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:

Now let's find the # of combinations where there's a BGB combination. Let the sequence of BG=A: BBAA BAAB BABA ABBA ABAB AABB

So there are 6 combinations here. But since BG can also be GB, we have to multiply by 2. We have to multiply by another 2 because there are 2 As. Therefore, there are 6x2x2=24 combinations.

We also have to include the instances where GG occurs at the end or the front, that is: GGBBBB BBBBGG (all the other instances such as BBGGBB have been taken care of in the calculation at the top)

So we have 24+2=26 possibilities of 4 boys and 2 girls.

The total number of possibilities is 2^6=64. Therefore, the probability is 26/64, which is 13/32. The answer is E, none of the above.

The probability that a family with 6 children has exactly four boys is:

A. 1/3 B. 1/64 C. 15/64 D. 3/8 E. none of the above

NOTE: If the probability of a certain event is \(p\) (\(\frac{1}{2}\) in our case), then the probability of it occurring \(k\) times (4 times in our case) in \(n\)-time (6 in our case) sequence is:

We are looking for the case BBBBGG, probability of each B or G is \(\frac{1}{2}\), hence \(\frac{1}{2^6}\). But BBBBGG can occur (these letters can be arranged) in \(\frac{6!}{4!2!}\) # of ways. So, the total probability would be \(\frac{6!}{4!2!}*\frac{1}{2^6}=\frac{15}{64}\).

Now let's find the # of combinations where there's a BGB combination. Let the sequence of BG=A: BBAA BAAB BABA ABBA ABAB AABB

So there are 6 combinations here. But since BG can also be GB, we have to multiply by 2. We have to multiply by another 2 because there are 2 As. Therefore, there are 6x2x2=24 combinations.

We also have to include the instances where GG occurs at the end or the front, that is: GGBBBB BBBBGG (all the other instances such as BBGGBB have been taken care of in the calculation at the top)

So we have 24+2=26 possibilities of 4 boys and 2 girls.

The total number of possibilities is 2^6=64. Therefore, the probability is 26/64, which is 13/32. The answer is E, none of the above.

Your solution can be simplified this way as well:

Total number of events : 2^6 as each position can be filled in exactly two ways Events in favor : (4 Boys and Two Girls; It is like 6 objects in which 4 are alike and 2 others are alike) so 6!/2!4!

If it were a probability question, I will follow Bunuel's aproach but the solution above is also important if it is counting problem.

Question: If we have n objects out f which p are alike, q are alike, how many possible number of comnitations are possible ? If it were a permutation problem, answer is n!/p!*q! as in the question above

For each child we've two options i.e., that child can either be a boy or a girl.Hence we have 2^6 = 64 total outcomes. Now out of 6 children any 4 of them can boys and this can be done in 6C4 = 15 ways. Now here I don't think that ordering does matter in this case.

So the probability comes out to be 15/64.

P.S. Please explain me if ordering really matters in this case as my understanding is quite naive in these problems.

For each child we've two options i.e., that child can either be a boy or a girl.Hence we have 2^6 = 64 total outcomes. Now out of 6 children any 4 of them can boys and this can be done in 6C4 = 15 ways. Now here I don't think that ordering does matter in this case.

So the probability comes out to be 15/64.

P.S. Please explain me if ordering really matters in this case as my understanding is quite naive in these problems.

This way of solving is also correct. You've already considered all possible arrangements for BBBBGG with 6C4, which is 6!/4!2!. If you look at my solution and at yours you'll see that both wrote the same formulas but with different approach.
_________________

hi bunuel could u plz provide a few application qs. to this formula? TIA

Bunuel wrote:

NOTE: If the probability of a certain event is \(p\) (\(\frac{1}{2}\) in our case), then the probability of it occurring \(k\) times (4 times in our case) in \(n\)-time (6 in our case) sequence is:

We are looking for the case BBBBGG, probability of each B or G is \(\frac{1}{2}\), hence \(\frac{1}{2^6}\). But BBBBGG can occur (these letters can be arranged) in \(\frac{6!}{4!2!}\) # of ways. So, the total probability would be \(\frac{6!}{4!2!}*\frac{1}{2^6}=\frac{15}{64}\).

Answer: C.

