Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

2

This post received KUDOS

8

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

75% (hard)

Question Stats:

54% (02:39) correct
46% (01:21) wrong based on 170 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

10

This post received KUDOS

Expert's post

2

This post was BOOKMARKED

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

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).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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
[#permalink]
23 Sep 2014, 10:14

On September 6, 2015, I started my MBA journey at London Business School. I took some pictures on my way from the airport to school, and uploaded them on...

When I was growing up, I read a story about a piccolo player. A master orchestra conductor came to town and he decided to practice with the largest orchestra...

Amy Cuddy, Harvard Business School professor, at TED Not all leadership looks the same; there is no prescribed formula for what makes a good leader. Rudi Gassner believed that...

We are thrilled to welcome the Class of 2017 to campus today, and data from the incoming class of students indicates that Kellogg’s community is about to reach a...