Question

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

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

Answer 1

The correct option is A $\frac{1}{4}$

A and B are two mutually exclusive events .So, $P(A\cap B)=0$
Because $S=A\cup B$ so: $P(A\cup B)=1$. It is a case of an Exhaustive Event too.
$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$
$1=P\left(A\right)+3P\left(A\right)-0$
$P\left(A\right)=\frac{1}{4}$