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})=\frac{7}{11} \\ & \Rightarrow \quad \mathrm{P}(\mathrm{~A} \cap \mathrm{~B})=1-\frac{7}{11}=\frac{4}{11} \end{aligned} $$

(ii) $\mathrm{P}(\mathrm{A} / \mathrm{B})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{\frac{4}{11}}{\frac{5}{11}}=\frac{4}{5}$.

(iii) $\mathrm{P}(\mathrm{B} / \mathrm{A})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}=\frac{\frac{4}{11}}{\frac{6}{11}}=\frac{2}{3}$.








Determine $\mathbf{P}(\mathbf{E} / \mathbf{F})$ in Exercises 6 to 9.

6. A coin is tossed three times, where (i) E: head on third toss, F: heads on first two tosses (ii) E: at least two heads, F: at most two heads (iii) E: at most two tails, F: at least one tail.
Solution

The sample space for the experiment ‘a coin is tossed three times’ is

$$ \begin{aligned} & \mathrm{S}=\{\mathrm{HHH}, \mathrm{HHT}, \mathrm{HTH}, \mathrm{THH}, \mathrm{HTT}, \mathrm{THT}, \mathrm{TTH}, \mathrm{TTT}\} \\ & \therefore \quad n(\mathrm{~S})=8 \end{aligned} $$

(i) E: head on third toss

$\Rightarrow \mathrm{E}=\{\mathrm{HHH}, \mathrm{HTH}, \mathrm{THH}, \mathrm{TTH}\}$

$$ \therefore \quad n(\mathrm{E})=4 $$

F: heads on first two tosses

$$ \begin{array}{ll} \Rightarrow \mathrm{F}=\{\mathrm{HHH}, \mathrm{HHT}\} & \\ \therefore \mathrm{E} \cap \mathrm{~F}=\{\mathrm{HHH}\} & \Rightarrow n(\mathrm{E} \cap \mathrm{~F})=1 \end{array} $$
$$ \text { Hence, } \begin{aligned} \mathrm{P}(\mathrm{E}) & =\frac{n(\mathrm{E})}{n(\mathrm{~S})}=\frac{4}{8}=\frac{1}{2} \\ \mathrm{P}(\mathrm{~F}) & =\frac{2}{8}=\frac{1}{4}, \quad \mathrm{P}(\mathrm{E} \cap \mathrm{~F})=\frac{1}{8} \end{aligned} $$

and hence $\mathrm{P}(\mathrm{E} / \mathrm{F})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}=\frac{\frac{1}{8}}{\frac{1}{4}}=\frac{1}{2}$.

(ii) E : at least two heads

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

(iii) E : at most two tails

$$ \begin{array}{lc} \Rightarrow & \mathrm{E}=\{\mathrm{TTH}, \mathrm{THT}, \mathrm{HTT}, \mathrm{THH}, \mathrm{HTH}, \mathrm{HHT}, \mathrm{HHH}\} \\ \therefore & n(\mathrm{E})=7 \end{array} $$

F : at least one tail

$\Rightarrow \mathrm{F}=\{\mathrm{THH}, \mathrm{HTH}, \mathrm{HHT}, \mathrm{TTH}, \mathrm{THT}, \mathrm{HTT}, \mathrm{TTT}\}$

$$ \therefore \quad n(\mathrm{~F})=7 $$

$\therefore \mathrm{E} \cap \mathrm{F}=\{\mathrm{TTH}, \mathrm{THT}, \mathrm{HTT}, \mathrm{THH}, \mathrm{HTH}, \mathrm{HHT}\}$

$$ \Rightarrow \quad n(\mathrm{E} \cap \mathrm{~F})=6 $$






Hence, $\mathrm{P}(\mathrm{E})=\frac{7}{8}, \mathrm{P}(\mathrm{F})=\frac{n(\mathrm{~F})}{n(\mathrm{~S})}=\frac{7}{8}, \mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{6}{8}=\frac{3}{4}$

and hence $\mathrm{P}(\mathrm{E} / \mathrm{F})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}=\frac{\frac{3}{4}}{\frac{7}{8}}=\frac{3}{4} \times \frac{8}{7}=\frac{6}{7}$.

7. Two cons are tossed once, where (i) E: tail appears on one coin F: one coin shows head (ii) E: no tail appears, F: no head appears.
Solution

The sample space for ‘two coins are tossed once’ is

$\mathrm{S}=\{\mathrm{HH}, \mathrm{HT}, \mathrm{TH}, \mathrm{TT}\}$

$$ \therefore \quad n(\mathrm{~S})=4 $$

(i) E : tail appears on one coin

$\Rightarrow \mathrm{E}=\{\mathrm{HT}, \mathrm{TH}\}$

$\therefore \quad n(\mathrm{E})=2$

F : one coin shows head

$\Rightarrow \mathrm{F}=\{\mathrm{HT}, \mathrm{TH}\}$

$\therefore \quad n(\mathrm{~F})=2$

$\therefore \quad \mathrm{E} \cap \mathrm{F}=\{\mathrm{HT}, \mathrm{TH}\}$

$\Rightarrow n(\mathrm{E} \cap \mathrm{F})=2$

Hence, $\mathrm{P}(\mathrm{E})=\frac{n(\mathrm{E})}{n(\mathrm{~S})}=\frac{2}{4}=\frac{1}{2}$,

$\mathrm{P}(\mathrm{F})=\frac{2}{4}=\frac{1}{2}, \mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{n(\mathrm{E} \cap \mathrm{F})}{n(\mathrm{~S})}=\frac{2}{4}=\frac{1}{2}$

and hence $\mathrm{P}(\mathrm{E} / \mathrm{F})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}=\frac{\frac{1}{2}}{\frac{1}{2}}=1$

(ii) E : no tail appears

$\Rightarrow \mathrm{E}=\{\mathrm{HH}\}$

$\therefore \quad n(\mathrm{E})=1$

F : no head appears

$\Rightarrow \mathrm{F}=\{\mathrm{TT}\}$

$\therefore \quad n(\mathrm{~F})=1$

$\therefore \quad \mathrm{E} \cap \mathrm{F}=\phi$

$\therefore n(\mathrm{E} \cap \mathrm{F})=0$

Hence, $\mathrm{P}(\mathrm{E})=\frac{1}{4}, \mathrm{P}(\mathrm{F})=\frac{n(\mathrm{~F})}{n(\mathrm{~S})}=\frac{1}{4}, \mathrm{P}(\mathrm{E} \cap \mathrm{F})=0$

and hence $\mathrm{P}(\mathrm{E} / \mathrm{F})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}=\frac{0}{\frac{1}{4}}=0$.

8. A die is thrown three times, E: 4 appears on the third toss, F: 6 and 5 appear respectively on first two tosses
Solution

The sample space for the random experiment that a die is thrown three times has $6 \times 6 \times 6=6-{ }^3=216$ points i.e., $n(S)=216$.

Next, E: 4 appears on third toss







$$ \begin{aligned} & =\{(1,1,4)(1,2,4) \ldots(1,6,4)(2,1,4)(2,2,4) \ldots(2,6,4) \\ & \quad(3,1,4)(3,2,4) \ldots(364)(4,1,4)(4,2,4) \ldots(4,6,4) \\ & \quad \begin{array}{l} (5,1,4)(5,2,4) \ldots(5,6,4)(6,1,4)(6,2,4) \ldots(6,6,4)\} \\ \mathrm{F}: 6 \text { and } 5 \text { appear respectively on first two tossess } \\ =\{(6,5,1)(6,5,2)(6,5,3)(6,5,4)(6,5,5)(6,5,6)\} \\ \mathrm{E} \cap \mathrm{~F}=\{(6,5,4)\} \end{array} \\ & \text { Therefore, } \mathrm{P}(\mathrm{~F})=\frac{6}{216} \text { and } \mathrm{P}(\mathrm{E} \cap \mathrm{~F})=\frac{1}{216} \\ & \text { Then, } \mathrm{P}(\mathrm{E} / \mathrm{F})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{~F})}{\mathrm{P}(\mathrm{~F})}=\frac{\frac{1}{216}}{\frac{6}{216}}=\frac{1}{6} \end{aligned} $$
9. Mother, father and son line up at random for a family picture E: son on one end, F: father in middle
Solution

Let $m, f$ and $s$ denote the mother, father and son respectively. The sample space is

$$ \begin{aligned} \mathrm{S} & =\{m f s, m s f, f m s, f s m, s m f, s f m\} \\ \therefore \quad n(\mathrm{~S}) & =6 \end{aligned} $$

E : son on one end $\Rightarrow \mathrm{E}=\{m f s, f m s, s m f, s f m\}$

$$ \Rightarrow n(\mathrm{E})=4 \Rightarrow \mathrm{P}(\mathrm{E})=\frac{4}{6} $$

F : father in middle $\Rightarrow \mathrm{F}=\{m f s, s f m\}$

$$ \begin{aligned} & \Rightarrow n(\mathrm{~F})=2 \Rightarrow \mathrm{P}(\mathrm{~F})=\frac{2}{6} \\ & \therefore \mathrm{E} \cap \mathrm{~F}=\{m f s, s f m\} \Rightarrow n(\mathrm{E} \cap \mathrm{~F})=2 \Rightarrow \mathrm{P}(\mathrm{E} \cap \mathrm{~F})=\frac{2}{6} \end{aligned} $$

Then, $P(E / F)=\frac{P(E \cap F)}{P(F)}=\frac{\frac{2}{6}}{\frac{2}{6}}=1$.

10. A black and a red dice 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 .
Solution

Let $x$ denote the outcome on black die and $y$ denote the outcome on red die. The sample space is

$$ \mathrm{S}=\{(x, y): x, y \in\{1,2,3,4,5,6\}\} \quad \Rightarrow \quad n(\mathrm{~S})=6 \times 6=36 $$

(a) Let $\mathrm{E}: \operatorname{sum} x+y>9 \Rightarrow x+y=10,11,12$

$$ \Rightarrow \quad E=\{(6,4),(6,5),(6,6),(5,5),(5,6),(4,6)\} $$

F : black die resulted in a 5.

$$ \begin{aligned} \Rightarrow \quad F & =\{(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\} \\ \therefore n(F) & =6 \Rightarrow P(F)=\frac{6}{36} \\ E \cap F & =\{(5,5),(5,6)\} \end{aligned} $$






$$ \begin{array}{ll} \therefore & n(E \cap F)=2 \Rightarrow P(E \cap F)=\frac{2}{36} \\ \therefore & P(E / F)=\frac{P(E \cap F)}{P(F)}=\frac{\frac{2}{36}}{\frac{6}{36}}=\frac{1}{3} \end{array} $$

(b) Let E : sum $x+y=8$

$$ \left.\begin{array}{l} \Rightarrow \quad \mathrm{E}=\{(2,6),(3,5),(4,4),(5,3),(6,2)\} \\ \mathrm{F}: \text { red die resulted in a number less than } 4 \\ \Rightarrow \quad \mathrm{~F}=\{(x, y): x \in\{1,2,3,4,5,6\} \text { and } y \in\{1,2,3\}\} \\ \quad=\{(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1), \\ \quad(3,2),(3,3),(4,1),(4,2),(4,3),(5,1),(5,2),(5,3), \\ \quad(6,1),(6,2),(6,3)\} \\ \quad n(\mathrm{~F})=6 \times 3=18 \Rightarrow \mathrm{P}(\mathrm{~F})=\frac{18}{36} \\ \quad \mathrm{E} \cap \mathrm{~F}=\{(5,3),(6,2)\} \end{array}\right\} \begin{aligned} \therefore \quad n(\mathrm{E} \cap \mathrm{~F})=2 \Rightarrow \mathrm{P}(\mathrm{E} \cap \mathrm{~F})=\frac{2}{36} \\ \therefore \quad \mathrm{P}(\mathrm{E} / \mathrm{F})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{~F})}{\mathrm{P}(\mathrm{~F})}=\frac{\frac{2}{36}}{\frac{18}{36}}=\frac{1}{9} . \end{aligned} . $$
11. A fair die is rolled. Consider events $\mathbf{E}=\{1,3,5\}$, $\mathbf{F}=\{2,3\}$ and $\mathbf{G}=\{2,3,4,5\}$. Find (i) $P(E / F)$ and $P(F / E)$ (ii) $P(E / G)$ and $P(G / E)$ (iii) $\mathbf{P}((\mathbf{E} \cup \mathbf{F}) / \mathbf{G})$ and $\mathbf{P}((\mathbf{E} \cap \mathbf{F}) / \mathbf{G})$.
Solution

Sample space $\mathrm{S}=\{1,2,3,4,5,6\} \Rightarrow n(\mathrm{~S})=6$

We are given that Event $\mathrm{E}=\{1,3,5\}, \quad \mathrm{F}=\{2,3\}, \mathrm{G}=\{2,3,4,5\}$

(i) $\therefore \quad \mathrm{E} \cap \mathrm{F}=\{3\}$

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

(ii) Further, we can observe that $\mathrm{E} \cap \mathrm{G}=\{3,5\}$

$$ \Rightarrow \quad n(E \cap G)=2 \Rightarrow P(E \cap G)=\frac{2}{6} $$





$$ \begin{aligned} & \therefore \quad \mathrm{P}(\mathrm{E} / \mathrm{G})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{G})}{\mathrm{P}(\mathrm{G})}=\frac{\frac{2}{6}}{\frac{4}{6}}=\frac{1}{2} \\ & \text { and } \mathrm{P}(\mathrm{G} / \mathrm{E})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{G})}{\mathrm{P}(\mathrm{E})}=\frac{\frac{2}{6}}{\frac{3}{6}}=\frac{2}{3} . \end{aligned} $$

