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 $P(A\cup B)$is equal to

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

Answer 1

The correct option is A

$\frac{89}{180}$

Explanation for the correct option.

Step 1: Determination of $\mathit{P}\left(A\cap B\right)$

From the question, we know that $P(B/A)=\frac{1}{15}$, $P\left(A\right)=\frac{1}{12},P\left(B\right)=\frac{5}{12}$.

As the formula for finding the probability of $B/A$ is

$\mathit{P}\left(B/A\right)\mathbf{=}\frac{\mathbf{}\mathbf{P}\left(A\cap B\right)}{\mathbf{P}\left(A\right)}$.

By the substitution of the given values in the formula we have,

$$\begin{gathered} \therefore \frac{1}{15}=\frac{P\left(A\cap B\right)}{{\displaystyle \frac{1}{12}}} \\ \Rightarrow \frac{1}{15}\times \frac{1}{12}=P\left(A\cap B\right) \\ \Rightarrow \frac{1}{180}=P\left(A\cap B\right) \end{gathered}$$

Step 2: Calculation for the $P(A\cup B)$

As the formula for the $P\left(A\cup B\right)$ is

$\mathit{P}\left(A\cup B\right)\mathbf{=}\mathit{P}\left(A\right)\mathbf{+}\mathit{P}\left(B\right)\mathbf{-}\mathit{P}\left(A\cap B\right)$, then

$\begin{array}{rcl}P(A\cup B)& =& \frac{1}{12}+\frac{5}{12}-\frac{1}{180}\\ & =& \frac{15+75-1}{180}\\ & =& \frac{89}{180}\end{array}$

Hence, option A is correct.