# Can you try your hands on this probability ques?

This question entails proper construction of the total and favorable number of cases. Here you assign 1 day to each person. For the number of cases you should consider that in how many ways each of the 4 persons may get a day from a total of 365 days. The favorable number of cases is obtained by sequentially eliminating a day from the year as you go on assigning birthdays to each person. Q. A company has four employees. find the probability that a. All of them have different birthdays. b. Exactly 2 of them have the same birthday. c. At least 2 of them have the same birthday.

a. 365 X 364 X 363 X 362 b. 365 X 364 X 363 c. 365 ^ 3 (?)

What are the probs? a) 365 X 364 X 363 X 362 / 365x365x365x365 ???

yes.

My guess is, a) All different birthdays - (365/365)(364/365)(363/365)(362/365) b) Exactly 2 have same birthday - 4C2 (365/365)(1/365)(364/365)(363/365) The Combination is required to identify which two share the same birthday Viz. A&B or A&C or B&C… etc c) Atleast 2 have same birthday is equal to (1- P(none having same birthday)) = (1 - (365/365)(364/365)(363/365)(362/365)) Can you post the answer for this?