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 500,000 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:

4 dices are thrown at the same time. What is the probability of getting ONLY 2 dices showing the same face?

A. 4/9 B. 5/9 C. 11/18 D. 7/9 E. none of the above

I suppose "only 2 dice showing the same face" means EXACTLY two? If so then:

Total # of outcomes = 6^4

Favorable outcomes = 4C2=6, choosing two dice which will provide the same face, these two dice can take 6 values, other two 5 and 4. So, favorable outcomes=4C2*6*5*4.

\(P=\frac{4C2*6*5*4}{6^4}=\frac{5}{9}\).

Answer: B.

If we have 4C2 aren't we assuming that the other two also have to be the same number? Or does it work for 2 the same number, the other two different?

I thought that we had to use 4!/2!1!1! because the last two terms were different.

4 dices are thrown at the same time. What is the probability of getting ONLY 2 dices showing the same face?

A. 4/9 B. 5/9 C. 11/18 D. 7/9 E. none of the above

I suppose "only 2 dice showing the same face" means EXACTLY two? If so then:

Total # of outcomes = 6^4

Favorable outcomes = 4C2=6, choosing two dice which will provide the same face, these two dice can take 6 values, other two 5 and 4. So, favorable outcomes=4C2*6*5*4.

\(P=\frac{4C2*6*5*4}{6^4}=\frac{5}{9}\).

Answer: B.

If we have 4C2 aren't we assuming that the other two also have to be the same number? Or does it work for 2 the same number, the other two different?

I thought that we had to use 4!/2!1!1! because the last two terms were different.

Could anybody please clarify Thanks Cheers J

We have 4 dice (A, B, C, D).

\(C^2_4\) is the ways to select which two dice out of 4, will provide the same face: (A, B), (A, C), (A, D), (B, C), (B, D), or (C, D) . Next, these two dice can take 6 values, the remaining two 5 and 4, respectively. So, the # of favorable outcomes is \(C^2_4*6*5*4\).

Re: 4 dices are thrown at the same time. What is the probability [#permalink]

Show Tags

06 Oct 2014, 05:11

Although I did get the answer right,I could not figure out the breakup of other possibilities.And it has been bugging me. Total possible outcomes = 6 * 6 * 6 * 6 = 1296

Possibility of having no.s on all dices same = 6 * 1 = 6

Possibility of having 3 same no.s = 6 * 5 * 4!/3! = 120

Possibility of having 2 same no.s and 2 different= 6 * 5C2 * 4!/2! = 6 * 10 * 12 = 720

Possibility of having 2 pairs of same no.s = 6 * 5 =30

Possibility of having all different no.s = 6 * 5 * 4* 3 = 360 Now these possibilities don't add up 6 + 120 + 720 + 30 + 360 = 1236(total is 1296 i.e. 60 more than this) Can anybody please tell which possibilities have i missed??? Thanks

Re: 4 dices are thrown at the same time. What is the probability [#permalink]

Show Tags

06 Oct 2014, 08:08

1

This post received KUDOS

Expert's post

arpitsharms wrote:

Although I did get the answer right,I could not figure out the breakup of other possibilities.And it has been bugging me. Total possible outcomes = 6 * 6 * 6 * 6 = 1296

Possibility of having no.s on all dices same = 6 * 1 = 6

Possibility of having 3 same no.s = 6 * 5 * 4!/3! = 120

Possibility of having 2 same no.s and 2 different= 6 * 5C2 * 4!/2! = 6 * 10 * 12 = 720

Possibility of having 2 pairs of same no.s = 6 * 5 =30

Possibility of having all different no.s = 6 * 5 * 4* 3 = 360 Now these possibilities don't add up 6 + 120 + 720 + 30 + 360 = 1236(total is 1296 i.e. 60 more than this) Can anybody please tell which possibilities have i missed??? Thanks

This should be 90: 3*(6*5). The number of ways to split 4 dice into 2 pairs is 3: AB - CD AC - BD AD - BC _________________

Re: 4 dices are thrown at the same time. What is the probability [#permalink]

Show Tags

11 Oct 2015, 21:28

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: 4 dices are thrown at the same time. What is the probability [#permalink]

Show Tags

18 Nov 2015, 05:06

Hello Bunuel

In such a question, is it imperative to assume that the dice will be different (not identical)?

Thank you

Bunuel wrote:

roshanaslam wrote:

4 dices are thrown at the same time. What is the probability of getting ONLY 2 dices showing the same face?

A. 4/9 B. 5/9 C. 11/18 D. 7/9 E. none of the above

I suppose "only 2 dice showing the same face" means EXACTLY two? If so then:

Total # of outcomes = 6^4

Favorable outcomes = 4C2=6, choosing two dice which will provide the same face, these two dice can take 6 values, other two 5 and 4. So, favorable outcomes=4C2*6*5*4.

Re: 4 dices are thrown at the same time. What is the probability [#permalink]

Show Tags

18 Nov 2015, 11:54

Expert's post

rsaahil90 wrote:

Hello Bunuel

In such a question, is it imperative to assume that the dice will be different (not identical)?

