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

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

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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. (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
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Re: Probability (700+ difficulty level) [#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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Re: Probability (700+ difficulty level) [#permalink]

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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: Probability (700+ difficulty level) [#permalink]

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New post 09 Oct 2011, 11:33
So if we solve the problem what Probability do we get? .0537?
Curious to know if 85c2/365c2 gonna give the same answer.

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Re: If there are 85 students in a statistics class and we assume [#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: Probability (700+ difficulty level) [#permalink]

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New post 25 May 2013, 23:18
Bunuel wrote:
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.



Hi Bunnel,

Please explain the difference between the below two arrangements:

=> no of ways a student can have a birthday = 365, so for 85 students total no ways to have birthdays is = 365^85
Now, a day can have a birthday in 86 ways i.e. it can have no birthday, 1 birthday....up till all 85 birthday = a total of 86 ways,
=> no of ways 365 days can have a birthday = 365^86

what kind of question can come based on the second case, I get confused between these two. Can you tell any trick how to differentiate b/w them.

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Re: Probability (700+ difficulty level) [#permalink]

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New post 26 May 2013, 01:20
Bunuel wrote:
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.



Hi Bunnel,

What is the difference between:

reverse prob. - = 1- 85/365*84/364...1/281, and the method you have given
reverse comb - = 1- fav/tot = 1- 365P85/365^85

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

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New post 04 Oct 2017, 03:06
Bunuel wrote:
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.


Hi Bunuel,
can you please explain why the number of days in the denominator does not decrease as in: (365/365)x(363/364)x(362/363) etc..

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Re: If there are 85 students in a statistics class and we assume   [#permalink] 04 Oct 2017, 03:06
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