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

Probability question

  1. Mar 5, 2005 #1
    Suppose a room contains n people. Assuming that days of the year are equally likely to be birthdays for each person, calculate the probability that at least two of the people have a common birthday.

    well I have the answer but I'm just curious as to the thought processes you go through to answer it.


    what ive done (incorrectly) is:

    if n = 2, P = 1/365^2
    if n = 3, P = 1/365^2 (person 1, 2) + 1/365^2 (person 1, 3) + 1/365^2 (person 2, 3)

    and generalized it to:
    ( 3 + (n sum i=3) (i)! / (i-1)! ) / 365^2

    there's no question that it's wrong but thats the process I take
     
  2. jcsd
  3. Mar 5, 2005 #2

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    work out the probability they all have different birthdays and subtract that from 1. Obviously if there are 366 people in the room the probabilit must be 1, mustn't it? (excludeing leap year things). Does n=366 give P=1 in your answer?
     
  4. Mar 5, 2005 #3
    probably not (npi), I know my answer is wrong, you recomend using conjugates (1 - P(all different)), I'll work on it some more, thanks for your response
     
  5. Mar 10, 2005 #4
    Ignoring leap years:

    P(bd2 = bd1) = 1/365
    P(bd2 != bd1) = 364/365
    P(bd3 = bd1) + P(bd3 = bd2) = 2/365, given that bd1 != bd2
    P(bd3 != bd1) AND P(bd3 != bd2) = 363/365

    Therefore P(bd1 != bd2 != bd3) = 1 - (363/365) * (364/365)

    So for any n,
    P(two birthdays are the same) =
    1 - (((365-(n-1))*(365-(n-2))...(364))/(365^n))
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Probability question
  1. Probability Question (Replies: 2)

  2. Probability question (Replies: 4)

  3. Probability Question (Replies: 21)

  4. Probability Question (Replies: 3)

  5. Probability questions (Replies: 1)

Loading...