Question

If $P\left(A\right)=0.4,P\left(B\right)=x,P\left(AUB\right)=0.7$, and the events $A$ and $B$ are mutually exclusive, then$x=?$

• A. $\frac{3}{10}$
• B. $\frac{1}{2}$
• C. $\frac{2}{5}$
• D. $\frac{1}{5}$

Answer 1

The correct option is A

$\frac{3}{10}$

Explanation for the correct option.

Given, $P\left(A\right)=0.4,P\left(B\right)=x,P\left(AUB\right)=0.7$.

Also $A\&B$are mutually exclusive it means, $P(A\cap B)=0$.

Apply this formula:

$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)....\left(1\right)$.

Substituting values in equation (1) we get

$\begin{array}{rcl}& \Rightarrow & 0.7=0.4+x-0\\ & \Rightarrow & 0.3=x\\ & \Rightarrow & \frac{3}{10}=x\end{array}$

Hence, option A is correct.