Question

If A and B are two mutually exclusive events such that $P\left(A\right)=\frac{1}{2}P\left(B\right)$ and $A\cup B=S$ the sample space, then $P\left(A\right)$

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

Answer 1

The correct option is B $\frac{1}{3}$

Given $P\left(A\right)=\frac{1}{2}P\left(B\right)$ and $A\cup B=S$ , the sample space A and B are mutually exclusive events which means $P(A\cap B)=0$

whereas $P(A\cup B)=S$ means $P(A\cup B)=1$

We know that $P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$

$\Rightarrow 1=P\left(A\right)+2P\left(A\right)-0\Rightarrow 1/3=P\left(A\right)$