Question

If A, B, C are three mutually exclusive and exhaustive events of an experiment such that 3P(A) = 2P(B) = P(C), then P(A) is equal to

• A. $\frac{1}{11}$
• B. $\frac{2}{11}$
• C. $\frac{5}{11}$
• D. $\frac{6}{11}$

Answer 1

The correct option is B

$\frac{2}{11}$

Let 3 P(A) = 2P(B) = P(C) = p. Then,

P(A) = $\frac{p}{3}$, P(B) = $\frac{p}{2}$ and P(C) = p

It is given that A, B, C are three mutually exclusive and exhaustive events.

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

$\left[P\right(A\cap B)=P(B\cap C)=P(C\cap A)=P(A\cap B\cap C)=0andP(A\cup B\cup C)=1]$

$\Rightarrow \frac{p}{3}+\frac{p}{2}+p=1$

$\Rightarrow \frac{11p}{6}=1$

$\Rightarrow p=\frac{6}{11}$

$\therefore P\left(A\right)=\frac{p}{3}=\frac{\frac{6}{11}}{3}=\frac{2}{11}$

Hence, the correct answer is option (b).