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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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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
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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: $$1-\frac{365!}{280!*365^{85}}$$.

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29 Mar 2010, 11:32
Thanks for the explanation !
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Re: Probability (700+ difficulty level) [#permalink]

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01 Apr 2010, 12:02
nice explanation! have you already done the GMAT Bunuel?
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Re: Probability (700+ difficulty level) [#permalink]

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11 Apr 2011, 08:31
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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

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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30 Apr 2011, 19:29
Interesting one indeed.
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Re: Probability (700+ difficulty level) [#permalink]

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27 Sep 2011, 23:32
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, 10:33
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, 12:23
great explanation.
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Re: Probability (700+ difficulty level) [#permalink]

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02 Feb 2012, 04:17
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, 04:38
Bumped for review and further discussion.
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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, 07:13
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, 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: $$1-\frac{365!}{280!*365^{85}}$$.

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, 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: $$1-\frac{365!}{280!*365^{85}}$$.

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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26 Jul 2014, 05:01
Hello from the GMAT Club BumpBot!

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

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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: $$1-\frac{365!}{280!*365^{85}}$$.

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

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03 May 2016, 01:09
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If there are 85 students in a statistics class and we assume   [#permalink] 03 May 2016, 01:09
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