Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Birthdays in the Same Month

  1. Nov 21, 2007 #1
    [SOLVED] Birthdays in the Same Month

    1. The problem statement, all variables and given/known data
    How many people have to be in a room in order that the probability that at least two of them celebrate their birthday in the same month is at least 1/2? Assume that all possible monthly outcomes are equally likely.

    2. Relevant equations
    Axioms and basic theorems of probability.

    3. The attempt at a solution
    Let E_n be the event that at least two of n people in a room celebrate their birthday in the same month. I'm basically asked to determine the value of n for which P(E_n) >= 1/2.

    It must be easier to calculate the complement of E_n, ~E_n, and from that calculate P(E_n). Note that ~E_n is the event that nobody celebrates their birthday in the same month.

    Now, ~E_n is the union of the events that nobody celebrates their birthday in the ith month of the year, which I will call F_i, i = 1 to 12. Since the F_i's are mutually exclusive, P(~E_n) = P(F_1) + ... + P(F_12).

    If month j has d days, then P(F_j) = (365 - d)^n / 365^n. For months 1, 3, 5, 7, 8, 10, 12 d = 31, for months 4, 6, 9, 11 d = 30, and for month 2 d = 28 (assuming no leap years).

    So P(~E_n) = 7 * 334^n / 365^n + 4 * 335^n / 365^n + 337^n / 365^n.

    P(E_n) = 1 - P(~E_n) >= 1/2. I don't know how to determine n analytically, but I did obtain it numerically (using a simple basic program) and got n = 37.

    The answer according to the book is 5. I must have done something wrong, but what?
  2. jcsd
  3. Nov 21, 2007 #2
    Aren't you supposed to assume that monthly outcomes are equally likely? So that you don't have to take february as 28 days and so on.
  4. Nov 21, 2007 #3
    Is that what that means? I always overcomplicate. Sigh.

    I think this way is much easier:

    If there are more than 12 people, then P(E_n) = 1 (by the pigeonhole principle). If there's only one person, then P(E_n) = 0. Thus 1 < n < 13. Assuming the latter, then P(~E_n) = 12 * ... * (12 - n) / 12^n.

    Of course, this doesn't make it easier to calculate P(E_n) analytically. Numerically I get that P(E_4) = 0.427083 and P(E_5) = 0.618056 so n = 5. Sweet!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook