Question

If m students born on independent days in 1991 are attending a lecture, show that the probability that at least two of them share a birthday is p = 1 - (365) !/{(365 - m) ! 365m }. Show that p > 1 when m = 23 .

Answer 1

1991 is a non-leap year. The number of days in 1991 is 365.

Assuming m < 365 and if all the students have different birthdays, number of possible ways = ³⁶⁵P_m

= (365)! / {(365 - m)! }

Number of possible birthdays of m students = 365^m    (Each student can have 365 different birthdays)

Probability that  all the students have different birthdays =

= Number of possible ways such that all the students have different birthdays / Total possible birthdays of m students

= (365)! / {(365 - m)!} / 365^m

= (365)! / {(365 - m)! 365^m }

Probability that at least two of them share a birthday = 1 - Probability that  all the students have different birthdays  

= 1 - (365)! / {(365 - m)! 365^m }

When m = 23,

Probability that at least two of them share a birthday = 1 - Probability that  all the students have different birthdays  

= 1 - (365)! / {(365 - 23)! 365²³ }

= 1 -  (365)! / {(342)! 365²³ }

= 1 - {(365 * 364 * ... * 342) / 365²³ }

= 1 - 1.9 x 10^-23

1

Thus, for m = 23, p > 1/2