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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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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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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
Re: If there are 85 students in a statistics class and we assume [#permalink]
26 Jul 2014, 05:01
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Re: If there are 85 students in a statistics class and we assume [#permalink]
23 Sep 2014, 10:14
cumulonimbus wrote:
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
bumping for an answer to this
gmatclubot
Re: If there are 85 students in a statistics class and we assume
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23 Sep 2014, 10:14
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