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 

Options

•  \[\frac{1}{11}\]

   

•  \[\frac{2}{11}\]

   

•   \[\frac{5}{11}\]

•  \[\frac{6}{11}\]

   

Answer 1

Let 3 P(A) = 2 P(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.

∴ P(A) + P(B) + P(C) = 1                               [P(A ∩ B) = P(B ∩ C) = P(C ∩ A) = P(A ∩ B ∩ C) = 0 and P(A ∪ B ∪ C) = 1]

\[\Rightarrow \frac{p}{3} + \frac{p}{2} + p = 1\]
\[ \Rightarrow \frac{11p}{6} = 1\]
\[ \Rightarrow p = \frac{6}{11}\]

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