Question

If $P\left(A\right)=\frac{1}{12},P\left(B\right)=\frac{5}{12},$ and $P\left(B|A\right)=\frac{1}{15}$ then $p\left(A\cup B\right)$ is equal to

• A. $\frac{89}{180}$
• B. $\frac{90}{180}$
• C. $\frac{91}{180}$
• D. $\frac{92}{180}$

Answer 1

Answer: (A)

Given
$P\left(A\right)=\frac{1}{12},P\left(B\right)=\frac{5}{12},P\left(B/A\right)=\frac{1}{15}$
We know that $P\left(B/A\right)=\frac{P\left(A\cap B\right)}{P\left(A\right)}$
$\Rightarrow \frac{1}{15}=\frac{P\left(A\cap B\right)}{1/12}$
$\Rightarrow P\left(A\cap B\right)=\frac{1}{15\times 12}=\frac{1}{180}$
But,
$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$
$\Rightarrow P\left(A\cup B\right)=\frac{1}{12}+\frac{5}{12}-\frac{1}{180}=\frac{89}{180}$