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Sid draws a card from a pack of cards, replaces it, shuffles the pack and then draws another card. What is the probability that the cards are both aces?

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P(ace) = \(\frac{4}{52}\) = \(\frac{1}{13}\) (\(\because\) there are 4 aces in a pack of 52 cards)

\(\therefore\) P(both aces) = P(first ace and second ace) = \(\frac{1}{13}\) x \(\frac{1}{13}\)\(\frac{1}{169}\)

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