# Five cards are dealt from a standard 52-card deck. What is the probability that we draw (a) 3 aces and 2 kings? (b) a "full house" (3 cards of one kind, 2 cards of another kind)?

**Solution:**

Given, 5 cards are drawn from a standard deck of 52 cards.

a) Probability that we draw 3 aces and 2 kings

Number of cards = 52

No of aces = 4

No of kings = 4

Number of cards drawn = 5

So, the probability becomes,

\(\frac{^{4}C_{3}\: \times \: ^{4}C_{2}}{^{52}C_{5}}\)

On simplification,

= 24 / 2598960

= \(9.23\times 10^{-6}\)

So the probability of drawing 3 aces and 2 kings is \(9.23\times 10^{-6}\)

b) a “full house” (3 cards of one kind and 2 cards of another kind)

Since there are 13 sets of each type, we have to select any 2 kinds of it.

The probability becomes \(\frac{2\times \, ^{13}C_{2}\: \times \, ^{4}C_{3}\, \times \, ^{4}C_{2}}{^{52}C_{5}}\)

On simplification,

= 3744 / 2598960

= 0.00144

So the probability of a “full house” is 0.00144.

Therefore, the probability of drawing 3 aces and 2 kings is \(9.23\times 10^{-6}\) and the probability of a “full house” is 0.00144.

## Five cards are dealt from a standard 52-card deck. What is the probability that we draw (a) 3 aces and 2 kings? (b) a "full house" (3 cards of one kind, 2 cards of another kind)?

**Summary:**

Five cards are dealt from a standard 52-card deck. The probability that we draw (a) 3 aces and 2 kings is \(9.23\times 10^{-6}\) (b) a "full house" (3 cards of one kind, 2 cards of another kind) is 0.00144.

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