Question

Let $A$ and $B$ be two events with $P\left(A^c\right)=0.55,P\left(B\right)=0.36,P(A\cup B)=\mathrm{0.60.}$ Then $P\left(A|B^c\right)$ is

• A. $\frac{1}{4}$
• B. $\frac{3}{8}$
• C. $\frac{1}{2}$
• D. $\frac{5}{8}$

Answer 1

The correct option is B $\frac{3}{8}$

$P\left(A\right)=1-0.55=0.45$
$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)\Rightarrow 0.60=0.45+0.36-P(A\cap B)\Rightarrow P(A\cap B)=0.21$
$P\left(\frac{A}{B^c}\right)=\frac{P(A\cap B^c)}{P\left(B^c\right)}\Rightarrow P\left(\frac{A}{B^c}\right)=\frac{P\left(A\right)-P(A\cap B)}{1-P\left(B\right)}$
$\Rightarrow P\left(\frac{A}{B^c}\right)=\frac{0.45-0.21}{1-0.36}\Rightarrow P\left(\frac{A}{B^c}\right)=\frac{0.24}{0.64}=\frac{3}{8}$