(iii) We can see that $\mathrm{E} \cup \mathrm{F}=\{1,2,3,5\}, \quad \mathrm{E} \cap \mathrm{F}=\{3\}$

$$ (\mathrm{E} \cup \mathrm{~F}) \cap \mathrm{G}=\{1,2,3,5\} \cap\{2,3,4,5\}=\{2,3,5\} $$
$$ \begin{aligned} & \Rightarrow \quad n((E \cup F) \cap G)=3 \Rightarrow P((E \cup F) \cap G)=\frac{3}{6} \\ & (E \cap F) \cap G=\{3\} \cap\{2,3,4,5\}=\{3\} \\ & \Rightarrow \quad n((E \cap F) \cap G)=1 \Rightarrow P((E \cap F) \cap G)=\frac{1}{6} \\ & \therefore \quad P((E \cup F) / G)=\frac{P((E \cup F) \cap G)}{P(G)}=\frac{\frac{3}{6}}{\frac{4}{6}}=\frac{3}{4} \\ & \text { and } \quad P((E \cap F) / G)=\frac{P((E \cap F) \cap G)}{P(G)}=\frac{\frac{1}{6}}{\frac{4}{6}}=\frac{1}{4} \end{aligned} $$
12. Assume that each born child 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?
Solution

Let the first (elder) child be denoted by capital letter and the second (younger) by a small letter. The sample space is

$$ \mathrm{S}=\{\mathrm{B} b, \mathrm{~B} g, \mathrm{G} b, \mathrm{G} g\} \quad \therefore n(\mathrm{~S})=4 $$

Let E : both children are girls, then $\mathrm{E}=\{\mathrm{G} g\}$

$$ \Rightarrow n(\mathrm{E})=1 \Rightarrow \mathrm{P}(\mathrm{E})=\frac{1}{4} $$

(i) Let F : the youngest (second) child is a girl, then

$$ \begin{gathered} \mathrm{F}=\{\mathrm{B} g, \mathrm{G} g\} \quad \therefore n(\mathrm{~F})=2 \quad \Rightarrow \mathrm{P}(\mathrm{~F})=\frac{n(\mathrm{~F})}{n(\mathrm{~S})}=\frac{2}{4} \\ \mathrm{E} \cap \mathrm{~F}=\{\mathrm{G} g\} \quad \therefore n(\mathrm{E} \cap \mathrm{~F})=1 \quad \Rightarrow \quad \mathrm{P}(\mathrm{E} \cap \mathrm{~F})=\frac{1}{4} \\ \therefore \mathrm{P}(\mathrm{E} / \mathrm{F})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{~F})}{\mathrm{P}(\mathrm{~F})}=\frac{\frac{1}{4}}{\frac{2}{4}}=\frac{1}{2} \end{gathered} $$

(ii) Let F : at least one (child) is a girl.

then $\mathrm{F}=\{\mathrm{B} g, \mathrm{G} b, \mathrm{G} g\}$






$$ \begin{aligned} & \therefore n(F)=3 \Rightarrow P(F)=\frac{n(F)}{n(S)}=\frac{3}{4} \\ & E \cap F=\{G g\} \quad \therefore n(E \cap F)=1 \Rightarrow P(E \cap F)=\frac{1}{4} \\ & \therefore P(E / F)=\frac{P(E \cap F)}{P(F)}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3} . \end{aligned} $$
13. An instructor has a question 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 question bank, what is the probability that it will be an easy question given that it is a multiple choice question?
Solution

Total number of questions $=300+200+500+400=1400$

$$ \therefore \quad n(\mathrm{~S})=1400 $$

Let E : selected question is easy

and F : selected question is a multiple choice question

then $\mathrm{E} \cap \mathrm{F}$ : selected question is an easy multiple choice question

$$ \begin{aligned} n(\mathrm{E} \cap \mathrm{~F}) & =500, \quad n(\mathrm{~F})=500+400=900 \\ \therefore \quad \mathrm{P}(\mathrm{E} \cap \mathrm{~F}) & =\frac{n(\mathrm{E} \cap \mathrm{~F})}{n(\mathrm{~S})}=\frac{500}{1400} \\ \text { and } \quad \mathrm{P}(\mathrm{~F}) & =\frac{n(\mathrm{~F})}{n(\mathrm{~S})}=\frac{900}{1400} \end{aligned} $$
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 $\boldsymbol{4}^{\prime}$.
Solution

Sample space for the random experiment of throwing two dice is

$$ \begin{aligned} \mathrm{S} & =\{(x, y): x, y \in\{1,2,3,4,5,6\}\} \\ n(\mathrm{~S}) & =6 \times 6=36 \end{aligned} $$

Let E : the sum of numbers on the dice is 4 .

$\Rightarrow \quad \mathrm{E}=\{(1,3),(2,2),(3,1)\}$

$$ \Rightarrow \quad n(\mathrm{E})=3 \Rightarrow \mathrm{P}(\mathrm{E})=\frac{n(\mathrm{E})}{n(\mathrm{~S})}=\frac{3}{36} $$

Let F : numbers appearing on the dice are different

$$ \begin{array}{ll} \Rightarrow & \mathrm{F}=\mathrm{S}-\{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\} \\ \therefore & n(\mathrm{~F})=36-6=30 \Rightarrow \mathrm{P}(\mathrm{~F})=\frac{30}{36} \end{array} $$




$$ \begin{aligned} & \text { Also, } \mathrm{E} \cap \mathrm{~F}=\{(1,3),(3,1)\} \quad \therefore \quad n(\mathrm{E} \cap \mathrm{~F})=2 \\ & \therefore \quad \mathrm{P}(\mathrm{E} \cap \mathrm{~F})=\frac{n(\mathrm{E} \cap \mathrm{~F})}{n(\mathrm{~S})}=\frac{2}{36} \\ & \therefore \quad \text { Required probability }=\mathrm{P}(\mathrm{E} / \mathrm{F})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{~F})}{\mathrm{P}(\mathrm{~F})}=\frac{\frac{2}{36}}{\frac{30}{36}}=\frac{1}{15} . \end{aligned} $$
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 $3^{\prime}$.
Solution

The sample space is given by

$$ \begin{aligned} \mathrm{S}= & \{(1, \mathrm{H}),(1, \mathrm{~T}),(2, \mathrm{H}),(2, \mathrm{~T}),(3,1),(3,2),(3,3),(3,4) \\ & (3,5),(3,6),(4, \mathrm{H}),(4, \mathrm{~T}),(5, \mathrm{H}),(5, \mathrm{~T}),(6,1),(6,2), \\ & (6,3),(6,4),(6,5),(6,6)\} \\ \Rightarrow \quad & n(\mathrm{~S})=20 \end{aligned} $$

Let E : the coin shows a tail

$$ \Rightarrow \mathrm{E}=\{(1, \mathrm{~T}),(2, \mathrm{~T}),(4, \mathrm{~T}),(5, \mathrm{~T})\} \Rightarrow n(\mathrm{E})=4 $$

Let F : at least one die shows a 3

$$ \begin{aligned} & \Rightarrow \quad \mathrm{F}=\{(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(6,3)\} \\ & \Rightarrow \quad n(\mathrm{~F})=7 \\ & \mathrm{E} \cap \mathrm{~F}=\phi \\ & \Rightarrow \quad n(\mathrm{E} \cap \mathrm{~F})=0 \Rightarrow \mathrm{P}(\mathrm{E} \cap \mathrm{~F})=\frac{n(\mathrm{E} \cap \mathrm{~F})}{n(\mathrm{~S})}=\frac{0}{20}=0 . \\ & \therefore \quad \text { Required probability }=\mathrm{P}(\mathrm{E} / \mathrm{F})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{~F})}{\mathrm{P}(\mathrm{~F})}=\frac{0}{\mathrm{P}(\mathrm{~F})}=0 . \end{aligned} $$

In each of the Exercises 16 and 17 choose the correct answer:

16. If $P(A)=\frac{1}{2}, P(B)=0$, then $P(A / B)$ is (A) 0 (B) $\frac{1}{2}$ (C) not defined (D) 1 .
Solution

$\mathrm{P}(\mathrm{A} / \mathrm{B})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{0}$ which is not defined

Hence, option (C) is the correct answer.

17. If $A$ and $B$ are events such that $P(A / B)=P(B / A)$, then (A) $\mathbf{A} \subset \mathbf{B}$ but $\mathbf{A} \neq \mathbf{B}$ (B) $\mathbf{A}=\mathbf{B}$ (C) $\mathbf{A} \cap \mathbf{B}=\phi$ (D) $\mathbf{P}(\mathbf{A})=\mathbf{P}(\mathbf{B})$.
Solution

We are given that $\quad \mathrm{P}(\mathrm{A} / \mathrm{B})=\mathrm{P}(\mathrm{B} / \mathrm{A})$

$$ \begin{aligned} \Rightarrow \frac{P(A \cap B)}{P(B)}= & \frac{P(B \cap A)}{P(A)} \Rightarrow \frac{P(A \cap B)}{P(B)}=\frac{P(A \cap B)}{P(A)} \\ & {[\because \quad A \cap B=B \cap A \quad \therefore \quad P(A \cap B)=P(B \cap A)] } \end{aligned} $$






Dividing both sides by $\mathrm{P}(\mathrm{A} \cap \mathrm{B}), \frac{1}{\mathrm{P}(\mathrm{B})}=\frac{1}{\mathrm{P}(\mathrm{A})}$

Cross-multiplying, $\mathrm{P}(\mathrm{A})=\mathrm{P}(\mathrm{B}) \quad \therefore$ The correct option is (D).

Exercise 13.2
1. If $P(A)=\frac{3}{5}$ and $P(B)=\frac{1}{5}$, find $P(A \cap B)$ if $A$ and $B$ are independent events.
Solution

Since A and B are independent events,

$$ \therefore \quad \mathrm{P}(\mathrm{~A} \cap \mathrm{~B})=\mathrm{P}(\mathrm{~A}) \mathrm{P}(\mathrm{~B})=\frac{3}{5} \times \frac{1}{5}=\frac{3}{25} . $$
2. Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.
Solution

Let $\mathrm{E}_1$ : first card drawn is black

$\mathrm{E}_2$ : second card drawn is black

then $\quad \mathrm{P}\left(\mathrm{E}_1\right)=\frac{26}{52}$

(There are 26 black cards in a standard pack of 52 cards.) After the first black card is removed (no replacement), 25 black cards remain among the 51 cards left in the pack.

$\mathrm{P}\left(\mathrm{E}_2, \mathrm{E}_1\right)$ i.e., probability that second card is black known that first card is black $=\frac{25}{51}$

Therefore, the required probability $=\mathrm{P}\left(\mathrm{E}_1 \cap \mathrm{E}_2\right)=\mathrm{P}\left(\mathrm{E}_1\right.$ and $\left.\mathrm{E}_2\right)$ i.e., probability that both cards are black.

$$ =\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{E}_2 \mathrm{E}_1\right)=\frac{26}{52} \times \frac{25}{51}=\frac{25}{102} . $$
3. A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.
Solution

We are given that The box contains a total of 15 oranges out of which 12 are good and 3 are bad. The box is approved for sale if all the three oranges drawn, (one by one), without replacement are good ones.

Let $\mathrm{E}_1$ : first orange drawn is good

$\mathrm{E}_2$ : second orange drawn is good

$\mathrm{E}_3$ : third orange drawn is good

then $\mathrm{P}\left(\mathrm{E}_1\right)=\frac{12}{15}, \mathrm{P}\left(\mathrm{E}_2 / \mathrm{E}_1\right)=\frac{11}{14}$ and $\mathrm{P}\left(\mathrm{E}_3 / \mathrm{E}_1 \mathrm{E}_2\right)$








i.e., the probability that the third orange is good, given the first two were also good $=\frac{10}{13}$

Therefore, the required probability $=\mathrm{P}\left(\mathrm{E}_1 \cap \mathrm{E}_2 \cap \mathrm{E}_3\right) =\mathrm{P}\left(\mathrm{E}_1\right.$ and $\mathrm{E}_2$ and $\left.\mathrm{E}_3\right)=\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{E}_2 / \mathrm{E}_1\right) \mathrm{P}\left(\mathrm{E}_3 / \mathrm{E}_1 \mathrm{E}_2\right)$

$$ =\frac{12}{15} \times \frac{11}{14} \times \frac{10}{13}=\frac{44}{91} . $$
4. A fair coin and an unbiased die are tossed. Let $\mathbf{A}$ be the event ‘head appears on the coin’ and B be the event ‘ 3 on the die’. Check whether A and B are independent events or not.
Solution

The sample space for ‘a fair coin and an unbiased die are tossed’ is

$$ \begin{aligned} \mathrm{S}= & \{(\mathrm{H}, 1),(\mathrm{H}, 2),(\mathrm{H}, 3),(\mathrm{H}, 4),(\mathrm{H}, 5),(\mathrm{H}, 6),(\mathrm{T}, 1),(\mathrm{T}, 2) \\ & (\mathrm{T}, 3),(\mathrm{T}, 4),(\mathrm{T}, 5),(\mathrm{T}, 6)\} \\ n(\mathrm{~S})= & 2 \times 6=12 \end{aligned} $$

A : head appears on the coin

and $\mathrm{B}: 3$ appears on the die

$$ \begin{aligned} & \therefore \quad \quad \mathrm{A}=\{(\mathrm{H}, 1),(\mathrm{H}, 2),(\mathrm{H}, 3),(\mathrm{H}, 4),(\mathrm{H}, 5),(\mathrm{H}, 6)\} \\ & \quad \mathrm{B}=\{(\mathrm{H}, 3),(\mathrm{T}, 3)\} \\ & \quad \mathrm{A} \cap \mathrm{~B}=\{(\mathrm{H}, 3)\} \\ & \therefore \quad n(\mathrm{~A})=6, n(\mathrm{~B})=2, n(\mathrm{~A} \cap \mathrm{~B})=1 \\ & \therefore \quad \mathrm{P}(\mathrm{~A})=\frac{n(\mathrm{~A})}{n(\mathrm{~S})}=\frac{6}{12}=\frac{1}{2}, \quad \mathrm{P}(\mathrm{~B})=\frac{2}{12}=\frac{1}{6}, \quad \mathrm{P}(\mathrm{~A} \cap \mathrm{~B})=\frac{1}{12} \end{aligned} $$

Since, $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{12}=\frac{1}{2} \times \frac{1}{6}=\mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B})$;

therefore the events A and B are independent.

