Birthday Probability

[moumi2013]

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!/65!)
D) 1- (365!/(280!*360^85))
E) 1-(85!/365^85)

OA:D

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Hi moumi2013,

What is the source of this question? This is considerably more complex than a typical GMAT probability question.

GMAT assassins aren't born, they're made,
Rich

[GMATGuruNY]

The answer choices should read as I've posted them below:

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��)
E. 1- (85!/365��)

P(at least 2 students have the same birthday) = 1 - (no 2 students have the same birthday).

P(no 2 students have the same birthday):
P(1st student has a viable birthday) = 365/365. (Any day of the year.)
P(2nd student has a birthday not already selected) = 364/365.
P(3rd student has a birthday not already selected) = 363/365.

This process will continue until we reach the 85th student.
Since the first 84 students will account for 84 of the 365 days of year, the number of favorable days for the 85th student = 365 - 84 = 281.
P(85th student has a birthday not already selected) = 281/365.

To combine the probabilities above, we multiply:
365/365 * 364/365 * 363/365 * ....* 283/365 * 282/365 * 281/365

= (365*364*363*....*283*282*281)/365��

= 365!/(280! * 365��).

Thus:
P(at least 2 students have the same birthday) = 1 - 365!/(280! * 365��).

The correct answer is D.