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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)?

Correct Answer:

**D**P(at least two students with the same birthday) = 1 – P(all birthdays are unique)

- Finding P(at least two students with the same birthday) directly will be extremely difficult. Consequently, we must look for a different way. If we find the complement of P(at least two students with the same birthday), we can find P(at least two students with the same birthday) since it will be equal to
*1 - P(complement)*.

P(A) = 1 - P(complement of A)

P(at least two students with the same birthday) = 1 – P(all birthdays are unique) - Find P(all birthdays are unique).
- If every birthday is unique, then with 85 students, there are 85 different days on which a birthday occurs.

Let N = Total Students Selected After Student S Selected

Let M = Total Students Selected Before Student S Selected

Let U = Unique Days of Year Remaining Before Student S Selected on Which Future Students Can Have Unique Birthday

Let P = P(Unique Birthday)

Let O = Students Remaining After Student S Selected

You can draw a table to help solve this problem.

N M U P O 1 0 365 = 365 - 0 365/365 84 = 85 - 1 2 1 364 = 365 - 1 364/365 83 = 85 - 2 3 2 363 = 365 - 2 363/365 82 = 85 - 3 4 3 362 = 365 - 3 362/365 81 = 85 - 4 83 82 283 = 365 - 82 283/365 2 = 85 - 83 84 83 282 = 365 - 83 282/365 1 = 85 - 84 85 84 281 = 365 - 84 281/365 0 = 85 - 85 =365 - M =U/365 =85 - N - For P(Unique Birthday with 1 student already selected [i.e., the M=1 row]): Since there are 365 possible days of the year yet only 364 days will satisfy the condition of all birthdays being unique since 1 day already has a birthday occupying it from the first student selected, P = 364/365
- P(all birthdays are unique) = P(unique birthday for 1st person selected) *P(unique birthday for 2nd person selected) *P(unique birthday for 3rd person selected) ... *P(unique birthday for 84th person selected) *P(unique birthday for 85th person selected)
- Translating this into math by multiplying the values in column P:

P(all birthdays are unique) = - At this stage, you should simplify the expression that equals P(all birthdays are unique).

From 365 to 281, there are (365-281)+1 = 85 numbers (remember that you need to add 1 since it is 365 to 281,*inclusive*). - In the denominator, you know that 365 will be multiplied together 85 times:

Denominator = 365^{85} - In the numerator, it is not as easy to simplify. However, you may notice that there appears to be something resembling a factorial. Specifically, the numerator is 365! truncated just before 280. If you take 365! and divide it by 280!, every term beneath 281 will cancel out, leaving you with the expression you are looking for in the numerator.

- Combine the numerator and denominator:

- Use the complement to find the answer to the original question:

P(at least two students with the same birthday) = 1 – P(all birthdays are unique)

P(at least two students with the same birthday) =