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June 2020 p53 q5
3101
A pair of fair coins is thrown repeatedly until a pair of tails is obtained. The random variable X denotes the number of throws required to obtain a pair of tails.
(a) Find the expected value of X. [1]
(b) Find the probability that exactly 3 throws are required to obtain a pair of tails. [1]
(c) Find the probability that fewer than 6 throws are required to obtain a pair of tails. [2]
Solution
(a) The probability of getting tails on both coins in one throw is \(\frac{1}{4}\). The expected number of throws to get a pair of tails is the reciprocal of this probability, which is \(\frac{1}{\frac{1}{4}} = 4\).
(b) The probability of getting exactly 3 throws to obtain a pair of tails is calculated by getting not tails in the first two throws and tails in the third throw. The probability is \(\left( \frac{3}{4} \right)^2 \times \frac{1}{4} = \frac{9}{64}\).
(c) The probability that fewer than 6 throws are required is the complement of the probability that 6 or more throws are required. The probability of not getting tails in the first 5 throws is \(\left( \frac{3}{4} \right)^5\). Therefore, the probability of fewer than 6 throws is \(1 - \left( \frac{3}{4} \right)^5 = 0.763\).
