Suppose that birthdays are uniformly distributed over the 365 days of the year.

Question

Suppose that birthdays are uniformly distributed over the 365 days of the year. (This is wrong on two counts there are 366 possible birthdays including February 29, and in the U.S. more children are born in the summer than in the winter. But, for simplicity, let’s assume that there are 365 days and each is equally likely.)

(a) Two people are chosen at random. What is the probability that they have different birthdays? What is the probability that they have the same birthday?

(b) Three people are chosen at random. What is the probability that they have different birthdays? What is the probability that at least two of them have the same birthday?

(c) Forty people are chosen at random. What is the probability that they have different birthdays? What is the probability that at least two of them have the same birthday?

Explanation / Answer

a) probability that they have different birthdays =P(first child has birthday on any day and second has brthday except that day)

=(365/365)*(364/365) =364/365 =0.9973

probability that they have the same birthday =1-0.9973 =0.0027

b) probability that they have different birthdays =(365/365)*(364/365)*(363/365) =0.9918

probability that at least two have same birthday =1- 0.9918 =0.0082

c) probability that they have different birthdays =(365/365)*(364/365)*(363/365)....(326/365) =0.1088

probability that at least two have same birthday = 1-0.1088 =0.8912

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