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It has been found that if A and B play a game 12 times, A wins 6 times, B wins 4 times and they draw twice. A and B take part in a series of 3 games. The probability that they will win alternately is A. 5 72 B. 5 36 C. 19 27 D. none of these Select the correct answer from above options

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Correct Answer - B Let E be the event that A wins the game and F be the event that B wins the game. Then, P(E)= 6 12 = 1 2 and P(F)= 4 12 = 1 3 ∴ Required probability =P{(E∩F∩E)∪(F∩E∩F)} =P(E∩F∩E)+P(F∩E∩F) =P(E)P(F)P(E)+P(F)P(E)P(F) = 1 2 × 1 3 × 1 2 + 1 3 × 1 2 + 1 3 × 1 2 × 1 3 = 5 36

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