Question

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

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

Answer 1

Answer: (A)

$\frac{1}{4}$

Answer 2

The correct answer is A:$\frac{1}{4}$
Given that;
$P(A^c)=0.3,P(B)=0.4$
$\therefore P(A)=1-P(A^c)=1-0.3=0.7$
$P(B)=1-0.4=0.6$
Now, $P\left(\frac{B}{A \cup B^{c}}\right)=\frac{P\left(B \cap\left(A \cup B^{c}\right)\right)}{P\left(A \cup B^{c}\right)}$
$=\frac{P(A \cap B)}{P(A)+P\left(B^{c}\right)-P\left(A \cap B^{c}\right)}$
$=\frac{P(A)-P\left(A \cap B^{c}\right)}{P(A)+P\left(B^{c}\right)-P\left(A \cap B^{c}\right)}$
$=\frac{0.7-0.5}{0.7+0.6-0.5}$
$=\frac{0.2}{0.8}=\frac{1}{4}$