Question

If $A,B$ and $C$ are mutually exclusive and exhaustive events of a random experiment such that $P\left(B\right)=\frac{3}{2}P\left(A\right)$ and $P\left(C\right)=\frac{1}{2}P\left(B\right)$, then $P(A\cup C)$ equals

• A. $\frac{10}{13}$
• B. $\frac{3}{13}$
• C. $\frac{6}{13}$
• D. $\frac{7}{13}$

Answer 1

The correct option is D

$\frac{7}{13}$

Explanation for the correct option:

Step 1. Find the value of $P(A\cup C)$:

Given that, $P\left(B\right)=\frac{3}{2}P\left(A\right)$ and $P\left(C\right)=\frac{1}{2}P\left(B\right)$

$\because A,B,$ and $C$ are mutually exclusive events, then

$P\left(A\right)+P\left(B\right)+P\left(C\right)=1$

Step 2. substitute the values of $P\left(B\right)$ and $P\left(C\right)$ in the formula:

$P\left(A\right)+\left(\frac{3}{2}\right)P\left(A\right)+\left(\frac{1}{2}\right)P\left(B\right)=1$

$\Rightarrow$$P\left(A\right)+\left(\frac{3}{2}\right)P\left(A\right)+\left(\frac{1}{2}\right)\left(\frac{3}{2}\right)P\left(A\right)=1$

$\Rightarrow$ $\left(\frac{13}{4}\right)P\left(A\right)=1$

$\Rightarrow$ $P\left(A\right)=\frac{4}{13}$

$P\left(C\right)=\left(\frac{1}{2}\right)\left(\frac{3}{2}\right)P\left(A\right)$

$\Rightarrow$$P\left(C\right)=\left(\frac{1}{2}\right)\left(\frac{3}{2}\right)\left(\frac{4}{13}\right)$

$\Rightarrow$$P\left(C\right)=\frac{3}{13}$

Step 3. For finding $P(A\cup C)$ we have

$\begin{array}{rcl}P(A\cap B)& =& P(B\cap C)\\ & =& P(C\cap A)\\ & =& 0\end{array}$ [$\because A,B,$ and $C$ are mutually exclusive events]

$\begin{array}{rcl}\therefore P(A\cup C)& =& P\left(A\right)+P\left(C\right)-0\\ & =& \frac{4}{13}+\frac{3}{13}\\ & =& \frac{\mathbf{7}}{\mathbf{13}}\end{array}$

Hence, Option ‘D’ is Correct.