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.

List-I		List-II
(A)	$P(E_{2})$	I.	$\frac{3}{16}$
(B)	$P\left(E_{1} \cup E_{2}\right)$	II.	$\frac{3}{29}$
(C)	$P\left(\bar{E}_{1} \mid \bar{E}_{2}\right)$	III.	$\frac{3}{32}$
(D)	$P\left(E_{1} \mid \bar{E}_{2}\right)$	IV.	$\frac{26}{29}$
		V.	$\frac{13}{32}$

The correct match is

• A. (A)-(I), (B)-(II), (C)-(IV), (D)-(II)
• B. (A)-(III), (B)-(I), (C)-(IV), (D)-(VI)
• C. (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
• D. (A)-(I), (B)-(II), (C)-(V), (D)-(IV)

Answer 1

Answer: (C)

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 $P\left(E_{2} \mid E_{1}\right)=\frac{1}{4}$
$\because\, P\left(E_{2} \mid E_{1}\right)=\frac{1}{4} $
$\Rightarrow \, \frac{P\left(E_{1} \cap E_{2}\right)}{P\left(E_{1}\right)}=\frac{1}{4}$
$\Rightarrow \,P\left(E_{1} \cap E_{2}\right)=\frac{1}{32}$
$\therefore \,P\left(E_{2}\right)=\frac{P\left(E_{1} \cap E_{2}\right)}{P\left(E_{1} \mid E_{2}\right)}=\frac{\overline{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) $
$=\frac{1}{8}+\frac{3}{32}-\frac{1}{32}=\frac{3}{16}$
$\because \, P\left(\bar{E}_{1}\right) =\frac{7}{8} \text { and } P\left(\bar{E}_{2}\right)=\frac{29}{32} $
and $P\left(\bar{E}_{1} \cap \bar{E}_{2}\right) =P\left(\overline{E_{1} \cup E_{2}}\right)$\
$=1-P\left(E_{1} \cup E_{2}\right)=\frac{13}{16}$
$\therefore \, P\left(\bar{E}_{1} \mid \bar{E}_{2}\right)=\frac{P\left(\bar{E}_{1} \cap \bar{E}_{2}\right)}{P\left(\bar{E}_{2}\right)}=\frac{\frac{13}{16}}{\frac{29}{32}}=\frac{26}{29}$
and $ 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}$