Question

Consider two events A and B that P(A) = $\frac{1}{4},P\left(\frac{B}{A}\right)=\frac{1}{2},P\left(\frac{A}{B}\right)=\frac{1}{4}.$ For each of the following statements, which is true
I. $P\left(\frac{A^c}{B^c}\right)=\frac{3}{4}$
II. The events A and B are mutually exclusive
III. $P\left(\frac{A}{B}\right)+P\left(\frac{A}{B^c}\right)=1$

• A. I only
• B. I and II
• C. I and III
• D. II and III

Answer 1

The correct option is A I only

$P\left(\frac{B}{A}\right)=\frac{P(A\cap B)}{P\left(A\right)}\Rightarrow \frac{1}{2}=\frac{P(A\cap B)}{\frac{1}{4}}$
$\Rightarrow P(A\cap B)=\frac{1}{8}$
Hence events A and B are not mutually exclusive
$\therefore$ Statement II is incorrect.
$P\left(\frac{B}{A}\right)=\frac{P(A\cap B)}{P\left(A\right)}\Rightarrow P\left(B\right)=\frac{1}{2}$
$\because$ $P(A\cap B)=\frac{1}{8}=P\left(A\right).P\left(B\right)$
$\therefore$ events A and B are independent events.
$P\left(\frac{A^c}{B^c}\right)=\frac{P(A^c\cap B^c)}{P\left(B^c\right)}=\frac{P\left(A^c\right)P\left(B^c\right)}{P\left(B^c\right)}$
$=\frac{1}{4}+\frac{P\left(A\right)-P(A\cap B)}{P\left(B^c\right)}=\frac{1}{4}+\frac{\frac{1}{4}-\frac{1}{8}}{\frac{1}{2}}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$
Hence statement III is incorrect.