Thank you

Bunuel wrote:

roshanaslam wrote:

4 dices are thrown at the same time. What is the probability of getting ONLY 2 dices showing the same face?

A. 4/9 B. 5/9 C. 11/18 D. 7/9 E. none of the above

I suppose "only 2 dice showing the same face" means EXACTLY two? If so then:

Total # of outcomes = 6^4

Favorable outcomes = 4C2=6, choosing two dice which will provide the same face, these two dice can take 6 values, other two 5 and 4. So, favorable outcomes=4C2*6*5*4.

\(P=\frac{4C2*6*5*4}{6^4}=\frac{5}{9}\).

Answer: B.

Yes. 3 on the first die, say on blue one, and 4 on the second die, say on red one, is a different case from 3 on the second die, on red die, and 4 on the first die, on blue one. _________________

Re: 4 dices are thrown at the same time. What is the probability [#permalink]

Show Tags

24 Jun 2016, 04:10

chetan2u wrote:

my take on this is.... choosing first no will have possibility of 1.... choosing the same on any of other three dice is 1/6 . third dice can have any of remaining 5 so probab is 5/6... fourth dice can have any of remaining 4 so probab is 4/6... so prob=1*1/6*5/6*4/6....but there are six different values so it becomes =1*1/6*5/6*4/6*6=5/9.. anyone if i have gone wrong

we multiply by 6 because there are 6 possibilities of having same numbers {1,1}.. {6,6} etc. we multiply by 4! for possible combinations in between the four dices, by 2! for two same numbers and again by 2! for possibilities between same & not-same numbers.

Can someone please explain why we are dividing again by 2!? Thanks!

Re: 4 dices are thrown at the same time. What is the probability [#permalink]

Show Tags

24 Jun 2016, 23:52

This is a good question. Earlier I guessed the right answer and later understood why that must be right. Step 1. Choose a side : let us say 2. First we need to choose to put 2 in any of the four places. That is 4C2 = 6. To each of these there is 5*4 possibilities of other two sides not exactly same as 2. Now to each of the possibilities above 6*5*4 we need to multiply by 6. Since we chose 2 above, we can choose any of the six. So now required possibilities are 6*6*5*4. Prob = 6*6*5*4/6*6*6*6. = 20/36 = 5/9. Choice B

Re: 4 dices are thrown at the same time. What is the probability [#permalink]

Show Tags

25 Jun 2016, 03:04

chetan2u wrote:

my take on this is.... choosing first no will have possibility of 1.... choosing the same on any of other three dice is 1/6 . third dice can have any of remaining 5 so probab is 5/6... fourth dice can have any of remaining 4 so probab is 4/6... so prob=1*1/6*5/6*4/6....but there are six different values so it becomes =1*1/6*5/6*4/6*6=5/9.. anyone if i have gone wrong

Nice explanation....I tried the same concept but faltered while calculating the probability of last dice (i considered 5/6 for that as well)... Your explanation just restored my faiths in my concepts (of course I made one of the classic error ....).

Re: 4 dices are thrown at the same time. What is the probability [#permalink]

Show Tags

26 Jun 2016, 01:37

Bunuel wrote:

roshanaslam wrote:

4 dices are thrown at the same time. What is the probability of getting ONLY 2 dices showing the same face?

A. 4/9 B. 5/9 C. 11/18 D. 7/9 E. none of the above

I suppose "only 2 dice showing the same face" means EXACTLY two? If so then:

Total # of outcomes = 6^4

Favorable outcomes = 4C2=6, choosing two dice which will provide the same face, these two dice can take 6 values, other two 5 and 4. So, favorable outcomes=4C2*6*5*4.

\(P=\frac{4C2*6*5*4}{6^4}=\frac{5}{9}\).

Answer: B.

The only thing I'd like to add, a potential shortcut, is that since the numerator has 5, and 5 is not divisible by 6, the final answer has to have a numerator divisible by 5, a condition that only answer choice B meets. _________________

Offering top quality GMAT tutoring service in Vietnam, Southeast Asia, and worldwide.

http://www.facebook.com/HanoiGMATtutor

gmatclubot

Re: 4 dices are thrown at the same time. What is the probability
[#permalink]
26 Jun 2016, 01:37

Excellent posts dLo saw your blog too..!! Man .. you have got some writing skills. And Just to make an argument = You had such an amazing resume ; i am glad...

So Much $$$ Business school costs a lot. This is obvious, whether you are a full-ride scholarship student or are paying fully out-of-pocket. Aside from the (constantly rising)...

I barely remember taking decent rest in the last 60 hours. It’s been relentless with submissions, birthday celebration, exams, vacating the flat, meeting people before leaving and of...

Rishabh from Gyan one services, India had a one to one interview with me where I shared my experience at IMD till now. http://www.gyanone.com/blog/life-at-imd-interview-with-imd-mba/ ...