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

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New post 29 Mar 2010, 11:46
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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)?


A. \(\frac{85}{365} * \frac{84}{364}\)

B. \(\frac{1}{365} * \frac{1}{364}\)

C. \(1 - \frac{85!}{365!}\)

D. \(1 - \frac{365!}{280!(365^{85})}\)

E. \(1 - \frac{85!}{(365^{85})}\)
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Re: If there are 85 students in a statistics class and we assume that ther  [#permalink]

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New post 29 Mar 2010, 12:20
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gmatprep09 wrote:
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! (365^85))
E) 1- (85!/(365^85))

Can someone please explain this in detail....Thanks


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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New post 19 Oct 2010, 16:27
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Re: If there are 85 students in a statistics class and we assume that ther  [#permalink]

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New post 19 Oct 2010, 16:31
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Yalephd wrote:
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) \(\frac{85}{365} * \frac{84}{364}\)

B) \(\frac{1}{365} * \frac{1}{364}\)

C) 1 - \(\frac{85!}{365!}\)

D) 1 - \(\frac{365!}{280!(365^8^5)}\)

E) 1 - \(\frac{85!}{365^8^5)}\)


______________________________________________
This seems like a rather difficult question for the GMAT.


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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Re: If there are 85 students in a statistics class and we assume that ther  [#permalink]

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New post 11 Apr 2011, 09:31
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gmatprep09 wrote:
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! (365^85))
E) 1- (85!/(365^85))

Can someone please explain this in detail....Thanks


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"
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Re: If there are 85 students in a statistics class and we assume that ther  [#permalink]

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New post 25 May 2013, 08:13
Interestling enough, for 60 students, the answer already comes close to 99% . Interesting, huh?
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Re: If there are 85 students in a statistics class and we assume that ther   [#permalink] 14 Sep 2018, 05:21
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