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Five cards are dealt from a standard deck of 52 cards. What is the probability that three or four of the cards are aces?

I know probabilities very well, but this one is stumping me up. Can you help?

By saying that you want to know the probability of 3 OR 4 aces out of 5 cards, you are asking for the probability of the UNION of the 2 events which we can denote by

3A = 3 cards are aces

4A = 4 cards are aces.

So we want Prob(3A ⋃ 4A). From probability theory, we have

(1) P(3A ⋃ 4A) = P(3A) + P(4A) - P(3A ⋂ 4A).

It should be clear that, in set notation, 4A ⊂ 3A, meaning that the set of 4 aces in a 5-card hand is a subset of having 3 aces in the hand (all 4 aces hands contain 3 aces but not all 3 ace hands have 4 aces). This means that

(2) P(3A ⋂ 4A) = P(4A)

which can be shown in a couple of ways, for instance, using conditional probabilities

P(3A ⋂ 4A) = P(4A)P(3A|4A) where P(3A|4A) = 1 since event 3A has to be true if 4A is true. Using eq. (2) in (1) gives

(3) P(3A ⋃ 4A) = P(3A).

So we only need to calculate P(3A). For this, we have

P(3A) = (# of ways to pick one of 4 aces out of 52 cards)･(one of 3 out of 51)･(2 out of 50)･(any 49 out of 49)･(48/48)･{# of combinations}

= (4/52)･(3/51)･(2/50)･(49/49)･(48/48)･{5!/(3!2!)}

= (24/132,600)･(10) =1.8x10^-3.

Voila!

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