Three fair coins are tossed. If both heads and tails appear, determine the probability that exactly one head appears.

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Solution

Three coins are tossed and hence the size of universal set is $2^{3} = 8$ . The number of outcomes where both heads and tails occur are 6 (event a) i.e $[H H T], H T H, [T H H], [T T H], [T H T], [H T T]$ .

The number of outcomes where exactly one head appears is (event b) $[T T H], [T H T], [H T T]$ . If both heads and tails appear, probability that one head appears is

$P (\frac{b}{a}) = \frac{P (b \cap a)}{P (a)} = \frac{\frac{3}{8}}{\frac{6}{8}} = \frac{1}{2}$

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