Given:A coin is biased so that a head is twice as likely to occur as a tail.The number of times the coin is tossed = 3Concept used:The sum of the probability of happening all events is 1.Calculation:P(H) + P(T) = 1 According to the question,? 2 × P(T) + P(T) = 1? 3 × P(T) = 1? P(T) = \(\frac{1}{3}\)Let the probability of getting a tail = \(\frac{1}{3}\)The probability of getting a head = \(\frac{2}{3}\)The sample space for two tails and one head.S = {(T, T, H), (T, H, T), (H, T, T)}For case 1: (T, T, H)The required probability, PTTH = \(\frac{1}{3} × \frac{1}{3} × \frac{2}{3} = \frac{2}{27} \)For case 2: (T, H, T)The required probability, PTHT = \(\frac{1}{3} × \frac{2}{3} × \frac{1}{3} = \frac{2}{27} \)For case 3: (H, T, T)The required probability, PHTT = \(\frac{2}{3} × \frac{1}{3} × \frac{1}{3} = \frac{2}{27} \)The probability of getting exactly two tails = 3 ×\(\frac{2}{27}\) = \(\frac{2}{9}\)? The probability of getting exactly two tails is \(\frac{2}{9}\).