what is the probability that at least 2 students in a class of 36 have the same birthday?
Do i punch this in to the calculator or how did you found this way or method for your solution:
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
THAT IS THE SOLUTION BUT I WANT TO UNDERSTAND THE STEPS FOR IT...CAN SOMEONE EXPLAIN THEM TO ME PLEASE...
Start with any student, he/she obviously will have a birthday (365/365 or probability is 1)
Now conider the second student. The probability that he/she will have a DIFFERENT birthday is (365/365)(364/365).
Now the third student's probability of having a DIFFERENT birthday is
(365/365)(364/365)(363/365)
Continuing this argument until we reach the last student to have a different birthday from all the rest we have
(329/365)(330/365)....(364/365)(365/365)
This is the probability that everybody will have a DIFFERENT birthday, so the the probability of NOT DIFFERENT, or at least somebody having the same birthday is 1 - the above multiplication.
I will let you finish the actual calculation.
yha um can you help me with my math
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