Question

What is the probability that in a group of three people at least two will have the same birth month? (Assume that all sequences of three birth months are equally likely.) (b) What is the probability that in a group of n people, n ≤ 12 , at least two will have the same birth month? (c) What is the probability that in a group of n people, n > 12 , at least two will have the same birth month? (d) What is the probability that at least two members of the U.S. House of Representatives have the same birth date?

Answer 1

(a) In a group of three people,

P(at least two have same birth month) = P(two have same birth month) + P(all three have same birth month)

= 1 - P(all have different birth month)

Let's find P(all have different birth month) .

Birth month of the first-person lie in any of the 12 month, so the birth month of the second person must fall in the remaining 11 months, and for third, it must be in the remaining 10 months.

P(all have different birth month) = 12/12*11/12*10/12 = 12! *3! / 9! * 3! * (12)^3 = 12C3*3! / (12)^3 = 12P3 / (12)^3

P(at least two have same birth month) = 1 - 12P3 / (12)^3

(b) Similarly proceeding the same way above, it can be shown that in a group of n people n <= 12,

P(at least two have same birth month) = 1 - 12Pn / (12)^n

(c) If number of people exceeds 12, then it is sure that at least two people will have the same birth month. Hence P(at least two have same birth month) = 1.

It can be answered by the pigeon-hole principle. It says that if there are n-pigeonholes and (n+1) pigeons, and we want to put every pigeon in one of the pigeonholes, then there will be at least one pigeonhole with more than one pigeon.

In this case, pigeons will be birth months, and pigeonholes will be months in a year.

(d) In the U.S. House of Representatives, there are 435 members. As here n>12,

P(at least two have same birth month) = 1.