Let ‘A’, ‘B’ and ‘C’ be three independent events with P(A)=13,P(B)=12 and P(C)=14. The probability of exactly 2 of these events occurring, is equal to:
A
14
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B
724
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C
34
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D
1724
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Solution
The correct option is A14 P (exactly two of A,B and C) = P (AB) + P (BC) + P (CA) - 3P (ABC) Since A, B, C are independent events, required probability = P(A) P(B) + P (B) P (C) + P(C) P (A) -3P (A) P(B) P(C) =13×12+12×14+14×13−3×13×12×14=624=14