_________________

--------------------------------------------------------------------------------------- If you think you can,you can If you think you can't,you are right.

hi bunuel could u plz provide a few application qs. to this formula? TIA

Bunuel wrote:

NOTE: If the probability of a certain event is \(p\) (\(\frac{1}{2}\) in our case), then the probability of it occurring \(k\) times (4 times in our case) in \(n\)-time (6 in our case) sequence is:

We are looking for the case BBBBGG, probability of each B or G is \(\frac{1}{2}\), hence \(\frac{1}{2^6}\). But BBBBGG can occur (these letters can be arranged) in \(\frac{6!}{4!2!}\) # of ways. So, the total probability would be \(\frac{6!}{4!2!}*\frac{1}{2^6}=\frac{15}{64}\).

Answer: C.

Check the Probability and Combinatorics chapters in Math Book (link in my signature): theory, examples and links to the problems.
_________________

Question: If we have n objects out of which p are alike, q are alike, and we are to choose r objects out of it, how many possible number of comnitations are possible ? A permutation problem, answer is nPr/p!*q! as in the discussion above , but in case of combination what is formula? ??

Lets say we have 10 books, out of which 3 are copies of same book, 4 are copies of another book in how many ways 4 books can be selected ?

so answer is : 5C4 I beleive i.e. 10-3-4 are different books i.e. 3 different books, 3 books of one type can be treated as one, 4 other different books can be treated as one again.. so total 5 books , out of which 4 can be selected.

@Bunel : While thinking of a question for you, I got the answer too . THANKS.

Question to Bunel : I have flexed my muscles in number roperties but need to sweat more, could you refer me some part over this site for basic and advance review of number properties, as well as well "hard but GOOD questions" .

Last edited by GMATMadeeasy on 09 Jan 2010, 16:29, edited 1 time in total.

Question: If we have n objects out of which p are alike, q are alike, and we are to choose r objects out of it, how many possible number of comnitations are possible ? A permutation problem, answer is nCr/p!*q! as in the discussion above , but in case of combination what is formula? ??

Can you please give an example of this, to understand correctly your question?
_________________

NOTE: If the probability of a certain event is \(p\) (\(\frac{1}{2}\) in our case), then the probability of it occurring \(k\) times (4 times in our case) in \(n\)-time (6 in our case) sequence is:

NOTE: If the probability of a certain event is \(p\) (\(\frac{1}{2}\) in our case), then the probability of it occurring \(k\) times (4 times in our case) in \(n\)-time (6 in our case) sequence is:

It's the same. Since n>=k, you can write as \(nCk\), \(C(n,k)\), \(C(k,n)\), \(C^n_k\), \(C^k_n\), it's clear what is meant. Actually in different books you can find different forms of writing this. Walker in his topic used \(C^n_k\), but \(C^k_n\) is also correct.
_________________

Re: The probability that a family with 6 children has exactly [#permalink]

Show Tags

27 Dec 2013, 00:40

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: The probability that a family with 6 children has exactly [#permalink]

Show Tags

02 Jan 2015, 13:23

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: The probability that a family with 6 children has exactly [#permalink]

Show Tags

12 Jan 2016, 15:43

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: The probability that a family with 6 children has exactly [#permalink]

Show Tags

19 Jan 2017, 19:40

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.
_________________

The probability that a family with 6 children has exactly four boys is:

A. 1/3 B. 1/64 C. 15/64 D. 3/8 E. none of the above

NOTE: If the probability of a certain event is \(p\) (\(\frac{1}{2}\) in our case), then the probability of it occurring \(k\) times (4 times in our case) in \(n\)-time (6 in our case) sequence is:

We are looking for the case BBBBGG, probability of each B or G is \(\frac{1}{2}\), hence \(\frac{1}{2^6}\). But BBBBGG can occur (these letters can be arranged) in \(\frac{6!}{4!2!}\) # of ways. So, the total probability would be \(\frac{6!}{4!2!}*\frac{1}{2^6}=\frac{15}{64}\).

Answer: C.[/Bunuel can you please explain the logic behind 'p' being 1/2]

The probability of having a girl = the probability of having a boy = 1/2.
_________________

Military MBA Acceptance Rate Analysis Transitioning from the military to MBA is a fairly popular path to follow. A little over 4% of MBA applications come from military veterans...

Best Schools for Young MBA Applicants Deciding when to start applying to business school can be a challenge. Salary increases dramatically after an MBA, but schools tend to prefer...

Marty Cagan is founding partner of the Silicon Valley Product Group, a consulting firm that helps companies with their product strategy. Prior to that he held product roles at...