elemprob-page5 - one of, so the answer is 4 4 13 · 12 4 1...

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Unformatted text preview: one of, so the answer is 4 4 13 · 12 4 1 . 52 5 Example. What is the probability that in a poker hand we get exactly 3 of a kind (and he other two cards are of different ranks)? Answer. The probability of 3 aces, 1 king and 1 queen is the rank we have 3 of and 12 2 4 3 4 1 4 1 52 . We have 13 choices for 5 choices for the other two ranks, so the answer is 12 13 2 4 3 4 1 52 5 4 1 . Example. In a class of 30 people, what is the probability everyone has a different birthday? (We assume each day is equally likely.) Answer. Let the first person have a birthday on some day. The probability that the second person has a different birthday will be 364 . The probability that the third person has a different birthday from the first 365 two people is 363 . So the answer is 365 364 363 336 1− ··· . 365 365 365 Example. Suppose 10 people put a key into a hat and then withdraw one randomly. What is the probability at least one person gets his/her own key? Answer. If Ei is the event that the ith person gets his/her own key, we want P(∪10 Ei ). One can show, i=1 either from a picture or an induction proof, that P(∪10 Ei ) = i=1 P(Ei1 ) − i1 P(Ei1 ∩ Ei2 ) + i1 <i2 P(Ei1 ∩ Ei2 ∩ Ei3 ) − · · · . i1 <i2 <i3 Now the probability that at least the 1st, 3rd, 5th, and 7th person gets his or her own key is the number of ways the 2nd, 4th, 6th, 8th, 9th, and 10th person can choose a key out of 6, namely 6!, divided by the number of ways 10 people can each choose a key, namely 10!. so P(E1 ∩ E3 ∩ E5 ∩ E7 ) = 6!/10!. There are 10 ways of selecting 4 people to have their own key out of 10, so 4 P(Ei1 ∩ Ei2 ∩ Ei3 ∩ Ei4 ) = i1 ,i2 ,i3 ,i4 10 4 6! 1 = 10! 4! The other terms are similar, and the answer is 1 1 1 1 − + − ··· − ≈ 1 − e−1 . 1! 2! 3! 10! 3. Conditional probability. 5 ...
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