5. A die marked $1,2,3$ in red and $4,5,6$ in green is tossed. Let $A$ be the event, ‘the number is even’, and $B$ be the event, ‘the number is red’. Are $A$ and $B$ independent?
Solution

The sample space is

$$ \mathrm{S}=\{\underbrace{1,2,3}_{\text {red }}, \underbrace{4,5,6}_{\text {green }}\} $$
$$ \therefore \quad n(\mathrm{~S})=6 $$

Event A : the number is even and $B$ : the number is red

$\Rightarrow \mathrm{A}=\{2,4,6\} \quad$ and $\mathrm{B}:\{1,2,3\} \quad \mathrm{A} \cap \mathrm{B}=\{2\}$

$\therefore \quad n(\mathrm{~A})=3, n(\mathrm{~B})=3, n(\mathrm{~A} \cap \mathrm{~B})=1$

$\therefore \quad \mathrm{P}(\mathrm{A})=\frac{3}{6}=\frac{1}{2}, \quad \mathrm{P}(\mathrm{B})=\frac{3}{6}=\frac{1}{2}, \quad \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{6}$

Since, $\frac{1}{6} \neq \frac{1}{2} \times \frac{1}{2}$, Therefore, $\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \neq \mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B})$.

Therefore, A and B are **not independent**.

6. Let $\mathbf{E}$ and $\mathbf{F}$ be events with $\mathbf{P}(\mathbf{E})=\frac{\mathbf{3}}{\mathbf{5}}, \mathbf{P}(\mathbf{F})=\frac{\mathbf{3}}{\mathbf{1 0}}$ and $P(E \cap F)=\frac{1}{5}$. Are $E$ and $F$ independent?
Solution

Here $\mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{1}{5}$ and $\mathrm{P}(\mathrm{E}) \mathrm{P}(\mathrm{F})=\frac{3}{5} \times \frac{3}{10}=\frac{9}{50}$

Since, $\frac{1}{5} \neq \frac{9}{50}$, therefore $\mathrm{P}(\mathrm{E} \cap \mathrm{F}) \neq \mathrm{P}(\mathrm{E}) \mathrm{P}(\mathrm{F})$.

Therefore, E and F are **not independent**.

7. Given that the events $A$ and $B$ are such that $P(A)=\frac{1}{2}$, $\mathbf{P}(\mathbf{A} \cup \mathbf{B})=\frac{\mathbf{3}}{\mathbf{5}}$ and $\mathbf{P}(\mathbf{B})=\boldsymbol{p}$. Find $\boldsymbol{p}$ if they are (i) mutually exclusive (ii) independent.
Solution

We are given that $\mathrm{P}(\mathrm{A})=\frac{1}{2}, \quad \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{3}{5}, \quad \mathrm{P}(\mathrm{B})=p$

We know that $\mathbf{P}(\mathbf{A} \cup \mathbf{B})=\mathbf{P}(\mathbf{A})+\mathbf{P}(\mathbf{B})-\mathbf{P}(\mathbf{A} \cap \mathbf{B})$

$$ \begin{aligned} & \therefore \quad \frac{3}{5}=\frac{1}{2}+p-\mathrm{P}(\mathrm{~A} \cap \mathrm{~B}) \\ & \Rightarrow \mathrm{P}(\mathrm{~A} \cap \mathrm{~B})=p+\frac{1}{2}-\frac{3}{5}=p-\frac{1}{10} \end{aligned} $$

(i) If A and B are mutually exclusive, then $\mathrm{A} \cap \mathrm{B}=\phi$ so that

$$ \mathrm{P}(\mathrm{~A} \cap \mathrm{~B})=0 \quad \Rightarrow p-\frac{1}{10}=0 \quad \therefore p=\frac{1}{10} $$

(ii) If A and B are independent, then $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B})$

$$ \begin{aligned} & \Rightarrow p-\frac{1}{10}=\frac{1}{2} p \Rightarrow p-\frac{1}{2} p=10 \Rightarrow \frac{1}{2} p=\frac{1}{10} \\ & \therefore p=\frac{1}{5} \end{aligned} $$
8. Let $\mathbf{A}$ and $\mathbf{B}$ be independent events with $\mathbf{P}(\mathbf{A})=0.3$ and $\mathbf{P}(\mathbf{B})=0.4$. Find (i) $\mathbf{P}(\mathbf{A} \cap \mathbf{B})$ (ii) $\mathbf{P}(\mathbf{A} \cup \mathbf{B})$ (iii) $\mathbf{P}(\mathbf{A} / \mathbf{B})$ (iv) $\mathbf{P}(\mathbf{B} / \mathbf{A})$.
Solution

We are given that $\mathrm{P}(\mathrm{A})=0.3$ and $\mathrm{P}(\mathrm{B})=0.4$

and $A$ and $B$ are independent events

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

(i) $\quad \therefore \quad \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B})=0.3 \times 0.4=0.12$.

(ii) and $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$

$$ \begin{aligned} & =\mathrm{P}(\mathrm{~A})+\mathrm{P}(\mathrm{~B})-\mathrm{P}(\mathrm{~A}) \mathrm{P}(\mathrm{~B}) \quad[\mathrm{By}(i)] \\ & =0.3+0.4-0.3 \times 0.4=0.7-0.12=0.58 . \end{aligned} $$

(iii) Also, $\mathrm{P}(\mathrm{A} / \mathrm{B})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{\mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B})}{\mathrm{P}(\mathrm{B})}[\mathrm{By}(i)]=\mathrm{P}(\mathrm{A})=0.3$.

(iv) Also $\mathrm{P}(\mathrm{B} / \mathrm{A})=\frac{\mathrm{P}(\mathrm{B} \cap \mathrm{A})}{\mathrm{P}(\mathrm{A})}=\frac{\mathrm{P}(\mathrm{B}) \mathrm{P}(\mathrm{A})}{\mathrm{P}(\mathrm{A})}[\mathrm{By}(i)]=\mathrm{P}(\mathrm{B})=0.4$.

9. If $A$ and $B$ are two events such that $P(A)=\frac{1}{4}$, $\mathbf{P}(\mathbf{B})=\frac{\mathbf{1}}{\mathbf{2}}$ and $\mathbf{P}(\mathbf{A} \cap \mathbf{B})=\frac{\mathbf{1}}{\mathbf{8}}$, find $\mathbf{P}($ not $\mathbf{A}$ and not $\mathbf{B})$.
Solution

We are given that $\mathrm{P}(\mathrm{A})=\frac{1}{4}, \mathrm{P}(\mathrm{B})=\frac{1}{2}$ and $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{8}$

Here $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{8}=\frac{1}{4} \times \frac{1}{2}=\mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B})$

$\Rightarrow \mathrm{A}$ and B are independent events

$\Rightarrow \mathrm{A}^{\prime}$ and $\mathrm{B}^{\prime}$ are independent events

$\Rightarrow \quad \mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)=\mathrm{P}\left(\mathrm{A}^{\prime}\right) \mathrm{P}\left(\mathrm{B}^{\prime}\right)$


Next, $\mathrm{P}($ not A and not B$)=\mathrm{P}\left(\mathrm{A}^{\prime}\right.$ and $\left.\mathrm{B}^{\prime}\right)=\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)$

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

[By (i)]

$=[1-\mathrm{P}(\mathrm{A})][1-\mathrm{P}(\mathrm{B})]$

$=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{2}\right)=\frac{3}{4} \times \frac{1}{2}=\frac{3}{8}$.

10. Events $A$ and $B$ are such that $P(A)=\frac{1}{2}, P(B)=\frac{7}{12}$ and $P($ not $A$ or not $B)=\frac{1}{4}$. State whether $A$ and $B$ are independent?
Solution

We are given that $\mathrm{P}(\mathrm{A})=\frac{1}{2}, \quad \mathrm{P}(\mathrm{B})=\frac{7}{12}$ and

$\mathrm{P}($ not A or not B$)=\frac{1}{4} \quad \Rightarrow \quad \mathrm{P}\left(\mathrm{A}^{\prime} \cup \mathrm{B}^{\prime}\right)=\frac{1}{4}$

$\Rightarrow \quad \mathrm{P}(\mathrm{A} \cap \mathrm{B})^{\prime}=\frac{1}{4} \quad$ (By De Morgan’s Law)

$\Rightarrow 1-\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{4} \quad\left[\because \mathrm{P}\left(\mathrm{A}^{\prime}\right)=1-\mathrm{P}(\mathrm{A})\right]$

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

Also, $\quad \mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B})=\frac{1}{2} \times \frac{7}{12}=\frac{7}{24}$

Since, $\mathrm{P}(\mathrm{A} \cap \mathrm{B})\left(=\frac{3}{4}\right) \neq \mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B})\left(=\frac{7}{24}\right)$, the events A and B are not independent.

11. Given two independent events $A$ and $B$ such that $P(A)=0.3$, $P(B)=0.6$. Find (i) $\mathrm{P}(\mathrm{A}$ and B$)$ (ii) $\mathbf{P}$ (A and not B) (iii) $\mathbf{P}(\mathbf{A}$ or $\mathbf{B})$ (iv) $\mathbf{P}$ (neither $\mathbf{A}$ nor $\mathbf{B}$ ).
Solution

We are given that A and B are independent events such that

$$ \mathrm{P}(\mathrm{~A})=0.3, \mathrm{P}(\mathrm{~B})=0.6 $$

(i) $\quad \therefore \quad \mathrm{P}(\mathrm{A}$ and B$)=\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B})=0.3 \times 0.6=0.18$

(ii) $\mathrm{P}(\mathrm{A}$ and not B$)=\mathrm{P}\left(\mathrm{A} \cap \mathrm{B}^{\prime}\right)=\mathrm{P}(\mathrm{A}) \mathrm{P}\left(\mathrm{B}^{\prime}\right)$

$\left[\because \mathrm{A}\right.$ and B are independent $\Rightarrow \mathrm{A}$ and $\mathrm{B}^{\prime}$ are independent $]$

$$ =\mathrm{P}(\mathrm{~A})[1-\mathrm{P}(\mathrm{~B})]=0.3(1-0.6)=0.3 \times 0.4=0.12 . $$

(iii) $\mathrm{P}(\mathrm{A}$ or B$)=\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$

$$ =\mathrm{P}(\mathrm{~A})+\mathrm{P}(\mathrm{~B})-\mathrm{P}(\mathrm{~A}) \mathrm{P}(\mathrm{~B}) $$

[ ∵ A and B are independent events]





$$ =0.3+0.6-0.3 \times 0.6=0.9-0.18=0.72 \text {. } $$

(iv) $\mathrm{P}($ neither A nor B$)=\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)=\mathrm{P}\left(\mathrm{A}^{\prime}\right) \mathrm{P}\left(\mathrm{B}^{\prime}\right)$

$\left[\because \mathrm{A}\right.$ and B are independent $\Rightarrow \mathrm{A}^{\prime}$ and $\mathrm{B}^{\prime}$ are independent $] =[1-\mathrm{P}(\mathrm{A})][1-\mathrm{P}(\mathrm{B})]=(1-0.3)(1-0.6)=0.7 \times 0.4=0.28$.

12. A die is tossed thrice. Find the probability of getting an odd number at least once.
Solution

Consider the sample space S for a single die toss.

Therefore, $\mathrm{S}=\{1,2,3,4,5,6\} \Rightarrow n(\mathrm{~S})=6$

Let E be the event of getting an odd number on a dice.

$$ \begin{aligned} \therefore & \mathrm{E} & =\{1,3,5\} \Rightarrow n(\mathrm{E})=3 \\ \Rightarrow & \mathrm{P}(\mathrm{E}) & =\frac{3}{6} \end{aligned} $$

Let event $\mathrm{E}_1$ : an odd number on first toss

$\mathrm{E}_2$ : an odd number on second toss

$\mathrm{E}_3$ : an odd number on third toss

then we know from common sense that $\mathrm{E}_1, \mathrm{E}_2, \mathrm{E}_3$ are independent events and hence $\mathrm{E}_1{ }^{\prime}, \mathrm{E}_2{ }^{\prime}, \mathrm{E}_3{ }^{\prime}$ are also independent.

Next, $\mathrm{P}\left(\mathrm{E}_1\right)=\mathrm{P}\left(\mathrm{E}_2\right)=\mathrm{P}\left(\mathrm{E}_3\right)=\frac{3}{6}=\frac{1}{2}$


[By (i)]


$\therefore \quad \mathrm{P}$ (an odd number at least once)

$$ \begin{aligned} & =1-\mathrm{P}(\text { an odd number on none of the three dice }) \\ & =1-\mathrm{P}\left(\mathrm{E}_1^{\prime} \text { and } \mathrm{E}_2^{\prime} \text { and } \mathrm{E}_3^{\prime}\right) \\ & =1-\mathrm{P}\left(\mathrm{E}_1^{\prime}\right) \mathrm{P}\left(\mathrm{E}_2^{\prime}\right) \mathrm{P}\left(\mathrm{E}_3^{\prime}\right) \\ & =1-\left(1-\frac{1}{2}\right)\left(1-\frac{1}{2}\right)\left(1-\frac{1}{2}\right)=1-\left(\frac{1}{2}\right)^3=\frac{7}{8} . \end{aligned} $$
13. Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that (i) both balls are red. (ii) first ball is black and second is red. (iii) one of them is black and other is red.
Solution

Since the two balls are drawn with replacement, the two draws are independent.

