Question

Two events $A$ and $B$ are such that $P\left(A\right)=\frac{1}{4},P\left(B\right)=\frac{1}{2}$ and $P\left(B|A\right)=\frac{1}{2}$
Consider the following statements
$\left(I\right)$ $P\left(\overline{A}|\overline{B}\right)=\frac{3}{4}$
$\left(II\right)$ $A$ and $B$ are mutually exclusive
$\left(III\right)$ $P\left(\overline{A}|\overline{B}\right)+P\left(A|\overline{B}\right)=1$
Then

• A. Only $I$ is correct
• B. Only $I$ and $II$ are correct
• C. Only $I$ and $III$ are correct
• D. Only $II$ and $III$ are correct

Answer 1

Answer: (C)

Only $I$ and $III$ are correct

$P\left(A\right)=\frac{1}{4},P\left(B\right)=\frac{1}{2},P\left(B|A\right)=\frac{1}{2}$

$P\left(B|A\right)=\frac{P(B\cap A)}{P\left(A\right)}\Rightarrow \frac{1}{2}=\frac{P(B\cap A)}{\frac{1}{4}}\Rightarrow P(B\cap A)=\frac{1}{8}=P\left(A\right)P\left(B\right)$

Hence, $A$ and $B$ are mutually independent events.

$\left(I\right)P\left(\overline{A}|\overline{B}\right)=\frac{P(\overline{A}\cap \overline{B})}{P\left(\overline{B}\right)}=\frac{P\left(\overline{A}\right)P\left(\overline{B}\right)}{1-P\left(B\right)}=\frac{(1-\frac{1}{4})(1-\frac{1}{2})}{1-\frac{1}{2}}=\frac{3}{4}$

Hence, $I$ is correct.

$\left(II\right)P(A\cap B)\ne 0$

Hence, $A$ and $B$ are not mutually exclusive. $II$ is not correct.

$\left(III\right)P\left(\overline{A}|\overline{B}\right)+P\left(A|\overline{B}\right)=\frac{P(\overline{A}\cap \overline{B})}{P\left(\overline{B}\right)}+\frac{P(A\cap \overline{B})}{P\left(\overline{B}\right)}=\frac{3}{4}+\frac{\frac{1}{4}\times (1-\frac{1}{2})}{1-\frac{1}{2}}=\frac{3}{4}+\frac{1}{4}=1$

Hence $III$ is also correct.

$\therefore$ Opt:$\left[C\right]$ is correct