Question

Two events A and B are such that $P($not $B)$ $=0.8$, $P(A\cup B)=0.5$ and $P(A\setminus B)=0.4$ Then $P\left(A\right)$ is equal to

• A. $0.28$
• B. $0.32$
• C. $0.38$
• D. None of the above

Answer 1

The correct option is C $0.38$

Solution:

Given: $P\left(\overline{B}\right)=0.8\therefore P\left(B\right)=1-P\left(\overline{B}\right)=1-0.8=0.2$

$P(A\cup B)=0.5$ and $P(A\setminus B)=0.4$

$\therefore P(A\setminus B)=\frac{P(A\cap B)}{P\left(B\right)}$

or, $0.4=\frac{P(A\cap B)}{0.2}$

or, $P(A\cap B)=0.08$

Hence, $P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$

or, $0.5=P\left(A\right)+0.2-0.08$

or, $P\left(A\right)=0.38$

Hence, C is the correct option.