Number of black balls $=10, \quad$ Number red balls $=8$

Total number of balls $=10+8=18$

(i) P (both balls are red)

$$ \begin{aligned} & =P(\text { first ball is red and second ball is red }) \\ & =P(\text { first ball is red }) \times P(\text { second ball is red }) \\ & =\frac{8}{18} \times \frac{8}{18}=\frac{4}{9} \times \frac{4}{9}=\frac{16}{81} \end{aligned} $$

(ii) P (first ball is black and second is red)

$=\mathrm{P}$ (first ball is black) $\times \mathrm{P}$ (second ball is red)

$$ =\frac{10}{18} \times \frac{8}{18}=\frac{5}{9} \times \frac{4}{9}=\frac{20}{81} . $$





(iii) P (one of the balls is black and other is red)

$$ \begin{aligned} &= \mathbf{P}(\text { first ball is black and second is red’ } \\ & \text { or ‘first ball is red and second is black’) } \\ &= \mathrm{P}(\text { first ball is black and second is red }) \\ &+\mathrm{P}(\text { first ball is red and second is black }) \\ &= \mathrm{P}(\text { first ball is black }) \times \mathrm{P}(\text { second ball is red }) \\ &+\mathrm{P}(\text { first ball is red }) \times \mathrm{P}(\text { second ball is black }) \\ &= \frac{10}{18} \times \frac{8}{18}+\frac{8}{18} \times \frac{10}{18}=\frac{5}{9} \times \frac{4}{9}+\frac{4}{9} \times \frac{5}{9} \\ &= \frac{20}{81}+\frac{20}{81}=\frac{40}{81} . \end{aligned} $$
14. Probability of solving specific problem independently by $\mathbf{A}$ and $B$ are $\frac{1}{2}$ and $\frac{1}{3}$ respectively. If both try to solve the problem independently, find the probability that (i) the problem is solved (ii) exactly one of them solves the problem.
Solution

Let event E: A solves the problem

and F : B solves the problem

then

$$ \mathrm{P}(\mathrm{E})=\frac{1}{2} \quad \text { and } \quad \mathrm{P}(\mathrm{~F})=\frac{1}{3} $$

(i) P (the problem is solved)

$=\mathrm{P}$ (at least one solves the problem)

$=1$ – Probability that neither A nor B solves the problem

$=1-\mathrm{P}\left(\mathrm{E}^{\prime} \cap \mathrm{F}^{\prime}\right)=1-\mathrm{P}\left(\mathrm{E}^{\prime}\right.$ and $\left.\mathrm{F}^{\prime}\right)=1-\mathrm{P}\left(\mathrm{E}^{\prime}\right) \mathrm{P}\left(\mathrm{F}^{\prime}\right)$

[ $\because$ Events E and F are independent (given)

$\Rightarrow \mathrm{E}^{\prime}$ and $\mathrm{F}^{\prime}$ are also independent]

$$ \begin{aligned} & =1-[1-\mathrm{P}(\mathrm{E})][1-\mathrm{P}(\mathrm{~F})] \\ & =1-\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)=1-\frac{1}{2} \times \frac{2}{3}=1-\frac{1}{3}=\frac{2}{3} . \end{aligned} $$

(ii) P (exactly one of them solves the problem)

$$ \begin{aligned} & =\mathrm{P}\left({ }^{\prime} \mathrm{A} \text { solves and } \mathrm{B}\right. \text { does not solve’ } \\ & \text { or ‘ } \mathrm{B} \text { solves and } \mathrm{A} \text { does not solve’ } \\ & =\mathrm{P}\left(\mathrm{E} \cap \mathrm{~F}^{\prime}\right)+\mathrm{P}\left(\mathrm{~F} \cap \mathrm{E}^{\prime}\right)=\mathrm{P}(\mathrm{E}) \mathrm{P}\left(\mathrm{~F}^{\prime}\right)+\mathrm{P}(\mathrm{~F}) \mathrm{P}\left(\mathrm{E}^{\prime}\right) \\ & (\because \quad \text { Events are independent }) \\ & =\frac{1}{2}\left(1-\frac{1}{3}\right)+\frac{1}{3}\left(1-\frac{1}{2}\right) \\ & =\frac{1}{2} \times \frac{2}{3}+\frac{1}{3} \times \frac{1}{2}=\frac{1}{3}+\frac{1}{6}=\frac{2+1}{6}=\frac{3}{6}=\frac{1}{2} . \end{aligned} $$
15. One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events $\mathbf{E}$ and $F$ independent? (i) E: ‘the card drawn is a spade’ F: ‘the card drawn is an ace’ (ii) E: ‘the card drawn is black’ F: ‘the card drawn is a king’ (iii) E: ‘the card drawn is a king or queen’ F: ‘the card drawn is a queen or jack’.
Solution

(i) E: the card is a spade

F : the card is an ace

$\Rightarrow \mathrm{E} \cap \mathrm{F}: \Rightarrow$ the common cards in E and F

⇒ the card is ace of spade $\Rightarrow n(\mathrm{E} \cap \mathrm{F})=1$

$\therefore \quad \mathrm{P}(\mathrm{E})=\frac{13}{52}=\frac{1}{4}, \mathrm{P}(\mathrm{F})=\frac{4}{52}=\frac{1}{13} \quad[\because \quad$ We know that in a pack of cards, there are 13 spade cards and 4 aces]

Next, $\mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{1}{52}=\frac{1}{4} \times \frac{1}{13}=\mathrm{P}(\mathrm{E}) \mathrm{P}(\mathrm{F})$

$\Rightarrow \mathrm{E}$ and F are independent.

(ii) E: the card is black


F : the card is a king

⇒ $\mathrm{E} \cap \mathrm{F}: \Rightarrow$ the common cards in E and F

⇒ the card is a black king $\Rightarrow n(\mathrm{E} \cap \mathrm{F})=2$

[ ∵ There are 2 black kings in a pack of 52 cards]

$$ \begin{aligned} & \therefore \quad \mathrm{P}(\mathrm{E})=\frac{26}{52}=\frac{1}{2}, \mathrm{P}(\mathrm{~F})=\frac{4}{52}=\frac{1}{13} \\ & \mathrm{P}(\mathrm{E} \cap \mathrm{~F})=\frac{2}{52}=\frac{1}{26}=\frac{1}{2} \times \frac{1}{13}=\mathrm{P}(\mathrm{E}) \mathrm{P}(\mathrm{~F}) \end{aligned} $$

$\Rightarrow \mathrm{E}$ and F are independent.

(iii) E : the card drawn is a king or queen


F : the card drawn is a queen or jack

$\Rightarrow \mathrm{E} \cap \mathrm{F}: \Rightarrow$ the common cards in E and $\mathrm{F} \Rightarrow$ the card drawn is a queen

$\therefore \quad \mathrm{P}(\mathrm{E})=\frac{4+4}{52}=\frac{2}{13}, \mathrm{P}(\mathrm{F})=\frac{4+4}{52}=\frac{2}{13}[\because$ We know that in a pack of 52 cards there are 4 jacks, 4 queens and 4 kings]

$$ \begin{aligned} & \mathrm{P}(\mathrm{E} \cap \mathrm{~F})=\frac{4}{52}=\frac{1}{13} \quad \text { Also, } \mathrm{P}(\mathrm{E}) \mathrm{P}(\mathrm{~F})=\frac{2}{13} \times \frac{2}{13}= \\ & \frac{4}{169} \end{aligned} $$

Since, $\mathrm{P}(\mathrm{E} \cap \mathrm{F}) \neq \mathrm{P}(\mathrm{E}) \mathrm{P}(\mathrm{F})$, the events E and F are not independent.

16. In a hostel, $60 \%$ of the students read Hindi news paper, $40 \%$ read English news paper and $20 \%$ read both Hindi and English news papers. A student is selected at random. (a) Find the probability that she reads neither Hindi nor English news papers. (b) If she reads Hindi news paper, find the probability that she reads English news paper. (c) If she reads English news paper, find the probability that she reads Hindi news paper.
Solution

Let Event A: a student reads Hindi newspaper







B: a student reads English newspaper

We are given that 60\% students read Hindi newspaper, 40\% read English newspaper and $20 \%$ read both.

Therefore, $\mathrm{P}(\mathrm{A})=\frac{60}{100}=\frac{3}{5}, \mathrm{P}(\mathrm{B})=\frac{40}{100}=\frac{2}{5}$,

$$ \mathrm{P}(\mathrm{~A} \cap \mathrm{~B})=\mathrm{P}(\mathrm{~A} \text { and } \mathrm{B})=\frac{20}{100}=\frac{1}{5} $$

(a) Required probability $=$ Probability that she reads neither newspaper $=\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)=\mathrm{P}(\mathrm{A} \cup \mathrm{B})^{\prime}$

(De-Morgan’s Law)

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

Remark: Here $\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right) \neq \mathrm{P}\left(\mathrm{A}^{\prime}\right) \mathrm{P}\left(\mathrm{B}^{\prime}\right)$ because events A and B are not independent

$$ \left(\because \mathrm{P}(\mathrm{~A}) \cdot \mathrm{P}(\mathrm{~B})=\frac{3}{5} \times \frac{2}{5}=\frac{6}{25} \neq \mathrm{P}(\mathrm{~A} \cap \mathrm{~B})\left(=\frac{1}{5}\right)\right) $$

and hence $\mathrm{A}^{\prime}$ and $\mathrm{B}^{\prime}$ are not independent.

(b) Required probability $=\mathrm{P}(\mathrm{B} / \mathrm{A})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}=\frac{\frac{1}{5}}{\frac{3}{5}}=\frac{1}{3}$.

(c) Required probability $=\mathrm{P}(\mathrm{A} / \mathrm{B})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{\frac{1}{5}}{\frac{2}{5}}=\frac{1}{2}$.


Choose the correct answer in Exercises 16 and 17:

17. The probability of obtaining an even prime number on each die, when a pair of dice is rolled is (A) 0 (B) $\frac{1}{3}$ (C) $\frac{1}{12}$ (D) $\frac{1}{36}$.
Solution

When a pair of dice is rolled, the sample space is

$\mathrm{S}=\{(x, y): x, y \in\{1,2,3,4,5,6\}\} . \therefore n(\mathrm{~S})=6 \times 6=36$

Let event E: an even prime number on each die

Since, 2 is the only even prime number, $\mathrm{E}=\{(2,2)\}$

$\therefore \quad n(\mathrm{E})=1$

Required probability $=\mathrm{P}(\mathrm{E})=\frac{1}{36} \Rightarrow$ (D) is the correct option.

18. Two events $A$ and $B$ will be independent, if (A) $A$ and $B$ are mutually exclusive (B) $\mathbf{P}\left(\mathbf{A}^{\prime} \mathbf{B}^{\prime}\right)=[\mathbf{1}-\mathbf{P}(\mathbf{A})][\mathbf{1}-\mathbf{P}(\mathbf{B})]$ (C) $\mathbf{P}(\mathbf{A})=\mathbf{P}(\mathbf{B})$ (D) $\mathbf{P}(\mathbf{A})+\mathbf{P}(\mathbf{B})=1$.
Solution

A and B are independent

$\Rightarrow \quad \mathrm{A}^{\prime}$ and $\mathrm{B}^{\prime}$ are independent






$\Rightarrow \quad \mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)=\mathrm{P}\left(\mathrm{A}^{\prime}\right) \mathrm{P}\left(\mathrm{B}^{\prime}\right)=[1-\mathrm{P}(\mathrm{A})][1-\mathrm{P}(\mathrm{B})]$

$\therefore \quad(\mathrm{B})$ is the correct option.

Exercise 13.3
1. An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?
Solution

We are given that Urn contains 5 red and 5 black balls

( $\Rightarrow \quad$ Total balls $=5+5=10$ )

Let $\mathrm{E}_1$ : first draw gives a red ball

$\mathrm{E}_2$ : first draw gives a black ball

(Here $\mathrm{E}_1$ and $\mathrm{E}_2$ form a mutually exclusive and exhaustive set of events.)

$$ \therefore \quad \mathrm{P}\left(\mathrm{E}_1\right)=\frac{5}{10}=\frac{1}{2} \quad \text { and } \quad \mathrm{P}\left(\mathrm{E}_2\right)=\frac{5}{10}=\frac{1}{2} $$

If the first draw is red, two extra red balls are added, making the urn contain $7$ red and $5$ black balls. If the first draw is black, two extra black balls are added, so the urn holds $5$ red and $7$ black balls.

Let A: second draw gives a red ball

Required probability $=\mathrm{P}(\mathrm{A})$

= P(first is red) × P(second is red | two extra red added) + P(first is black) × P(second is red | two extra black added)

$=\frac{1}{2} \times \frac{7}{12}+\frac{1}{2} \times \frac{5}{12}=\frac{7}{24}+\frac{5}{24}=\frac{12}{24}=\frac{1}{2}$.

2. A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.
Solution

Let event $\mathrm{E}_1$ : first bag is selected

$\mathrm{E}_2$ : second bag is selected

(Here $\mathrm{E}_1$ and $\mathrm{E}_2$ form a mutually exclusive and exhaustive set of events.)

$$ \therefore \quad \mathrm{P}\left(\mathrm{E}_1\right)=\mathrm{P}\left(\mathrm{E}_2\right)=\frac{1}{2} $$

Let A : ball drawn is red.

