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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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 Spring '09
 wen
 Probability

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