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

$\begin {array} {1 1} (A)\;\large\frac{1}{4} & \quad (B)\;\large\frac{1}{2} \\ (C)\;\large\frac{3}{4} & \quad (D)\;\large\frac{3}{8} \end {array}$

Can you answer this question?

$P(A \cup B) = P(S)$
$\Rightarrow P(A)+P(B)=1$
$\Rightarrow \large\frac{1}{3} P(B) +P(B)=1$
$\Rightarrow P(B) = \large\frac{3}{4}$
$\therefore P(A) = \large\frac{1}{3} \: \: \: P(B) = \large\frac{1}{4}$
answered Jan 6, 2014