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 $P\left(E_1\mid {\bar{E}}_2\right)=1-P\left({\bar{E}}_1\mid {\bar{E}}_2\right)=1-\frac{26}{29}=\frac{3}{29}$