1.

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 toA.  \(\frac{1}{11}\)B.  \(\frac{2}{11}\)C. \(\frac{5}{11}\)D. \(\frac{6}{11}\)

Answer»

3P(A) = 2P(B) = P(C) = k(say) 

∴ P(A) = \(\frac{k}{3}\)

P(B) = \(\frac{k}{2}\)

And P(C) = k 

As events A,B and C are mutually exclusive and exhaustive, 

∴ P(A ∪ B ∪ C) = P(A) + P(B) + P(C) = 1 

⇒ 1 = \(\frac{k}{3} + \frac{k}{2} + {k}\)

⇒ 1 = \(\frac{2k+3k+6k}{6}\) = \(\frac{11k}{6}\) 

⇒ k = \(\frac{6}{11}\) 

As, P(A) =\( \frac{k}{3}\) = \(\frac{1}{3}\times\frac{6}{11}\) = \(\frac{2}{11}\)

Our answer matches with option (b) 

∴ Option (b) is the only correct choice 


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