Question

Let $A$ and $B$ be two events with $P\left(A^C\right)=0.3,P\left(B\right)=0.4$ and $P(A\cap B^C)=0.5$. Then, $P(B|A\cup B^C)$ is equal to

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

Answer 1

Answer: (A)

$\frac{1}{4}$

Given, $P\left(A^C\right)=0.3,P\left(B\right)=0.4$ and $P(A\cap B^C)=0.5$

$⟹P\left(A\right)=1-0.3=0.7,P\left(B\right)=1-0.4=0.6$

Now,

$P\left(\frac{B}{A\cup B^C}\right)=\frac{P(B\cap (A\cup B^C\left)\right)}{P(A\cup B^C)}$

$=\frac{P(A\cap B)}{P\left(A\right)+P\left(B^C\right)-P(A\cap B^C)}$

$=\frac{P\left(A\right)-P(A\cap B^C)}{P\left(A\right)+P\left(B^C\right)-P(A\cap B^C)}$

$=\frac{0.7-0.5}{0.7+0.6-0.5}=\frac{0.2}{0.8}=\frac{1}{4}$