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If there are 85 students in a statistics class and we assume [#permalink]
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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. (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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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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29 Mar 2010, 12:32
Thanks for the explanation !



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Re: Probability (700+ difficulty level) [#permalink]
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01 Apr 2010, 13:02
nice explanation! have you already done the GMAT Bunuel?



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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 \(P^{365}_{85}=\frac{365!}{(36585)!}=\frac{365!}{280!}\) 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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30 Apr 2011, 20:29
Interesting one indeed.
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Re: Probability (700+ difficulty level) [#permalink]
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28 Sep 2011, 00:32
tough one.... will this much tough questions come for the real exam??!!!
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Re: Probability (700+ difficulty level) [#permalink]
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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: Probability (700+ difficulty level) [#permalink]
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19 Oct 2011, 13:23
great explanation.



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Re: Probability (700+ difficulty level) [#permalink]
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02 Feb 2012, 05:17
Thanks Fluke for the simple explanation of the solution. Actually your explanation clarified my doubt that I had in the solution.
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Re: If there are 85 students in a statistics class and we assume [#permalink]
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23 May 2013, 05:38



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Re: If there are 85 students in a statistics class and we assume [#permalink]
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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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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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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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23 Sep 2014, 11:14
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



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