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If there are 85 students in a statistics class and we assume [#permalink]
29 Mar 2010, 10:46

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Difficulty:

45% (medium)

Question Stats:

52% (02:44) correct
47% (00:59) wrong based on 71 sessions

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

Re: Probability (700+ difficulty level) [#permalink]
29 Mar 2010, 11:20

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Expert's post

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}}.

Re: Probability (700+ difficulty level) [#permalink]
11 Apr 2011, 08:31

2

This post received KUDOS

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)

Re: Probability (700+ difficulty level) [#permalink]
25 May 2013, 22:18

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.

Re: Probability (700+ difficulty level) [#permalink]
26 May 2013, 00:20

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

gmatclubot

Re: Probability (700+ difficulty level)
[#permalink]
26 May 2013, 00:20