gmatprep09 wrote:
If there are 85 students in a statistics class and we assume that there are 365 days in a year, what is the probability that at least two students in the class have the same birthday (assuming birthdays are distributed independently)?
A) (85/365)* (84/364)
B) (1/365)* (1/364)
C) 1- (85!/365!)
D) 1- (365!/ 280! (365^85))
E) 1- (85!/(365^85))
Can someone please explain this in detail....Thanks
As Bunuel already explained:
P(At least two students have same birthday) = 1 - P(At most 0 students have the same birthday)
= 1 - P(All students have different birthdays)
How to choose 85 different birthdays out of 365 days OR choose 85 different days without repetition out of 365?
It is \(P^{365}_{85}=\frac{365!}{(365-85)!}=\frac{365!}{280!}\)
Total possibilities= (365)^85 as every student can choose from 365 days.
P(All students have different birthdays) \(=\frac{365!}{280!*(365)^{85}}\)
P(At least two students have same birthday) = 1 - P(All students have different birthdays)
\(P=1-\frac{365!}{280!*(365)^{85}}\)
Ans: "D"