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

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a

13

b

25

c

38

d

611

answer is D.

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Detailed Solution

A∪B∪C=S, A∩B=B∩C=C∩A=ϕ P(A)+P(B)+P(C)=1 P(A)=2P(B)=3P(C)=k ⇒ k+k2+k3 =1 ⇒ k= 611 ∴ P(A)=611

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