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Question -

Consider the following-

  1. It is equi-probable to have a birthday in any month of the year.
  2. 6 random people are put into a room.

Q1. Find the probability that at least two people have a birthday in the same month.

Q2. Find the probability that three of the people have birthdays in one month, and the other three have birthdays in another month.

My Approach -

A1. 77.71%

P( all 6 people have different birthday months) = (12 * 11 * 10 * 9 * 8 * 7) / (12)^6 = 665280 / 2985984 = 385 / 1728

P( at least 2 people have the same birthday month) = 1 - P( all 6 people have different birthday months) = 1 - 385 / 1728 = 1343 / 1728 = 0.7771

A2. 12.96%

P( at least one triple birthday month)≈1 − exp(−C(6,3) / (12)^2 )= 0.1296

I am unsure about my answers and my approach, especially for Q2. Any and all help will be extremely helpful.

Solved with reference to this post - Probability of 3 people in a room of 30 having the same birthday

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  • $\begingroup$ Do you want an exact answer, or good upper/lower bounds? $\endgroup$
    – user432944
    Commented Jun 29, 2021 at 2:13
  • $\begingroup$ The formula you used for the triple birthday is an approximation; a much better approximation is $0.1141.$ But one triple birthday is not good enough; you need two triple birthdays, which is much less likely. $\endgroup$
    – David K
    Commented Jun 29, 2021 at 2:23
  • $\begingroup$ I don't exactly know how to tackle this Q2. Would appreciate any help in figuring out the solution. $\endgroup$
    – Raul Martz
    Commented Jun 29, 2021 at 3:08

1 Answer 1

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For Q2,

You want to

  • split the six people into a group of three to share a birthday-month, leaving another group of three to share another month,
  • choose two of the twelve months to be those months, putting the first chosen group into the earlier of the months, and
  • and then divide by the total number of equally likely cases, so something like $$\dfrac{{6 \choose 3}{12 \choose 2}}{12^6} \approx 0.000442$$
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  • $\begingroup$ I think the final $\binom21$ is wrong here. When you count $\binom 63$ you already get two distinguishable groups -- namely the 3 people you chose and the 3 you didn't -- so we can just decide to assign the 3 people we chose to the numerically smallest of the two chosen months, and get all the possibilities that way. $\endgroup$
    – Troposphere
    Commented Jul 10, 2021 at 16:00
  • $\begingroup$ @Troposphere That is a fair point - edited - thank you $\endgroup$
    – Henry
    Commented Jul 10, 2021 at 16:25

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