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All the jacks, queens, kings, and aces of a regular 52 cards deck are taken out. The 16 cards are thoroughly shuffled and my opponent, a person who always tells the truth, simultaneously draws two cards at random and says, “I hold at least one ace”. The probability that he holds two aces is

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2/8

4/9

2/3

1/9

answer is D.

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Detailed Solution

A: exactly one ace

B: both aces

$\begin{array}{l} E : A \cup B \\ P (B / A \cup B) = \frac{^{4} C_{2}}{^{4} {C_{1}}^{12} C_{1} +^{4} C_{2}} = \frac{6}{54} = \frac{1}{9} \end{array}$

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