Question

a deck of playing cards contains 52 cards, four of which are aces. (round your answers to four decimal places.) (a) what is the probability that the deal of a five-card hand provides a pair of aces? (b) what is the probability that the deal of a five-card hand provides exactly one ace? (c) what is the probability that the deal of a five-card hand provides no aces? (d) what is the probability that the deal of a five-card hand provides at least one ace?

Answer 1

Answer: a)0.0399, b)0.2995, c)0.6588, d)0.3412

Explanation:

It is the same exact formula as the only other user here made, it's just that their final answer is wrong. Just put it in your calculator (the formulas of the other users) and these are the answers you should be getting

Answer 2

(a) 0.0399

(b) 0.2995

(c) 0.6588

(d) 0.3412

Explanation:

You want the probability distribution in 5-card hands for 2, 1, 0, and not 0 aces.

Probability

The probability of some number of aces is the product of the ways that number of aces can be drawn from the 4 in the deck, multiplied by the number of ways the remaining cards in the hand can be drawn from the 48 non-aces in the deck, all divided by the number of possible 5-card hands.

P(2 aces)

P(2 aces) = 4C2 · 48C3 / 52C5 ≈ 0.0399

P(1 ace)

P(1 ace) = 4C1 · 48C4 / 52C5 ≈ 0.2995

P(0 aces)

P(0 aces) = 48C5 / 52C5 ≈ 0.6588

P(>0 aces)

P(>0 aces) = 1 -P(0 aces) = 1 -0.6588 = 0.3412

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Additional comment

nCk = n!/(k!(n-k)!) . . . the number of ways k can be chosen from n

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