If there are 85 students in a statistics class and we assume that ther

The easiest way to solve this problem is to calculate opposite probability and subtract this value from 1:

The opposite probability is that all students have the birthdays on different days: \(\frac{365}{365}*\frac{364}{365}*\frac{363}{365}*...*\frac{281}{365}=\frac{365*364*363*...*281}{365^{85}}=\frac{365!}{280!*365^{85}}\) total 85 birthdays (first student can have birthday on any day =1=365/365, the probability that the second student will have the birthday on another day is 364/365, the probability that the third student will have the birthday not on this two days is 363/365, and so on).

So, the probability that at least two students in the class have the same birthday is: \(1-\frac{365!}{280!*365^{85}}\).

Answer: D.
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As Bunuel already explained:

P(At least two students have same birthday) = 1 - P(At most 0 students have the same birthday)
= 1 - P(All students have different birthdays)

How to choose 85 different birthdays out of 365 days OR choose 85 different days without repetition out of 365?
It is \(P^{365}_{85}=\frac{365!}{(365-85)!}=\frac{365!}{280!}\)

Total possibilities= (365)^85 as every student can choose from 365 days.

P(All students have different birthdays) \(=\frac{365!}{280!*(365)^{85}}\)

P(At least two students have same birthday) = 1 - P(All students have different birthdays)

\(P=1-\frac{365!}{280!*(365)^{85}}\)

Ans: "D"

22. Probability

• Theory
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  Probability Made Easy!
  GMAT Tip of the Week: 99 Problems But Probability Ain't One
  
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    Probability and Geometry Problems
    Advanced Topic: Probability and Combinations Questions With Solutions
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For more:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread.[/quote]
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Probability(Atleast 2 share a bday) = 1 - Probability(All 85 have distinct birthdays)

total ways to pick 85 birthdays = 365^85
Ways to pick distinct 85 out of the 365 & distribute it to 85 people = C(365,85)*85!

Hence Probability = \(1-\frac{365!}{280!365^{85}}\)

ANswer is (d)
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Interestling enough, for 60 students, the answer already comes close to 99% . Interesting, huh?

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