Then $\mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)=$ Probability that a red ball is chosen from bag first $=\frac{4}{4+4}=\frac{4}{8} \quad$ and Similarly, $\mathrm{P}\left(\mathrm{A} / \mathrm{E}_2\right)=\frac{2}{8}$

We have to find $\mathrm{P}\left(\mathrm{E}_1 / \mathrm{A}\right)$.

i.e., Probability (that the ball is from bag I given that it is red).






We know that $\mathrm{P}\left(\mathrm{E}_1 / \mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)}{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)+\mathrm{P}\left(\mathrm{E}_2\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_2\right)}$

(applying Bayes’ Theorem)

$$ \text { Putting values, }=\frac{\frac{1}{2} \times \frac{4}{8}}{\frac{1}{2} \times \frac{4}{8}+\frac{1}{2} \times \frac{2}{8}} $$
$$ \text { Multiplying by L.C.M. }=8=\frac{4}{4+2}=\frac{4}{6}=\frac{2}{3} \text {. } $$
3. Of the students in a college, it is known that $\mathbf{6 0 \%}$ reside in hostel and 40\% are day scholars (not residing in hostel). Previous year results report that $30 \%$ of all students who reside in hostel attain A grade and 20\% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is a hostlier?
Solution

Let event $\mathrm{E}_1$ : a student is residing in hostel

$\mathrm{E}_2$ : a student is a day scholar (i.e., not residing in hostel) Here $\mathrm{E}_1$ and $\mathrm{E}_2$ form a mutually exclusive and exhaustive set of events.

We are given that $60 \%$ students reside in hostel and $40 \%$ don’t reside in hostel (i.e., are day scholars)

$$ \therefore \quad \mathrm{P}\left(\mathrm{E}_1\right)=\frac{60}{100}=\frac{3}{5} \text { and } \mathrm{P}\left(\mathrm{E}_2\right)=\frac{40}{100}=\frac{2}{5} $$

Let event E: a student attains A grade

We are given that 30\% of hostelers get A grade and 20\% day scholars get A grade.

$\therefore \mathrm{P}\left(\mathrm{E} / \mathrm{E}_1\right)$ i.e., probability that a hostlier gets A grade

$$ =\frac{30}{100}=\frac{3}{10} \quad \text { and } \quad \mathrm{P}\left(\mathrm{E} / \mathrm{E}_2\right)=\frac{20}{100}=\frac{2}{10} $$

We have to find $\mathrm{P}\left(\mathrm{E}_1 / \mathrm{E}\right)$. i.e., $\mathrm{P}(\mathrm{a}$ student getting A grade resides in hostel)

We know that

$$ \begin{aligned} \mathrm{P}\left(\mathrm{E}_1 / \mathrm{E}\right) & =\frac{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{E} / \mathrm{E}_1\right)}{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{E} / \mathrm{E}_1\right)+\mathrm{P}\left(\mathrm{E}_2\right) \mathrm{P}\left(\mathrm{E} / \mathrm{E}_2\right)} \quad \text { (applying Bayes’ Theorem) } \\ & =\frac{\frac{3}{5} \times \frac{3}{10}}{\frac{3}{5} \times \frac{3}{10}+\frac{2}{5} \times \frac{2}{10}} \end{aligned} $$

Multiplying each term by $50 gives =\frac{9}{9+4}=\frac{9}{13}$.

4. In answering a question on a multiple choice test, a student either knows the answer or guesses. Let $\frac{3}{4}$ be the probability that he knows the answer and $\frac{1}{4}$ be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability $\frac{1}{4}$. What is the probability that the student knows the answer given that he answered it correctly?
Solution

Let event $\mathrm{E}_1$ : the student knows the answer $\mathrm{E}_2$ : the student guesses the answer Here $\mathrm{E}_1$ and $\mathrm{E}_2$ form a mutually exclusive and exhaustive set of events.

We are given that $\mathrm{P}\left(\mathrm{E}_1\right)=\frac{3}{4} \quad$ and $\quad \mathrm{P}\left(\mathrm{E}_2\right)=\frac{1}{4}$

Let event A: the student answered correctly.

Then $P\left(A / E_1\right)=1 \quad(\because$ When he knows the answer, he answers correctly is a sure event)

and $\mathrm{P}\left(\mathrm{A} / \mathrm{E}_2\right)$ i.e., $\mathrm{P}($ he answers correctly when he guesses the answer) $=\frac{1}{4}$

(given)

We have to find $\mathrm{P}\left(\mathrm{E}_1 / \mathrm{A}\right)$.

Required probability $=$ Probability that the student knows the answer given that he answered it correctly

$\mathrm{P}\left(\mathrm{E}_1 / \mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)}{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)+\mathrm{P}\left(\mathrm{E}_2\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_2\right)} \quad$ (applying Bayes’ Theorem) Substituting values, $

=\frac{\frac{3}{4} \times 1}{\frac{3}{4} \times 1+\frac{1}{4} \times \frac{1}{4}}=\frac{\frac{3}{4}}{\frac{3}{4}+\frac{1}{16}}=\frac{\frac{3}{4}}{\frac{12+1}{16}}=\frac{\frac{3}{4}}{\frac{13}{16}}=\frac{3}{4} \times \frac{16}{13}=\frac{12}{13} .

5. A laboratory blood test is $99 \%$ effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for $0.5 \%$ of the healthy person tested (i.e., if a healthy person is tested, then, with probability 0.005 , the test will imply he has the disease). If 0.1 per cent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?
Solution

Let event $\mathrm{E}_1$ : the person has the disease and $\mathrm{E}_2=\mathrm{E}_1^{\prime}$ : the person is healthy Here $\mathrm{E}_1$ and $\mathrm{E}_2$ form a mutually exclusive and exhaustive set of events.

We are given that 0.1 per cent (i.e., $0.1 \%$ ) of the population actually has the disease.

$$ \therefore \quad \mathrm{P}\left(\mathrm{E}_1\right)=\frac{0.1}{100}=\frac{1}{1000}, $$

and therefore $\mathrm{P}\left(\mathrm{E}_2\right)=\mathrm{P}\left(\mathrm{E}_1{ }^{\prime}\right)=1-\mathrm{P}\left(\mathrm{E}_1\right)=\frac{999}{1000}$

Let A : test result is positive

We are given that $\quad \mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)=\mathrm{P}$ (Test result of a person having disease is







$$ \text { positive })=99 \%=\frac{99}{100} \text { and } \mathrm{P}\left(\mathrm{~A} / \mathrm{E}_2\right)=0.5 \%=\frac{0.5}{100}=\frac{5}{1000} . $$

We have to find $\mathrm{P}\left(\mathrm{E}_1 / \mathrm{A}\right)$.

i.e., the probability that a person truly has the disease given a positive test result.

We know that $\mathrm{P}\left(\mathrm{E}_1 / \mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)}{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)+\mathrm{P}\left(\mathrm{E}_2\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_2\right)}$

(applying Bayes’ Theorem)

$$ =\frac{\frac{1}{1000} \times \frac{99}{100}}{\frac{1}{1000} \times \frac{99}{100}+\frac{999}{1000} \times \frac{5}{1000}} $$

Multiplying every term by $1000 \times 1000$,

$$ =\frac{990}{990+4995}=\frac{990}{5985}=\frac{198}{1197}=\frac{22}{133} . $$
6. There are three coins. One is a two headed coin (having head on both faces), another is a biased coin that comes up heads 75\% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?
Solution

Let event $\mathrm{E}_1$ : the chosen coin is two headed

$\mathrm{E}_2$ : the chosen coin is biased

$\mathrm{E}_3$ : the chosen coin is unbiased

then $\mathrm{E}_1, \mathrm{E}_2, \mathrm{E}_3$ are mutually exclusive and exhaustive events.

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

(Each of the three coins has an equal probability of being selected.) Let event A : the tossed coin shows head. Then $\mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)$ i.e., $\mathrm{P}(\mathrm{A}$ coin having head on both faces shows head)

$$ \begin{aligned} & =\frac{2}{2}=1 \\ \text { We are given that } \mathrm{P}\left(\mathrm{~A} / \mathrm{E}_2\right) & =\frac{75}{100}=\frac{3}{4}, \mathrm{P}\left(\mathrm{~A} / \mathrm{E}_3\right)=\frac{1}{2} \end{aligned} $$

(The third coin is a fair/unbiased coin.)

We have to find $\mathrm{P}\left(\mathrm{E}_1 / \mathrm{A}\right)$. i.e., probability (that the coin chosen is two headed coin given that the coin shows heads)

We know that

$$ \mathrm{P}\left(\mathrm{E}_1 / \mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{~A} / \mathrm{E}_1\right)}{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{~A} / \mathrm{E}_1\right)+\mathrm{P}\left(\mathrm{E}_2\right) \mathrm{P}\left(\mathrm{~A} / \mathrm{E}_2\right)+\underset{(\mathrm{By} \text { Baye’s Theorem })}{\mathrm{P}\left(\mathrm{E}_3\right) \mathrm{P}\left(\mathrm{~A} / \mathrm{E}_3\right)}} $$
$$ \text { Putting values, }=\frac{\frac{1}{3} \times 1}{\frac{1}{3} \times 1+\frac{1}{3} \times \frac{3}{4}+\frac{1}{3} \times \frac{1}{2}} $$







Multiplying every term by $3, we get =\frac{1}{1+\frac{3}{4}+\frac{1}{2}}=\frac{1}{\frac{9}{4}}=\frac{4}{9}$.

7. An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accidents are $0.01,0.03$ and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?
Solution

Let event $\mathrm{E}_1$ : the insured person is a scooter driver

$\mathrm{E}_2$ : the insured person is a car driver

$\mathrm{E}_3$ : the insured person is a truck driver

then $\mathrm{E}_1, \mathrm{E}_2, \mathrm{E}_3$ are mutually exclusive and exhaustive events.

The total number of insured persons

$$ \begin{aligned} & =2000+4000+6000=12000 \\ \therefore \quad \mathrm{P}\left(\mathrm{E}_1\right)=\frac{n\left(\mathrm{E}_1\right)}{n(\mathrm{~S})} & =\frac{2000}{12000}=\frac{1}{6}, \quad \mathrm{P}\left(\mathrm{E}_2\right)=\frac{4000}{12000}=\frac{1}{3}, \\ \mathrm{P}\left(\mathrm{E}_3\right)=\frac{6000}{12000} & =\frac{1}{2} \end{aligned} $$

Let event A: insured person meets with an accident.

We are given that $\mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)$ i.e., $\mathrm{P}(\mathrm{An}$ insured scooter driver meets with an accident $)=0.01=\frac{1}{100}, \quad \mathrm{P}\left(\mathrm{A} / \mathrm{E}_2\right)=0.03=\frac{3}{100}$,

and

$$ \mathrm{P}\left(\mathrm{~A} / \mathrm{E}_3\right)=0.15=\frac{15}{100} $$

We have to find $\mathrm{P}\left(\mathrm{E}_1 / \mathrm{A}\right)$. i.e., $\mathrm{P}($ The person is a scooter driver given that an insured person has met with an accident).

We know that

$$ \mathrm{P}\left(\mathrm{E}_1 / \mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{~A} / \mathrm{E}_1\right)}{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{~A} / \mathrm{E}_1\right)+\mathrm{P}\left(\mathrm{E}_2\right) \mathrm{P}\left(\mathrm{~A} / \mathrm{E}_2\right)+\mathrm{P}\left(\mathrm{E}_3\right) \mathrm{P}\left(\mathrm{~A} / \mathrm{E}_3\right)} $$

(applying Bayes’ Theorem)

$$ =\frac{\frac{1}{6} \times \frac{1}{100}}{\frac{1}{6} \times \frac{1}{100}+\frac{1}{3} \times \frac{3}{100}+\frac{1}{2} \times \frac{15}{100}} $$

Multiplying every term by $600, we get =\frac{1}{1+6+45}=\frac{1}{52}$.

8. A factory has two machines $A$ and $B$. Past record shows that machine A produced $60 \%$ of the items of output and machine $B$ produced $40 \%$ of the items. Further, $2 \%$ of the items produced by machine A and $1 \%$ produced by machine $B$ were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine $B$ ?
Solution

Let event $\mathrm{E}_1$ : the item is produced by machine A and $\mathrm{E}_2$ : the item is produced by machine B then $\mathrm{E}_1, \mathrm{E}_2$ are mutually exclusive and exhaustive events.

We are given that $\quad \mathrm{P}\left(\mathrm{E}_1\right)=60 \%=\frac{60}{100}=\frac{3}{5}, \quad \mathrm{P}\left(\mathrm{E}_2\right)=40 \%=\frac{40}{100}=\frac{2}{5}$

Let $D$ : the chosen item is defective

We are given that $\mathrm{P}\left(\mathrm{D} / \mathrm{E}_1\right)$ i.e., $\mathrm{P}($ an item produced by machine A is defective $)=\frac{2}{100}, \mathrm{P}\left(\mathrm{D} / \mathrm{E}_2\right)=\frac{1}{100}$

We have to find $\mathrm{P}\left(\mathrm{E}_2 / \mathrm{D}\right)=\mathrm{P}($ An item is produced by machine B given that it is defective)

We know that

$$ \begin{aligned} \mathrm{P}\left(\mathrm{E}_2 / \mathrm{D}\right) & =\frac{\mathrm{P}\left(\mathrm{E}_2\right) \mathrm{P}\left(\mathrm{D} / \mathrm{E}_2\right)}{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{D} / \mathrm{E}_1\right)+\mathrm{P}\left(\mathrm{E}_2\right) \mathrm{P}\left(\mathrm{D} / \mathrm{E}_2\right)} \\ & =\frac{\frac{2}{5} \times \frac{1}{100}}{\frac{3}{5} \times \frac{2}{100}+\frac{2}{5} \times \frac{1}{100}} \end{aligned} $$

Multiplying every term by $500, we get =\frac{2}{6+2}=\frac{2}{8}=\frac{1}{4}$.

9. Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 , if the second group wins. Find the probability that the new product introduced was by the second group.
Solution

Let event $\mathrm{E}_1$ : first group wins

and event $\mathrm{E}_2$ : second group wins

then $\mathrm{E}_1, \mathrm{E}_2$ are mutually exclusive and exhaustive events.

We are given that $\mathrm{P}\left(\mathrm{E}_1\right)=0.6=\frac{6}{10}, \quad \mathrm{P}\left(\mathrm{E}_2\right)=0.4=\frac{4}{10}$

Let A: the new product is introduced

We are given that $\mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)=\mathrm{P}$ (New product being introduced if first group wins) $=0.7=\frac{7}{10}$, and $\mathrm{P}\left(\mathrm{A} / \mathrm{E}_2\right)=0.3=\frac{3}{10}$.

We have to find $\mathrm{P}\left(\mathrm{E}_2 / \mathrm{A}\right)$. (i.e., probability that second group wins given that the new product was introduced)

We know that

$$ \mathrm{P}\left(\mathrm{E}_2 / \mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{E}_2\right) \mathrm{P}\left(\mathrm{~A} / \mathrm{E}_2\right)}{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{~A} / \mathrm{E}_1\right)+\mathrm{P}\left(\mathrm{E}_2\right) \mathrm{P}\left(\mathrm{~A} / \mathrm{E}_2\right)} \quad(\text { By Bayes’ Theorem }) $$





$$ =\frac{\frac{4}{10} \times \frac{3}{10}}{\frac{6}{10} \times \frac{7}{10}+\frac{4}{10} \times \frac{3}{10}} $$

Multiplying every term by 100 ,

$$ =\frac{12}{42+12}=\frac{12}{54}=\frac{2}{9} . $$
10. Suppose a girl throws a die. If she gets a 5 or 6 , she tosses a coin three times and notes the number of heads. If she gets $1,2,3$ or 4 , she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw $1,2,3$ or 4 with the die?
Solution

Let event $\mathrm{E}_1$ : the die shows $1,2,3$ or $4 \Rightarrow n\left(\mathrm{E}_1\right)=4$ and $\mathrm{E}_2$ : the die shows 5 or $6 \Rightarrow n\left(\mathrm{E}_2\right)=2$ then $\mathrm{E}_1, \mathrm{E}_2$ are mutually exclusive and exhaustive events.

$$ \therefore \quad \mathrm{P}\left(\mathrm{E}_1\right)=\frac{4}{6}=\frac{2}{3}, \mathrm{P}\left(\mathrm{E}_2\right)=\frac{2}{6}=\frac{1}{3} $$

$[\because$ We know that sample space on tossing a dice is $=\{1,2,3,4$, 5, 6\} and has 6 points]

Let A: the girl obtained exactly one head

then $\mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)=\mathrm{P}($ exactly one head when a coin is tossed once $)=\frac{1}{2}$ and $\mathrm{P}\left(\mathrm{A} / \mathrm{E}_2\right)=\mathrm{P}$ (exactly one head when a coin is tossed three times $)=\frac{3}{8}$

$[\because$ the sample space when a coin is tossed three times is $\mathrm{S}=\{\mathrm{HHH}, \mathrm{HHT}, \mathrm{HTH}, \mathrm{THH}, \mathrm{HTT}, \mathrm{THT}, \mathrm{TTH}, \mathrm{TTT}\}$ Let event E: exactly one head appears, then E $=\{\mathrm{HTT}, \mathrm{THT}, \mathrm{TTH}\}$

and $\mathrm{P}(\mathrm{E})=\left(\mathrm{P}\left(\mathrm{A} / \mathrm{E}_2\right)\right.$ here in this question $\left.)=\frac{3}{8}\right]$

We have to find $\mathrm{P}\left(\mathrm{E}_1 / \mathrm{A}\right)$. i.e., $\mathrm{P}(\mathrm{A}$ dice shows $1,2,3,4$ given that she gets exactly one head).

We know that

$$ \begin{aligned} \mathrm{P}\left(\mathrm{E}_1 / \mathrm{A}\right) & =\frac{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{~A} / \mathrm{E}_1\right)}{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{~A} / \mathrm{E}_1\right)+\mathrm{P}\left(\mathrm{E}_2\right) \mathrm{P}\left(\mathrm{~A} / \mathrm{E}_2\right)} \quad \text { (applying Bayes’ Theorem) } \\ & =\frac{\frac{2}{3} \times \frac{1}{2}}{\frac{2}{3} \times \frac{1}{2}+\frac{1}{3} \times \frac{3}{8}}=\frac{\frac{1}{3}}{\frac{1}{3}+\frac{1}{8}} \end{aligned} $$

Multiplying by L.C.M. $=24,=\frac{8}{8+3}=\frac{8}{11}$.

11. A manufacturer has three machine operators $A, B$ and $C$. The first operator A produces $1 \%$ defective items, where as the other two operators $B$ and $C$ produce $5 \%$ and $7 \%$ defective items respectively. A is on the job for $\mathbf{5 0 \%}$ of the time, $B$ is on the job for $30 \%$ of the time and $C$ is on the job for $20 \%$ of the time. A defective item is produced, what is the probability that it was produced by $A$ ?
Solution

Let event $\mathrm{E}_1$ : operator A is on job

$\mathrm{E}_2$ : operator B is on job

$\mathrm{E}_3$ : operator C is on job

then $\mathrm{E}_1, \mathrm{E}_2, \mathrm{E}_3$ are mutually exclusive and exhaustive.

We are given that $\quad P\left(E_1\right)=50 \%=\frac{50}{100}, \quad P\left(E_2\right)=\frac{30}{100}, \quad P\left(E_3\right)=\frac{20}{100}$

Let D: a defective item is produced

We are given that $\mathrm{P}\left(\mathrm{D} / \mathrm{E}_1\right)=\mathrm{P}($ an item produced by operator A on the job is defective $)=\frac{1}{100}, \quad \mathrm{P}\left(\mathrm{D} / \mathrm{E}_2\right)=\frac{5}{100}, \quad \mathrm{P}\left(\mathrm{D} / \mathrm{E}_3\right)=\frac{7}{100}$

We have to find $\mathrm{P}\left(\mathrm{E}_1 / \mathrm{D}\right) .=\mathrm{P}($ An item was produced by A given that it is defective)

We know that

$$ \mathrm{P}\left(\mathrm{E}_1 / \mathrm{D}\right)=\frac{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{D} / \mathrm{E}_1\right)}{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{D} / \mathrm{E}_1\right)+\mathrm{P}\left(\mathrm{E}_2\right) \mathrm{P}\left(\mathrm{D} / \mathrm{E}_2\right)+\mathrm{P}\left(\mathrm{E}_3\right) \mathrm{P}\left(\mathrm{D} / \mathrm{E}_3\right)} $$

(applying Bayes’ Theorem)

Putting values, $=\frac{\frac{50}{100} \times \frac{1}{100}}{\frac{50}{100} \times \frac{1}{100}+\frac{30}{100} \times \frac{5}{100}+\frac{20}{100} \times \frac{7}{100}}$

Multiplying every term by $100 \times 100$

$$ =\frac{50}{50+150+140}=\frac{50}{340}=\frac{5}{34} . $$
12. A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.
Solution

Let $\mathrm{E}_1$ : the lost card is a diamond and $\mathrm{E}_2=\mathrm{E}_1{ }^{\prime}$ : the lost card is not a diamond then $\mathrm{E}_1, \mathrm{E}_2$ are mutually exclusive and exhaustive.

We know that there are 13 diamond cards in a pack of 52 cards.

$$ \therefore \quad \mathrm{P}\left(\mathrm{E}_1\right)=\frac{13}{52}=\frac{1}{4}, \quad \mathrm{P}\left(\mathrm{E}_2\right)=\mathrm{P}\left(\mathrm{E}_1^{\prime}\right)=1-\mathrm{P}\left(\mathrm{E}_1\right)=1-\frac{1}{4}=\frac{3}{4} $$

Let event A: two cards drawn from the remaining pack are diamonds

then $\mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)=\mathrm{P}$ (drawing two diamond cards when the lost

card is a diamond card) $=\frac{{ }^{12} \mathrm{C}_2}{{ }^{51} \mathrm{C}_2}=\frac{\frac{12 \times 11}{2 \times 1}}{\frac{51 \times 50}{2 \times 1}}=\frac{132}{2550}$








[ ∵ The lost card is a diamond, therefore, there are 12 diamond cards in the remaining pack of 51 cards]

and

$$ \mathrm{P}\left(\mathrm{~A} / \mathrm{E}_2\right)=\frac{{ }^{13} \mathrm{C}_2}{{ }^{51} \mathrm{C}_2}=\frac{\frac{13 \times 12}{2 \times 1}}{\frac{51 \times 50}{2 \times 1}}=\frac{156}{2550} $$

[ ∵ The lost card is not a diamond, therefore, there are 13 diamond cards in the remaining pack of 51 cards]

We have to find $\mathrm{P}\left(\mathrm{E}_1 / \mathrm{A}\right)$ i.e., P (lost card is a diamond card given that the two cards drawn from the remaining pack of 51 cards are diamonds)

We know that

$$ \begin{aligned} \mathrm{P}\left(\mathrm{E}_1 / \mathrm{A}\right) & =\frac{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{~A} / \mathrm{E}_1\right)}{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{~A} / \mathrm{E}_1\right)+\mathrm{P}\left(\mathrm{E}_2\right) \mathrm{P}\left(\mathrm{~A} / \mathrm{E}_2\right)} \text { (applying Bayes’ Theorem) } \\ & =\frac{\frac{1}{4} \times \frac{132}{2550}}{\frac{1}{4} \times \frac{132}{2550}+\frac{3}{4} \times \frac{156}{2550}} \end{aligned} $$

Multiplying every term by $4 \times 2550, we get =\frac{132}{132+468}=\frac{132}{600}=\frac{11}{50}$.

13. Probability that $A$ speaks truth is $\frac{4}{5}$. A coin is tossed. $A$ reports that a head appears. The probability that actually there was head is (A) $\frac{\mathbf{4}}{\mathbf{5}}$ (B) $\frac{1}{2}$ (C) $\frac{1}{5}$ (D) $\frac{2}{5}$.
Solution

Let event $\mathrm{E}_1$ : a head appears on a coin.

and $\mathrm{E}_2=\mathrm{E}_1^{\prime}$ : a head does not appear

then $\mathrm{E}_1, \mathrm{E}_2$ are mutually exclusive and exhaustive events

$$ \mathrm{P}\left(\mathrm{E}_1\right)=\mathrm{P}\left(\mathrm{E}_2\right)=\frac{1}{2} $$

Let event H: (Person) A reports that a head appears

We are given that $\mathrm{P}\left(\mathrm{H} / \mathrm{E}_1\right)=\mathrm{P}$ (Person A reports that a head appears when actually there is head $)=\mathrm{P}(\mathrm{A}$ speaks truth $)=\frac{4}{5}$

and hence $\mathrm{P}\left(\mathrm{H} / \mathrm{E}_2\right)=\mathrm{P}(\mathrm{A}$ tells a lie $)=1-\frac{4}{5}=\frac{1}{5}$

we have to find $\mathrm{P}\left(\mathrm{E}_1 / \mathrm{H}\right)=\mathrm{P}(\mathrm{A}$ head (actually) appears; reported that a head has appeared)

We know that

$\mathrm{P}\left(\mathrm{E}_1 / \mathrm{H}\right)=\frac{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{H} / \mathrm{E}_1\right)}{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{H} / \mathrm{E}_1\right)+\mathrm{P}\left(\mathrm{E}_2\right) \mathrm{P}\left(\mathrm{H} / \mathrm{E}_2\right)} \quad$ (applying Bayes’ Theorem)







Substituting values, $=\frac{\frac{1}{2} \times \frac{4}{5}}{\frac{1}{2} \times \frac{4}{5}+\frac{1}{2} \times \frac{1}{5}}$

Multiplying by L.C.M. $=10,=\frac{4}{4+1}=\frac{4}{5}$.

Hence, option (A) is the correct answer.

14. If $A$ and $B$ are two events such that $A \subset B$ and $P(B) \neq 0$, then which of the following is correct? (A) $\mathbf{P}(\mathbf{A} / \mathbf{B})=\frac{\mathbf{P}(\mathbf{B})}{\mathbf{P}(\mathbf{A})}$ (B) $\mathbf{P}(\mathbf{A} / \mathbf{B})<\mathbf{P}(\mathbf{A})$ (C) $\mathbf{P ( A / B )} \geq \mathbf{P ( A )}$ (D) None of these.
Solution

$\mathrm{A} \subset \mathrm{B} \Rightarrow \mathrm{A} \cap \mathrm{B}=\mathrm{A}$

$$ \therefore \quad \mathrm{P}(\mathrm{~A} / \mathrm{B})=\frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~B})}=\frac{\mathrm{P}(\mathrm{~A})}{\mathrm{P}(\mathrm{~B})} $$

Since, $\mathrm{P}(\mathrm{B}) \neq 0$,

$$ \begin{gathered} \therefore \quad 0<\mathrm{P}(\mathrm{~B}) \leq 1 \Rightarrow \frac{1}{\mathrm{P}(\mathrm{~B})} \geq 1 \\ \therefore \quad \text { From }(i), \mathrm{P}(\mathrm{~A} / \mathrm{B})=\mathrm{P}(\mathrm{~A}) \times \frac{1}{\mathrm{P}(\mathrm{~B})} \geq \mathrm{P}(\mathrm{~A}) \end{gathered} $$

Hence, the correct option is (C).

