Question

Two events $A$ and $B$are such that $P(notB)$ $=0.8$, $P(A\cup B)=0.5$ and $P\left(\frac{A}{B}\right)=0.4.$ 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

$0.38$

Find the required probability:

Given,

$P(notB)$$=0.8$

$P(A\cup B)=0.5$

$P\left(\frac{A}{B}\right)=0.4$

Now, $P\left(B\right)=1-0.8=0.2$

So, the $P\left(\frac{A}{B}\right)=\frac{P(A\cap B)}{P\left(B\right)}$

$$\begin{gathered} \Rightarrow P(A\cap B)=P\left(\frac{A}{B}\right)\times P\left(B\right) \\ \Rightarrow P(A\cap B)=0.4\times 0.2 \\ \Rightarrow P(A\cap B)=0.08 \end{gathered}$$

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

$$\begin{gathered} \Rightarrow 0.5=P\left(A\right)+0.2-0.08 \\ \Rightarrow P\left(A\right)=0.38 \end{gathered}$$

Hence, the correct option is (C).