# If $P(A)=\large\frac{1}{12}\: P(B)=\large\frac{5}{12}\: and \: P(B/A) = \large\frac{1}{15}$ then $P(A \cup B )$ is equal to

$\begin {array} {1 1} (A)\;\large\frac{89}{180} & \quad (B)\;\large\frac{90}{180} \\ (C)\;\large\frac{91}{180} & \quad (D)\;\large\frac{92}{180} \end {array}$

$\because P(A)=\large\frac{1}{12}$
$P(B) = \large\frac{5}{12}$
$P(B/A) = \large\frac{1}{15}$
We know that $P(B/A) = \large\frac{P(A \cap B )}{P(A)}$
$= \large\frac{1}{15}$
$= \large\frac{P(A \cap B )}{\large\frac{1}{12}}$
$\Rightarrow P(A \cap B ) = \large\frac{1}{180}$
Also, $P(A \cup B ) = P(A)+P(B)-P(A \cap B )$
$= -\large\frac{1}{12}+\large\frac{5}{12}-\large\frac{1}{180}$
$= \large\frac{15+75-1}{180}$
$= \large\frac{89}{180}$
Ans : (A)