I need help or hint on the setup:
what is the probability that at least 2 students in a class of 36 have the same birthday?
This is 1 minus the probability that all students have different birthdays. Suppose that the birthday of a student is completely random. Then we can evaluate the probability by counting in how many ways we can assign different birthdays to the 36 students and divide that by the total number of ways a birthday can be assigned to a student.
Clearly, the number of ways you can assign different birthdays to the students is:
365!/(365-36)!
This is:
365*364*...*330
For the first student you are assigning a birthday to, you have 365 choices, for the second there are 364 choices left, etc.
The total number of ways bithdays can be assigned is, of course:
365^36
The probability that all students have different birthdays is thus:
365!/[(365-36)! 365^36]
The probaility that this is not the case is therefore:
1 - 365!/[(365-36)! 365^36]
If it is not the case that all students have different birthdays then at least two students must have the same birthday, so this is the desired probability.
You can use Stirling's formula to numerically evaluate the expression:
Log(N!) = N Log (N) - N
+ 1/2 Log(2 pi N) + 1/(12 N) + ...
Log(365!) = 1792.3316
Log(329!) = 1581.7202
36 Log(365) = 212.3963
And therefore:
365!/[(365-36)! 365^36] =
Exp[1803.9383 - 1581.7202 - 212.3963] =
Exp[-1.7849] = 0.16781
1 - 0.16781 = 0.83219
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