Question

If $A,B$ and $C$ are mutually exclusive and exclusive 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 $P(A\cup C)=$

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

Answer 1

Answer: (D)

$\frac{7}{13}$

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

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

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

$P\left(C\right)=\frac{3}{4}P\left(A\right)$

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

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

$P(A\cup C)=P\left(A\right)+P\left(C\right)$

$=\frac{4}{13}+\frac{3}{4}\times \frac{4}{13}$

$P(A\cup C)=\frac{7}{13}$