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Question

In tossing a fair coin twice, find the probability of getting exactly one tail


Solution

When a fair coin is tossed twice, $$S = \left \{HH, TT, HT, TH\right \} \therefore n(S) = 4$$
Let $$D$$ be the event of getting exactly one tail
$$D = \left \{HT, TH\right \}\therefore n(D) = 2$$
$$\therefore P(D) = \dfrac {n(D)}{n(S)} = \dfrac {2}{4}$$

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