If $E_1$ and $E_2$ are two events of a random experiment such that $P\left(E_1\right)=\frac{1}{8}, P\left(E_1 \mid E_2\right)=\frac{1}{3}$, $P\left(E_2 \mid E_1\right)=\frac{1}{4}$, then match the items of List-I with the items of List-II. $\begin{array}{lllc} \hline & \text { List-I } & & \text { List-II } \ \hline \text { (A) } & P\left(E_2\right) & \text { I. } & \frac{3}{16} \ \hline \text { (B) } & P\left(E_1 \cup E_2\right) & \text { II. } & \frac{3}{29} \ \hline \text { (C) } & P\left(\bar{E}_1 \mid \bar{E}_2\right) & \text { III. } & \frac{3}{32} \ \hline \text { (D) } & P\left(E_1 \mid \bar{E}_2\right) & \text { IV. } & \frac{26}{29} \ \hline & & \text { V. } & \frac{13}{32} \ \hline \end{array}$ The correct match is

Question

If $E_1$ and $E_2$ are two events of a random experiment such that $P\left(E_1\right)=\frac{1}{8}, P\left(E_1 \mid E_2\right)=\frac{1}{3}$, $P\left(E_2 \mid E_1\right)=\frac{1}{4}$, then match the items of List-I with the items of List-II. $\begin{array}{lllc} \hline & \text { List-I } & & \text { List-II } \ \hline \text { (A) } & P\left(E_2\right) & \text { I. } & \frac{3}{16} \ \hline \text { (B) } & P\left(E_1 \cup E_2\right) & \text { II. } & \frac{3}{29} \ \hline \text { (C) } & P\left(\bar{E}_1 \mid \bar{E}_2\right) & \text { III. } & \frac{3}{32} \ \hline \text { (D) } & P\left(E_1 \mid \bar{E}_2\right) & \text { IV. } & \frac{26}{29} \ \hline & & \text { V. } & \frac{13}{32} \ \hline \end{array}$ The correct match is, $\begin{array}{cc} & A & B & C & D \ & I & III & IV & II \end{array}$, $\begin{array}{cc} & A & B & C & D \ & III & I & IV & V \end{array}$, $\begin{array}{cc} & A & B & C & D \ & III & I & IV & II \end{array}$, $\begin{array}{cc} & A & B & C & D \ & I & II & V & IV \end{array}$

MARKS App · Accepted Answer

For two given events $E_1$ and $E_2$, the given information are $P\left(E_1\right)=\frac{1}{8}, P\left(E_1 \mid E_2\right)=\frac{1}{3}$ and $\begin{aligned} & P\left(E_2 \mid E_1\right)=\frac{1}{4} \ & \because P\left(E_2 \mid E_1\right)=\frac{1}{4} \Rightarrow \frac{P\left(E_1 \cap E_2\right)}{P\left(E_1\right)}=\frac{1}{4} \ & \Rightarrow \quad P\left(E_1 \cap E_2\right)=\frac{1}{32} \ & \therefore \quad P\left(E_2\right)=\frac{P\left(E_1 \cap E_2\right)}{P\left(E_1 \mid E_2\right)}=\frac{\frac{1}{32}}{\frac{1}{3}}=\frac{3}{32} \ & \therefore P\left(E_1 \cup E_2\right)=P\left(E_1\right)+P\left(E_2\right)-P\left(E_1 \cap E_2\right) \ & =\frac{1}{8}+\frac{3}{32}-\frac{1}{32}=\frac{3}{16} \end{aligned}$ $\begin{aligned} \because \quad P\left(\bar{E}_1\right) & =\frac{7}{8} \text { and } P\left(\bar{E}_2\right)=\frac{29}{32} \ \text {and } P\left(\bar{E}_1 \cap \bar{E}_2\right) & =P\left(\overline{E_1 \cup E_2}\right) \ & =1-P\left(E_1 \cup E_2\right)=\frac{13}{16} \ \therefore \quad P\left(\bar{E}_1 \mid \bar{E}_2\right) & =\frac{P\left(\bar{E}_1 \cap \bar{E}_2\right)}{P\left(\bar{E}_2\right)}=\frac{\frac{13}{16}}{\frac{29}{32}}=\frac{26}{29} \ \text {and } \quad P\left(E_1 \mid \bar{E}_2\right) & =1-P\left(\bar{E}_1 \mid \bar{E}_2\right)=1-\frac{26}{29}=\frac{3}{29} \end{aligned}$ Hence, option (c) is correct.

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