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
To calculate the probability that at least 2 students in a class of 36 have the same birthday, we can use the concept of complementary probability.
First, let's calculate the probability that all students have different birthdays:
The total number of ways to assign birthdays to the students is 365^36 because each student can have any of the 365 possible birthdays.
The number of ways to assign different birthdays to the students is given by 365!/(365-36)!, which is equivalent to 365x364x...x330.
Therefore, the probability that all students have different birthdays is given by (365!/[(365-36)! x 365^36]).
To find the probability that at least 2 students share the same birthday, we subtract this probability from 1 because it is the complementary event.
So, the desired probability is 1 - (365!/[(365-36)! x 365^36]).
To numerically evaluate this expression, you can use Stirling's formula:
Log(N!) = N Log (N) - N + 1/2 Log(2 pi N) + 1/(12 N) + ...
Using Stirling's formula, calculate Log(365!) and Log(329!). Also, calculate 36 Log(365) to simplify the expression.
Then, substitute these values into the expression: Exp[1803.9383 - 1581.7202 - 212.3963] = Exp[-1.7849].
Finally, evaluate the exponential value, which is approximately 0.16781.
Therefore, the probability that at least 2 students in a class of 36 have the same birthday is approximately 1 - 0.16781 = 0.83219.