Question

If $A$ and $B$ are mutually exclusive events with $P(A)=\frac{1}{2}\times P(B)$ and $A\cup B=S,$ (total sample space) then $P(A)$ is equal to

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

Answer 1

Answer: (B)

Given, $P(A)=\frac{1}{2}P(B)$
$\Rightarrow$ $P(B)=2P(A)$
and $A\cup B=S$
$\therefore$ $P(A\cup B)=P(S)=1$
$\Rightarrow$ $P(A)+P(B)=1$
$[\because P(A\cap B)=0]$
$\Rightarrow$ $P(A)+2P(A)=1$
$\Rightarrow$ $P(A)=\frac{1}{3}$