Two fair coins are flipped. As a result of this, tails and heads runs occurred where a tail run is a consecutive occurrence of at least one head. Determine the probability function of number of tail runs.(a) \(\frac{1}{2}\)(b) \(\frac{5}{6}\)(c) \(\frac{32}{19}\)(d) \(\frac{6}{73}\)The question was posed to me during an internship interview.This interesting question is from Probability Distribution topic in section Discrete Probability of Discrete Mathematics

Two fair coins are flipped. As a result of this, tails and heads runs

1.	Two fair coins are flipped. As a result of this, tails and heads runs occurred where a tail run is a consecutive occurrence of at least one head. Determine the probability function of number of tail runs.(a) \(\frac{1}{2}\)(b) \(\frac{5}{6}\)(c) \(\frac{32}{19}\)(d) \(\frac{6}{73}\)The question was posed to me during an internship interview.This interesting question is from Probability Distribution topic in section Discrete Probability of Discrete Mathematics
Answer» Correct answer is (a) \(\frac{1}{2}\) For explanation: The sample space of the EXPERIMENT is S = {HH, HT, TH, TT}. Let X is the number of tails and It takes up the VALUES 0, 1 and 2. Now,P(no TAIL) = p(0) = \(\frac{1}{4}\), P(one tail) = p(1) = \(\frac{2}{4}\) and P(two tails) = p(2) = \(\frac{1}{4}\). So, X is the number of tail runs and it takes up the values 0 and 1. P(X = 0) = p(0) = \(\frac{2}{4} = \frac{1}{4}\).

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