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Question
A pack of cards contains 4 aces, 4 kings, 4 queens and 4 jacks. Two cards are drawn at random. The probability that at least one of them is an ace is
Options
1/5
3/16
9/20
1/9
Solution
\[\frac{9}{20}\] We have:
P(both are aces) = \[\frac{{}^4 C_2}{{}^{16} C_2} =\]
\[\frac{4}{16} \times \frac{3}{15} = \frac{1}{20}\] P(one is ace) = \[\frac{{}^4 C_1 \times^{12} C_1}{C {{}^{16}}_2} = \frac{2}{5}\]
∴ P(at least one is ace) = \[\frac{1}{20} + \frac{2}{5} = \frac{9}{20}\]
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