The Birthday Problem; Queueing at a BankDate: 08/26/99 at 09:52:06 From: Sophia Subject: Probability I am having an awful time figuring out probability! Can you please help? 1. In a gathering of 23 people, what is the probability that at least two people have the same birthday? Assume a 365-day year. I figure that I have (365)^23 possible outcomes. But the problem I am having trouble with is the numerator. I don't know how to say that _at least_ two people have the same birthday. Any help? 2. Forty customers at a bank are waiting for the doors to open, whereupon they will each choose, at random, one of five tellers (and after choosing they do not change lines). If no more customers after the forty arrive, then... a) What is the probability that teller #2 has no customers? I _think_ I may have this one. I think the total possible outcomes is (5)^40, and the possibility that teller #2 is not selected by everyone is (4/5)^40. So I think the answer is: [(4/5)^40]/[(5)^40]. Am I right? But I am stuck and really confused by the others... b) What is the most likely number of customers that teller #3 is to serve? I am guessing that I need to find out which number between 0-40 is most likely to occur in a teller line. But I don't know how to do that. c) What is the probability that every teller has at least one customer? d) What is the probability that exactly two tellers do not have any customers? I know it seems like I haven't made much progress, and I guess I haven't. But I have really been racking my brain on this. I just need a good explanation of how to think about these problems. I would be so grateful for any help! Thanks, Sophia Date: 08/26/99 at 14:41:38 From: Doctor Anthony Subject: Re: Probability >1. In a gathering of 23 people, what is the probability that at least > two people have the same birthday? Assume a 365-day year. They chose 23 people because that is the number above which there is a better than 50% probability of at least two sharing a birthday. If no two share a birthday then the first person can choose a birthday in 365 ways, the second person in 364 ways, the third in 363 ways, and so on. If there are n people in the room, the probability that no two share a birthday is 365 x 364 x 363 x ... x (365-n +1) P(365,n) ---------------------------------- = -------- 365^n 365^n So the probability that two at least share a birthday is: P(365,n) 1 - -------- 365^n P(365,n) If this must be >= 0.5 then we require -------- <= 0.5 365^n (Here, P(365,n) is the number of permutations of n things that can be made from 365 different things) There is no easy way to solve the equation for n, but with a calculator which gives permutations and combinations, trial and error will quickly establish that n must be around 23. P(365,23) --------- = 0.4927 365^23 and 1 - 0.4927 = 0.5073 and so 23 people are sufficient to make the probability of a shared birthday greater than 50%. >2. Forty customers at a bank are waiting for the doors to open... >a) What is the probability that teller #2 has no customers? You can think of this as binomial probability. For any teller the probability of 'success' with a particular customer is 1/5. So n = 40, p = 1/5, and q = 4/5 P(0) = (4/5)^40 = 0.0001329 >b) What is the most likely number of customers that teller #3 is to > serve? The mean number at any teller is np = 40 x 1/5 = 8 and the most likely number at any teller MUST lie within 1 of the mean. So we test P(7), P(8), P(9) and find which of these is greatest. P(7) = C(40,7)(1/5)^7 (4/5)^33 = 0.15125 P(8) = C(40,8)(1/5)^8 (4/5)^32 = 0.15598 P(9) = C(40,9)(1/5)^9 (4/5)^31 = 0.13865 and so P(8) is the greatest probability and we conclude that the most likely number of customers at ANY position is 8. >c) What is the probability that every teller has at least one > customer? This is equivalent to distributing 40 numbered balls into 5 numbered urns such that no urn is empty, and we require the coefficient of x^40/40! in the expansion of [e^x - 1]^5 = e^5x - 5e^4x + 10e^3x - 10e^2x + 5e^x - 1 and the coefficient of x^40/40! is (x^40/40!).[5^40 - 5.(4^40) + 10.(3^40) - 10.(2^40) + 5] = (x^40/40!).[9.08890251 x 10^27] 9.08890251 x 10^27 and the required probability = ------------------- 5^40 = 0.999335 >d) What is the probability that exactly two tellers do not have any > customers? You can choose 2 tellers from 5 in C(5,2) = 10 ways You then place 40 balls at random into 3 cells so that no cell is empty. Again we must deal with equiprobable events and the number of ways is given by the coefficient of x^40/40! in the expansion of [e^x - 1]^3 = e^3x - 3.e^2x + 3.e^x - 1 The term in x^40/40! is (x^40/40!).[3^40 - 3(2^40) + 3] = (x^40/40!).[1.2157662 x 10^19] 10 x 1.2157662 x 10^19 and the required probability = ----------------------- 5^40 = 1.33675 x 10^-8 - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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