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Homework Help: Another Birthday Probability Problem

  1. Nov 21, 2007 #1
    [SOLVED] Another Birthday Probability Problem

    1. The problem statement, all variables and given/known data
    Given 20 people, what is the probability that among the 12 months in the year there are 4 months containing exactly 2 birthdays and 4 containing exactly three birthdays.

    2. Relevant equations
    Axioms and basic theorems of probability.

    3. The attempt at a solution
    The sample space of this problem is a set of 20-tuples (n_1, ..., n_20) where n_i is an integer from 1 to 12 representing the month person i was born in. The sample space has u = 12^20 elements.

    Divide the people into 8 distinct groups, 4 of 2 and 4 of 3. This can be done in

    [tex]v = \frac{20!}{2!^4 3!^4}[/tex]

    ways. Now each of these 8 groups needs to be assigned to a distinct month. Pick any ordering of the groups. The first group has 12 months to choose from, the second has 11 months to choose from, etc. There are thus w = 12 * 11 * ... * 5 = 12!/4! assignments.

    The probability sought is vw/u. Unfortunately, when I calculate this I get a different answer than the book's 1.0604 x 10^-3. Where could I have gone wrong?
  2. jcsd
  3. Nov 21, 2007 #2


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    Your counting of the ways to assign the groups to months is not correct. Four months get assigned 4 people each (call these months "red"), four get assigned 2 people each ("blue"), and four get assigned 0 people each ("green"). How many distinct ways are there to "color" the 12 months?

    When I use this count for w, I get your book's answer.
  4. Nov 21, 2007 #3
    OK. I know what the problem with my counting is: In v, I'm counting permutations of possible groups. For example, if A, B, C, D are four 3-person groups, v is counting all possible 4-permutations of those groups as being distinct. v should actually be

    [tex]\frac{20!}{2!^4 3!^4 4!^2}[/tex]

    and that will produce the correct answer.

    If I leave v unchanged however and consider the orderings of the groups, then I would have to ignore certain group-month assignments. For example, if groups A, B, C, D are assigned to months 1, 2, 3, 4, then I would have to ignore the assignment of D, C, B, A to months 4, 3, 2, 1 which I'm incorrectly counting in w. To ignore these, w needs to equal C(12,4) * C(8,4). That will produce the correct answer.

    Thanks for the pointer Avodyne. I will mark this thread as solved now.
  5. Dec 9, 2011 #4
    Re: [SOLVED] Another Birthday Probability Problem

    Sorry, I am new to these....But I have a quick question..
    If two people from the same group (same month) is picked and switched.
    That would be considered as another situation in u isnt? because u have order
    But that would not count on v? because v is combination?
    Then how can this be true?

    THanks for any advice
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