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What is the probability that the ace of spades is at the bottom of a standard deck of 52 cards given that the ace of hearts is not at the top?

I asked my older brother, and he said it should be $\frac{50}{51} \cdot \frac{1}{51}$ because that's $$\mathbb{P}(A\heartsuit \text{ not at top}) \times \mathbb{P}(A\spadesuit \text{ at bottom}),$$ but I'm not sure if I agree. Shouldn't the $\frac{50}{51}$ be $\frac{50}{52}$?

Thanks you!

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    The number is correct but the reason isn't (or is unclear). It may have been a misprint. I would say it is $(50/51)(1/51)$=(probability that the ace of hearts is not at the bottom)$\times$(probability that the ace is at the bottom, out of the 51 non-ace-of-hearts positions).2012-01-09

3 Answers 3