Sharing a Zodiac SignDate: 04/24/2001 at 23:57:15 From: Katrin Subject: Probability In a class of 35 students what is the probability that every student in the class has the same zodiac sign as at least one other classmate? Date: 04/26/2001 at 17:01:23 From: Doctor Mitteldorf Subject: Re: Probability Dear Katrin, This is an advanced question, involving the "inclusion-exclusion principle" for its solution. Fasten your seatbelt... To begin with, we'll assume that all 12 signs are equally likely, even though we know in fact that more babies are born in the spring than in the fall. We'll calculate the probability of at least one sign having exactly one student "stranded" in it. (Not 0, not 2, but exactly 1.) Then we'll subtract that number from unity to get the answer they're asking for. One more preliminary: the number of ways to distribute 35 students in 12 signs is 12^35. This is our denominator, the universe of all possibilities. Now, the first step: How many ways are there to have NO students at all in Capricorn? Answer: There are 11 possible placements for 35 students, so the count is 11^35. The probability of this happening is 11^35/[denominator], which is (11/12)^35 = 0.0476. We didn't really need that number, but we're building toward the next calculation, which we do need: How many ways are there to distribute the students with EXACTLY ONE person in Capricorn? The number of ways of distributing the other 34 students among 11 months is 11^34. Multiply this by 35, because there are 35 different students who could be stranded in Capricorn. So our numerator is 35*11^34, and (dividing by the same denominator) we calculate a probability 0.1514. That's the probability that there's exactly one student in Capricorn. But there are 11 other signs. Each of them has the same probability of stranding exactly one student. So we calculate a probability of 12*0.1514, which is 1.816. But how did we end up with a number bigger than 1 for a probability? The catch is that we've double-counted some possibilities. There are some cases in which there's one student stranded in Capricorn AND one student stranded in Aries. We counted those cases twice, once for Capricorn and once for Aries. We should only have counted them once. So we need to correct our answer by subtracting the number of cases in which there are two months, each stranding a student. There are 35 students who could be stranded in Capricorn and 34 left for Aries, for a factor of 35*34 possibilities; the other 33 students could be distributed among the other 10 signs in 10^33 ways. So the set of possibilities we've overcounted has 35*34*10^33 members for each pair of months, and there are 12*11/2 = 66 pairs of months. We've overcounted by 66*35*34*10^33, which, when divided by our denominator, yields 1.3297. We're down to 0.4869 for our probability, which is a reasonable number since it's less than one. Done? Not so fast. Now we've overcounted the overcount. The reason is that there are some possibilities in which THREE signs have stranded students. We need to count these possibilites and add them back. 35 possible students for Capricorn * 34 for Aries * 33 for Sagittarius; the other students could be distributed in 9^32 different ways; there are 12*11*10/6 = 220 different triples of months. So we've overcounted the overcount by 220*35*34*33*9^32 which, when divided by our denominator, yields 0.5022. Adding to 0.4869, we have .9891. By now you know that we're going to have to correct the correction of the correction as well, subtracting and adding and subtracting and adding all the way down to eleven-plets. I've done this on a computer, and for a final answer I get 0.896. You can read more about inclusion-exclusion calculations in these answers from our archives: Probability of Matching Envelopes and Letters http://mathforum.org/dr.math/problems/dey3.3.98.html Letters and Envelopes and the Inclusion-Exclusion Principle http://mathforum.org/dr.math/problems/jarrad03.27.99.html - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/ |
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