Question

If S is the sample space and P(A) = $\frac{1}{3}$ P(B) and S = $A\cup B$, where A and B are two mutually exclusive events, then P(A) =

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $\frac{3}{4}$
• D. $\frac{3}{8}$

Answer 1

The correct option is A

$\frac{1}{4}$

Let P(B) = p

Then P(A) = $\frac{1}{3}p$

Since A and B are two mutually exclusive events, we have :

$A\cup B=S$

$\Rightarrow P(A\cup B)=P\left(S\right)$

$\Rightarrow P(A\cup B)=1$

$\Rightarrow P\left(A\right)+\left(B\right)=1$

$\Rightarrow \frac{1}{3}p+p=1$

$\Rightarrow \frac{4p}{3}=1$

$\Rightarrow p=\frac{3}{4}$

$\therefore P\left(A\right)=\frac{1}{3}p=\frac{1}{3}\times \frac{3}{4}=\times 14$