Miscellaneous Exercise
1. $\mathbf{A}$ and $\mathbf{B}$ are two events such that $\mathbf{P}(\mathbf{A}) \neq 0$. Find $\mathbf{P}(\mathbf{B} / \mathbf{A})$, if (i) A is a subset of B (ii) $\mathbf{A} \cap \mathbf{B}=\phi$.
Solution

We know that

$$ \mathrm{P}(\mathrm{~B} / \mathrm{A})=\frac{\mathrm{P}(\mathrm{~B} \cap \mathrm{~A})}{\mathrm{P}(\mathrm{~A})}=\frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~A})} $$

(i) A is a subset of B (given)

$\therefore \mathrm{A} \cap \mathrm{B}$ i.e., set of common points of A and

B is A i.e., $\mathrm{A} \cap \mathrm{B}=\mathrm{A}$ (See adjoining figure)

Putting $\mathrm{A} \cap \mathrm{B}=\mathrm{A}$ in ( $i$ ), $\mathrm{P}(\mathrm{B} / \mathrm{A})=\frac{\mathrm{P}(\mathrm{A})}{\mathrm{P}(\mathrm{A})}=1$.

(ii) $\mathrm{A} \cap \mathrm{B}=\phi$ (given)


Putting $\mathrm{A} \cap \mathrm{B}=\phi$ in $(i), \mathrm{P}(\mathrm{B} / \mathrm{A})=\frac{\mathrm{P}(\phi)}{\mathrm{P}(\mathrm{A})}=\frac{0}{\mathrm{P}(\mathrm{A})}=0$.

2. A couple has two children, (i) Find the probability that both children are males, if it is known that at least one of the children is male. (ii) Find the probability that both children are females, if it is known that the elder child is a female.
Solution

Let the first (elder) child be denoted by capital letter and the second (younger) child by small letter. The sample space is

$$ \mathrm{S}=\{\mathrm{B} b, \mathrm{~B} g, \mathrm{G} b, \mathrm{G} g\} \quad \Rightarrow n(\mathrm{~S})=4 $$

(i) Let E : both children are males $\Rightarrow \mathrm{E}=\{\mathrm{B} b\}$

$$ \Rightarrow n(\mathrm{E})=1 \quad \Rightarrow \mathrm{P}(\mathrm{E})=\frac{n(\mathrm{E})}{n(\mathrm{~S})}=\frac{1}{4} $$

F : at least one child is male $\quad \Rightarrow \mathrm{F}=\{\mathrm{B} b, \mathrm{~B} g, \mathrm{G} b\}$

$$ \Rightarrow n(\mathrm{~F})=3 \quad \Rightarrow \mathrm{P}(\mathrm{~F})=\frac{n(\mathrm{~F})}{n(\mathrm{~S})}=\frac{3}{4} $$

then $\mathrm{E} \cap \mathrm{F}=\{\mathrm{B} b\}$

$$ \therefore n(\mathrm{E} \cap \mathrm{~F})=1 \quad \Rightarrow \mathrm{P}(\mathrm{E} \cap \mathrm{~F})=\frac{1}{4} $$

Therefore, the required probability

$$ =\mathrm{P}(\mathrm{E} / \mathrm{F})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{~F})}{\mathrm{P}(\mathrm{~F})}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3} $$

(ii) Let E : both children are females $\Rightarrow \mathrm{E}=\{\mathrm{Gg}\}$

$$ \Rightarrow \quad n(\mathrm{E})=1 $$
$$ \begin{aligned} & \Rightarrow \quad \mathrm{P}(\mathrm{E})=\frac{n(\mathrm{E})}{n(\mathrm{~S})}=\frac{1}{4} \\ & \Rightarrow \quad \mathrm{~F}=\{\mathrm{G} b, \mathrm{G} g\} \\ & \Rightarrow \quad \mathrm{P}(\mathrm{~F})=\frac{n(\mathrm{~F})}{n(\mathrm{~S})}=\frac{2}{4} \end{aligned} $$

F: elder child is a female

$$ \Rightarrow \quad n(\mathrm{~F})=2 $$

then $\mathrm{E} \cap \mathrm{F}=\{\mathrm{G} g\}$

$$ \therefore \quad n(\mathrm{E} \cap \mathrm{~F})=1 $$
$$ 1 \quad \Rightarrow \mathrm{P}(\mathrm{E} \cap \mathrm{~F})=\frac{1}{4} $$

Therefore, the required probability

$$ =\mathrm{P}(\mathrm{E} / \mathrm{F})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{~F})}{\mathrm{P}(\mathrm{~F})}=\frac{\frac{1}{4}}{\frac{2}{4}}=\frac{1}{2} . $$
3. Suppose that $5 \%$ of men and $0.25 \%$ of women have grey hair. A grey haired person is selected at random. What is the probability of this person being male? Assume that there are equal number of males and females.
Solution

Let $\mathrm{E}_1$ : selected person is male

$\mathrm{E}_2$ : selected person is a female

then $\mathrm{E}_1$ and $\mathrm{E}_2$ are mutually exclusive and exhaustive.

$$ \mathrm{P}\left(\mathrm{E}_1\right)=\mathrm{P}\left(\mathrm{E}_2\right)=\frac{1}{2} $$






[ ∵ We are given that There are equal number of males and females]

Let A : selected person is grey haired.

We are given that $\mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)$ i.e., $\mathrm{P}(\mathrm{A}$ male person is grey haired $)$

$$ =5 \%=\frac{5}{100}=\frac{1}{20} ; $$

and $\mathrm{P}\left(\mathrm{A} / \mathrm{E}_2\right)=0.25 \%=\frac{0.25}{100}=\frac{\frac{25}{100}}{100}=\frac{1}{400}$

We have to find $\mathrm{P}\left(\mathrm{E}_1 / \mathrm{A}\right)$

i.e., $\mathrm{P}(\mathrm{A}$ person is a male given that the person is grey haired).

We know that $\mathrm{P}\left(\mathrm{E}_1 / \mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)}{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)+\mathrm{P}\left(\mathrm{E}_2\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_2\right)}$

(applying Bayes’ Theorem)

$$ =\frac{\frac{1}{2} \times \frac{1}{20}}{\frac{1}{2} \times \frac{1}{20}+\frac{1}{2} \times \frac{1}{400}} $$

Multiplying every term by $800, we get =\frac{20}{20+1}=\frac{20}{21}$.

4. Suppose that $90 \%$ of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed?
Solution

Let X denote the number of right-handed persons. Here, $n=10$

$$ \begin{aligned} & \quad p=\mathrm{P}(\text { a person is right-handed })=90 \%=\frac{90}{100}=\frac{9}{10}=0.9 \\ & \therefore \quad q=1-p=1-0.9=0.1 \\ & \mathrm{P}(\text { at most } 6 \text { persons are right-handed })=\mathrm{P}(\mathrm{X} \leq 6) \\ & \quad=1-\mathrm{P}(7 \leq \mathrm{X} \leq 10) \\ & \quad=1-[\mathrm{P}(\mathrm{X}=7)+\mathrm{P}(\mathrm{X}=8)+\mathrm{P}(\mathrm{X}=9)+\mathrm{P}(\mathrm{X}=10)] \end{aligned} $$

Putting $x=7,8,9,10$ in $\mathrm{P}(x)={ }^n \mathrm{C}_x p^x q^{n-x}$ (Here $n=10$ )

$$ \begin{aligned} & ={ }^{10} \mathrm{C}_x p^x q^{10-x} \\ & =1-\left[{ }^{10} \mathrm{C}_7 p^7 q^3+{ }^{10} \mathrm{C}_8 p^8 q^2+{ }^{10} \mathrm{C}_9 p^9 q+{ }^{10} \mathrm{C}_{10} p^{10}\right] \\ & =1-\sum_{x=7}^{10}{ }^{10} \mathrm{C}_x p^x q^{10-x}=1-\sum_{x=7}^{10}{ }^{10} \mathrm{C}_x(0.9)^x(0.1)^{10-x} . \end{aligned} $$
5. If a leap year is selected at random, what is the chance that it will contain 53 Tuesdays?
Solution

We know that a leap year has 366 days i.e., 52 weeks +2 additional days. Thus, everyday of the weak occurs 52 times from Jan. 1 to Dec. 29. The last two consecutive days (Dec. 30 and Dec. 31) can be

(i) Sunday and Monday

(ii) Monday and Tuesday

(iii) Tuesday and Wednesday

(iv) Wednesday and Thursday

(v) Thursday and Friday

(vi) Friday and Saturday

(vii) Saturday and Sunday


For a leap year to have 53 Tuesdays, there are two favourable possibilities (ii) and (iii) out of a total of 7 .

Therefore, the required probability $=\frac{2}{7}$.

6. Suppose we have four boxes $A, B, C$ and $D$ containing coloured marbles as given below: \begin{tabular}{|c|c|c|c|} \hline Box & \multicolumn{3}{|c|}{ Marble colour } \\ \hline & Red & White & Black \\ \hline A & 1 & 6 & 3 \\ \hline B & 6 & 2 & 2 \\ \hline C & 8 & 1 & 1 \\ \hline D & 0 & 6 & 4 \\ \hline \end{tabular} One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A?, box B?, box C?
Solution

Let event $\mathrm{E}_1$ : box A is selected, $\quad \mathrm{E}_2$ : box B is selected

$\mathrm{E}_3$ : box C is selected, $\mathrm{E}_4$ : box D is selected then $\mathrm{E}_1, \mathrm{E}_2, \mathrm{E}_3, \mathrm{E}_4$ are mutually exclusive and exhaustive.

Also, $\mathrm{P}\left(\mathrm{E}_1\right)=\mathrm{P}\left(\mathrm{E}_2\right)=\mathrm{P}\left(\mathrm{E}_3\right)=\mathrm{P}\left(\mathrm{E}_4\right)=\frac{1}{4}$

[ ∵ Each of the four boxes have equal chance of being selected] Let event E : marble drawn is red

From Given Table: $\mathrm{P}\left(\mathrm{E} / \mathrm{E}_1\right)=\mathrm{P}($ selecting a red marble from box A$)$

$$ \begin{aligned} & =\frac{1}{1+6+3} \text { (From first row of given table) }=\frac{1}{10} \\ \mathrm{P}\left(\mathrm{E} / \mathrm{E}_2\right) & =\frac{6}{6+2+2}=\frac{6}{10} \\ \mathrm{P}\left(\mathrm{E} / \mathrm{E}_3\right) & =\frac{8}{8+1+1}=\frac{8}{10} \\ \mathrm{P}\left(\mathrm{E} / \mathrm{E}_4\right) & =\frac{0}{0+6+4}=0 \end{aligned} $$

$\therefore \quad \mathrm{P}$ (The box selected is A given that the marble drawn is red) $=\mathrm{P}\left(\mathrm{E}_1 / \mathrm{E}\right)$

$$ =\frac{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{E} / \mathrm{E}_1\right)}{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{E} / \mathrm{E}_1\right)+\mathrm{P}\left(\mathrm{E}_2\right) \mathrm{P}\left(\mathrm{E} / \mathrm{E}_2\right)+\mathrm{P}\left(\mathrm{E}_3\right) \mathrm{P}\left(\mathrm{E} / \mathrm{E}_3\right)+\mathrm{P}\left(\mathrm{E}_4\right) \mathrm{P}\left(\mathrm{E} / \mathrm{E}_4\right)} $$

(applying Bayes’ Theorem)

$$ =\frac{\frac{1}{4} \times \frac{1}{10}}{\frac{1}{4} \times \frac{1}{10}+\frac{1}{4} \times \frac{6}{10}+\frac{1}{4} \times \frac{8}{10}+\frac{1}{4} \times 0} $$






Multiplying every term by $40, we get =\frac{1}{1+6+8+0}=\frac{1}{15}$

P (The box selected is B given that the marble drawn is red)

$$ \begin{aligned} & =\mathrm{P}\left(\mathrm{E}_2 / \mathrm{E}\right) \\ & =\frac{\mathrm{P}\left(\mathrm{E}_2\right) \mathrm{P}\left(\mathrm{E} / \mathrm{E}_2\right)}{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{E} / \mathrm{E}_1\right)+\mathrm{P}\left(\mathrm{E}_2\right) \mathrm{P}\left(\mathrm{E} / \mathrm{E}_2\right)+\mathrm{P}\left(\mathrm{E}_3\right) \mathrm{P}\left(\mathrm{E} / \mathrm{E}_3\right)+\mathrm{P}\left(\mathrm{E}_4\right) \mathrm{P}\left(\mathrm{E} / \mathrm{E}_4\right)} \end{aligned} $$

(applying Bayes’ Theorem)

$$ =\frac{\frac{1}{4} \times \frac{6}{10}}{\frac{1}{4} \times \frac{1}{10}+\frac{1}{4} \times \frac{6}{10}+\frac{1}{4} \times \frac{8}{10}+\frac{1}{4} \times 0} $$

Multiplying every term by $40, we get =\frac{6}{1+6+8+0}=\frac{6}{15}=\frac{2}{5}$

$\mathrm{P}($ Box selected is C given that the marble is red $)=\mathrm{P}\left(\mathrm{E}_3 / \mathrm{E}\right)$

$$ \begin{aligned} & =\frac{\mathrm{P}\left(\mathrm{E}_3\right) \mathrm{P}\left(\mathrm{E} / \mathrm{E}_3\right)}{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{E} / \mathrm{E}_1\right)+\mathrm{P}\left(\mathrm{E}_2\right) \mathrm{P}\left(\mathrm{E} / \mathrm{E}_2\right)+\mathrm{P}\left(\mathrm{E}_3\right) \mathrm{P}\left(\mathrm{E} / \mathrm{E}_3\right)+\mathrm{P}\left(\mathrm{E}_4\right) \mathrm{P}\left(\mathrm{E} / \mathrm{E}_4\right)} \\ & =\frac{\frac{1}{4} \times \frac{8}{10}}{\frac{1}{4} \times \frac{1}{10}+\frac{1}{4} \times \frac{6}{10}+\frac{1}{4} \times \frac{8}{10}+\frac{1}{4} \times 0} \end{aligned} $$

Multiplying every term by $40, we get =\frac{8}{1+6+8+0}=\frac{8}{15}$.

