1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Avodyne

    User Avatar
    Science Advisor

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Another Birthday Probability Problem
  1. Another problem (Replies: 2)

Loading...