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[QUOTE="WWCY, post: 6048385, member: 608494"] [h2]Homework Statement [/h2] 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?[h2]Homework Equations[/h2] [h2]The Attempt at a Solution[/h2] 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! [/QUOTE]
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