A biased coin is tossed repeatedly until a tail appears for the first time. The head is 3 times likely to appear as the tail. The probability that the number of tosses required will be more than 4 , given that in first two toss no tail has occur, is

Step-by-step solution

P H = 3 4 , P T = 1 4 Let, x be the number of tosses required P x ≥ 5 = 3 4 4 1 4 + 3 4 5 1 4 + 3 4 6 1 4 + . … = 3 4 4 1 4 1 - 3 4 = 3 4 4 P x ≥ 3 = 3 4 2 1 4 + 3 4 3 1 4 + 3 4 4 1 4 + . … = 3 4 2 1 4 1 - 3 4 = 3 4 2 P x ≥ 5 x ≥ 3 = P x ≥ 5 ∩ x ≥ 3 P x ≥ 3 = P x ≥ 5 P x ≥ 3 = 3 4 4 3 4 2 = 9 16