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## Homework Statement

Hi all, could someone give my working a quick skim to see if it checks out? Many thanks in advance.

Suppose that 5 cards are dealt from a 52-card deck. What is the probability of drawing at least two kings given that there is at least one king?

## Homework Equations

## The Attempt at a Solution

Let ##B## denote the event that at least 2 kings are drawn, and ##A## the event that at least 1 king is drawn. Because ##B## is a strict subset of ##A##,

$$P(B|A) = P(A \cap B)/P(A) = P(B)/P(A)$$

Compute ##P(A)##, ##P(A^c )## denotes the probability of not drawing a single king.

$$P(A) = 1 - P(A^c) = 1 - \frac{48 \choose 5}{52 \choose 5} \approx 1 - 0.6588$$

Compute ##P(B)##, ##P(B^c)## denotes the probability of not drawing at least 2 kings, which is the sum of probabilities of drawing 1 king ##P(1)## and the probability of not drawing a single king ##P(A^c)##.

$$P(B) = 1 - P(B^c) = 1 - (P(1) - P(A^c))$$

$$P(1) = \frac{5 \times {4\choose 1} \times 48 \times 47 \times 46 \times 45 }{52 \times 51 \times 50 \times 49 \times 48} \approx 0.299$$

where the numerator is the number of ways one can have a hand of 5 containing a single king.

$$P(B) \approx 1 - (0.299 + 0.6588) \approx 0.0422$$

finally,

$$P(B|A) = P(B)/P(A) = 0.0422 / (1-0.6588) \approx 0.1237$$

I'm pretty sure that there is a quicker way to do all of this even if my work checks out, I'd appreciate if someone could demonstrate a more efficient calculation!