A bag contains (N+1) coins - N fair coins, and one coin with 'Head' on both sides. A coin is selected at random and tossed. If the probability of getting 'Head' is 9 16 , then N is equal to:

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Correct answer: B. 7

Let E_1 be the event of selecting a fair coin and E_2 be the event of selecting the two-headed coin. P(E_1) = N N+1 P(E_2) = 1 N+1 Let H be the event of getting a Head. P(H|E_1) = 1 2 P(H|E_2) = 1 Using the theorem of total probability: P(H) = P(E_1)P(H|E_1) + P(E_2)P(H|E_2) P(H) = ( N N+1 ) ( 1 2 ) + ( 1 N+1 )(1) P(H) = N+2 2(N+1) Given P(H) = 9 16 : N+2 2(N+1) = 9 16 N+2 N+1 = 9 8 8N + 16 = 9N + 9 N = 7 Answer: 7

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