7. Assume that the chances of a patient having a heart attack is $40 \%$. It is also assumed that a meditation and yoga course reduce the risk of heart attack by $30 \%$ and prescription of certain drug reduces its chances by $25 \%$. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga?
Solution

Let event $\mathrm{E}_1$ : patient follows meditation and yoga course

$\mathrm{E}_2$ : patient follows prescription of certain drug then $\mathrm{E}_1, \mathrm{E}_2$ are mutually exclusive and exhaustive.

$$ \mathrm{P}\left(\mathrm{E}_1\right)=\mathrm{P}\left(\mathrm{E}_2\right)=\frac{1}{2} $$

(We are given that A patient can choose any one of the two options with equal probabilities)

Let event A: patient suffers a heart attack

We are given that $\mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)$ (A patient following Meditation and Yoga has a heart attack $)=\frac{40}{100}-\frac{30}{100} \times \frac{40}{100}=\frac{40}{100}-\frac{12}{100}$








[Given $40 \%-($ reduced by $30 \%$ of $40 \%)]=\frac{28}{100}$

Similarly, $\mathrm{P}\left(\mathrm{A} / \mathrm{E}_2\right)=\frac{40}{100}-\frac{25}{100} \times \frac{40}{100}=\frac{40}{100}-\frac{10}{100}=\frac{30}{100}$

We have to find $\mathrm{P}\left(\mathrm{E}_1 / \mathrm{A}\right)$ i.e., $\mathrm{P}($ a patient followed a course of Meditation and Yoga given that the patient had a heart attack)

We know that $\mathrm{P}\left(\mathrm{E}_1 / \mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)}{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)+\mathrm{P}\left(\mathrm{E}_2\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_2\right)}$

(applying Bayes’ Theorem)

$$ =\frac{\frac{1}{2} \times \frac{28}{100}}{\frac{1}{2} \times \frac{28}{100}+\frac{1}{2} \times \frac{30}{100}} $$

Multiplying every term by $2 \times 100, we get =\frac{28}{28+30}=\frac{28}{58}=\frac{14}{29}$.

8. If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability $\frac{1}{2}$ ).
Solution

We know that a second order determinant $\left|\begin{array}{ll}a & b \\ c & d\end{array}\right|$ has four entries.

It is given that each entry is either zero or one i.e., each entry can be filled in two ways.

∴ Number of determinants with entries 0 and 1

$$ =2 \times 2 \times 2 \times 2=16 $$

∴ Sample space $\mathrm{S}=\left\{\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\left|,\left|\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right|,\left|\begin{array}{ll}0 & 0 \\ 1 & 0\end{array}\right|,\left|\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right|\right.\right.$,

$$ \begin{aligned} & \left|\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right|,\left|\begin{array}{ll} 0 & 1 \\ 0 & 1 \end{array}\right|,\left|\begin{array}{ll} 1 & 0 \\ 1 & 0 \end{array}\right|,\left|\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right|,\left|\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right|,\left|\begin{array}{ll} 0 & 0 \\ 1 & 1 \end{array}\right|,\left|\begin{array}{ll} 1 & 1 \\ 0 & 0 \end{array}\right|, \\ & \left.\left|\begin{array}{ll} 0 & 1 \\ 1 & 1 \end{array}\right|,\left|\begin{array}{ll} 1 & 0 \\ 1 & 1 \end{array}\right|,\left|\begin{array}{ll} 1 & 1 \\ 1 & 0 \end{array}\right|,\left|\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right|,\left|\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right|\right\} \end{aligned} $$

$\therefore n(\mathrm{~S})=16$

Let E be the event ( $\subset \mathrm{S}$ ) having determinants of S such that the value of determinant is positive.

$$ \begin{aligned} & \therefore \quad \mathrm{E}=\left\{\left|\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right|,\left|\begin{array}{ll} 1 & 0 \\ 1 & 1 \end{array}\right|,\left|\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right|\right\} . \\ & n(\mathrm{E})=3 \\ & \therefore \text { Required probability }=\mathrm{P}(\mathrm{E})=\frac{n(\mathrm{E})}{n(\mathrm{~S})}=\frac{3}{16} . \end{aligned} $$
9. An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known: $\mathbf{P}(\mathbf{A}$ fails $)=0.2$ $\mathrm{P}(\mathrm{B}$ fails alone $)=0.15$ $P(A$ and $B$ fail $)=0.15$ Evaluate the following probabilities (i) P(A fails/B has failed) (ii) P(A fails alone)
Solution

We are given that $\mathrm{P}(\mathrm{A}$ fails $)=0.2, \mathrm{P}(\mathrm{B}$ fails alone $)=0.15$

and $\mathrm{P}(\mathrm{A}$ and B fail $)=0.15$

$\therefore \quad \mathrm{P}(\mathrm{B}$ fails $)=\mathrm{P}(\mathrm{B})$ fails alone $)+\mathrm{P}(\mathrm{A}$ and B fail)


$$ =0.15+0.15=0.30 $$

(i) $\mathrm{P}(\mathrm{A}$ fails $/ \mathrm{B}$ has failed)

$$ =\mathrm{P}(\mathrm{~A} / \mathrm{B})=\frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~B})}=\frac{\mathrm{P}(\mathrm{~A} \text { and } \mathrm{B} \text { fail })}{\mathrm{P}(\mathrm{~B} \text { fails })}=\frac{0.15}{0.30}=\frac{1}{2}=0.5 $$

(ii) $\mathrm{P}(\mathrm{A}$ fails alone $)=\mathrm{P}(\mathrm{A}$ fails $)-\mathrm{P}(\mathrm{A}$ and B fail $)$


$$ =0.2-0.15=0.05 \text {. } $$
10. Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.
Solution

Let event $\mathrm{E}_1$ : a red ball is transferred from bag I to bag II and $\mathrm{E}_2$ : a black ball is transferred from bag I to bag II

then $\mathrm{E}_1$ and $\mathrm{E}_2$ are mutually exclusive and exhaustive.

$$ \mathrm{P}\left(\mathrm{E}_1\right)=\mathrm{P}(\text { Red ball drawn from bag } \mathrm{I})=\frac{3}{3+4}=\frac{3}{7}, $$

and $\quad \mathrm{P}\left(\mathrm{E}_2\right)=\frac{4}{3+4}=\frac{4}{7}$


Let event A represent drawing a red ball from bag II.

then $\quad \mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)=\mathrm{P}$ (drawing a red ball from bag II when transferred ball is red) $=\frac{4+1}{(4+1)+5}=\frac{5}{10}$


Similarly, $

\mathrm{P}\left(\mathrm{~A} / \mathrm{E}_2\right)=\frac{4}{4+(5+1)}=\frac{4}{10}

$$ We have to find $\mathrm{P}\left(\mathrm{E}_2 / \mathrm{A}\right)$ i.e., $\mathrm{P}($ Transferred ball is black given that ball drawn from Bag II after transfer is red) We know that $\mathrm{P}\left(\mathrm{E}_2 / \mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{E}_2\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_2\right)}{\mathrm{P}\left(\mathrm{E}_1\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)+\mathrm{P}\left(\mathrm{E}_2\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_2\right)}$ (applying Bayes’ Theorem) $$

=\frac{\frac{4}{7} \times \frac{4}{10}}{\frac{3}{7} \times \frac{5}{10}+\frac{4}{7} \times \frac{4}{10}}

$$ Multiplying every term by 70 $$

=\frac{16}{15+16}=\frac{16}{31}

$$ Choose the correct answer in each of the following: $$
11. If $\mathbf{A}$ and $\mathbf{B}$ are two events such that $\mathbf{P}(\mathbf{A}) \neq 0$ and $\mathbf{P}(\mathbf{B} / \mathbf{A})=1$, then (A) $\mathbf{A} \subset \mathbf{B}$ (B) $\mathbf{B} \subset \mathbf{A}$ (C) $\mathrm{B}=\phi$ (D) $\mathbf{A}=\phi$.
Solution

$\mathrm{P}(\mathrm{B} / \mathrm{A})=1 \Rightarrow \frac{\mathrm{P}(\mathrm{B} \cap \mathrm{A})}{\mathrm{P}(\mathrm{A})}=1$

Cross-multiplying, $

\Rightarrow P(B \cap A)=P(A) \quad \Rightarrow B \cap A=A \quad \Rightarrow A \subset B

$$ Hence, option (A) is the correct answer. $$
12. If $\mathbf{P}(\mathbf{A} / \mathbf{B})>\mathbf{P}(\mathbf{A})$, then which of the following is correct: (A) $\mathbf{P}(\mathbf{B} / \mathbf{A})<\mathbf{P}(\mathbf{B})$ (B) $\mathbf{P}(\mathbf{A} \cap \mathbf{B})<\mathbf{P}(\mathbf{A}) \cdot \mathbf{P}(\mathbf{B})$ (C) $\mathbf{P}(\mathbf{B} / \mathbf{A})>\mathbf{P}(\mathbf{B})$ (D) $\mathbf{P}(\mathbf{B} / \mathbf{A})=\mathbf{P}(\mathbf{B})$.
Solution

We are given that $\mathrm{P}(\mathrm{A} / \mathrm{B})>\mathrm{P}(\mathrm{A})$

$\Rightarrow \frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}>\mathrm{P}(\mathrm{A})$

Multiplying both sides by $\mathrm{P}(\mathrm{B}), \mathrm{P}(\mathrm{A} \cap \mathrm{B})>\mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B})$

Dividing both sides by $\mathrm{P}(\mathrm{A})$,

$$ \frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~A})}>\mathrm{P}(\mathrm{~B}) \quad \text { or } \quad \mathrm{P}(\mathrm{~B} / \mathrm{A})>\mathrm{P}(\mathrm{~B}) $$

Therefore, option (C) is the correct choice.

OR

Interchanging A and B in ( $i$ ), $\mathrm{P}(\mathrm{B} / \mathrm{A})>\mathrm{P}(\mathrm{B})$.

13. If $A$ and $B$ are any two events such that $ \mathbf{P}(\mathbf{A})+\mathbf{P}(\mathbf{B})-\mathbf{P}(\mathbf{A} \text { and } \mathbf{B})=\mathbf{P}(\mathbf{A}) \text {, then } $ (A) $\mathrm{P}(\mathrm{B} / \mathrm{A})=1$ (B) $\mathrm{P}(\mathrm{A} / \mathrm{B})=1$ (C) $\mathbf{P}(\mathbf{B} / \mathbf{A})=\mathbf{0}$ (D) $\mathbf{P}(\mathbf{A} / \mathbf{B})=\mathbf{0}$.
Solution

$\quad \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A}$ and B$)=\mathrm{P}(\mathrm{A})$

$$ \begin{array}{lr} \Rightarrow & \mathrm{P}(\mathrm{~A})+\mathrm{P}(\mathrm{~B})-\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})=\mathrm{P}(\mathrm{~A}) \\ \Rightarrow & \mathrm{P}(\mathrm{~B})-\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})=0 \\ \Rightarrow & \mathrm{P}(\mathrm{~B})=\mathrm{P}(\mathrm{~A} \cap \mathrm{~B}) \\ \Rightarrow & \frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~B})}=1 \\ & \Rightarrow \mathrm{P}(\mathrm{~A} / \mathrm{B})=1 \end{array} $$

$\therefore \quad$ The correct option is (B).

Test Your Mathematical Logic

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Probability Class 12 (121301)

General Instruction:

1. There are 10 MCQ's in the Test.

2. Passing %age is 50.

3. After time gets over, test will be submitted itself.

1 / 10

The probability distribution of a random variable gives:

A) The possible values of the random variable.

B) The probabilities of the possible values.

C) The mean of the random variable.

D) The standard deviation of the random variable.

2 / 10

In a bag, there are 6 red balls and 4 blue balls. If two balls are drawn at random without replacement, what is the probability of getting two red balls?

A) 3/8

B) 1/3

C) 5/12

D) 1/4

3 / 10

Two dice are rolled. What is the probability of getting a sum of 7?

A) 1/6

B) 1/12

C) 1/36

D) 1/4

4 / 10

In a class, 70% of the students wear glasses. If a student is chosen at random, what is the probability that they do not wear glasses?

A) 0.3

B) 0.7

C) 0.4

D) 0.6

5 / 10

In a bag, there are 4 red balls and 5 blue balls. If two balls are drawn at random without replacement, what is the probability of getting a red ball and then a blue ball?

A) 4/45

B) 4/9

C) 5/18

D) 2/9

6 / 10

A card is drawn from a well-shuffled deck and not replaced. Then a second card is drawn. What is the probability of getting a heart and then a spade?

A) 1/52

B) 1/26

C) 1/104

D) 1/13

7 / 10

If two events are independent, what can you say about their probabilities?

A) The probabilities are equal.

B) The probabilities are different.

C) The probabilities are multiplied.

D) The probabilities are added.

8 / 10

A bag contains 4 red balls and 6 blue balls. If one ball is drawn at random, what is the probability of drawing a red ball?

A) 4/10

B) 6/10

C) 1/4

D) 3/5

9 / 10

If two events are mutually exclusive, what can you say about their probabilities?

A) The probabilities are equal.

B) The probabilities are different.

C) The probabilities are multiplied.

D) The probabilities are added.

10 / 10

If two events are dependent, what can you say about their probabilities?

A) The probabilities are equal.

B) The probabilities are different.

C) The probabilities are multiplied.

D) The probabilities are added.

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