Question

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

Let A, B and C be three mutually exclusive events
given 3P(A) = 2P(B) = P(C)
Since A, B, C are mutually exclusive
⇒ P(A∩B) = P(B∩C) = P(C∩A) = P(A∩B∩C) = 0 and P(A∪B∪C) = 1
∴ P(A∪B∪C) = P(A) + P(B) + P(C) − P(A∩B) − P(B∩C) − P(A∩C) − P(A∩B∩C)
$$\begin{gathered} \mathrm{i}.\mathrm{e}1=P\left(A\right)+\frac{3}{2}P\left(A\right)+3P\left(A\right) \\ \mathrm{i}.\mathrm{e}1=\frac{2P\left(A\right)+3P\left(A\right)+6P\left(A\right)}{2} \\ \mathrm{i}.\mathrm{e}2=11P\left(A\right) \\ \mathrm{i}.\mathrm{e}P\left(A\right)=\frac{2}{11}=\frac{2}{11} \end{gathered}$$
Hence, the correct answer is option B.