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:

Three pairs of siblings, each pair consisting of one girl [#permalink]

Show Tags

05 Aug 2012, 03:32

11

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

42% (02:58) correct
58% (01:49) wrong based on 248 sessions

HideShow timer Statistics

Three pairs of siblings, each pair consisting of one girl and one boy, are randomly seated at a table. What is the probability that all three girls are seated on the left of their boy siblings?

Three pairs of siblings, each pair consisting of one girl and one boy, are randomly seated at a table. What is the probability that all three girls are seated on the left of their boy siblings?

Notice that we need a girl to be to the left of her sibling, but not necessarily right to the left of him (meaning that if B and G are siblings, then GB arrangement as well as for example G*B arrangement is possible).

Now, the probability that one particular sibling is seated that way is 1/2 (a girl can be either to the left of her sibling or to the right), the probability that two siblings are seated that way is 1/2*1/2 and the probability that all three siblings are seated that way is 1/2*1/2*1/2=1/8.

Re: Three pairs of siblings, each pair consisting of one girl [#permalink]

Show Tags

05 Aug 2012, 04:35

Bunuel wrote:

EvaJager wrote:

Three pairs of siblings, each pair consisting of one girl and one boy, are randomly seated at a table. What is the probability that all three girls are seated on the left of their boy siblings?

Notice that we need a girl to be to the left of her sibling, but not necessarily right to the left of him (meaning that if B and G are siblings, then GB arrangement as well as for example G*B arrangement is possible).

Now, the probability that one particular sibling is seated that way is 1/2 (a girl can be either to the left of her sibling or to the right), the probability that two siblings are seated that way is 1/2*1/2 and the probability that all three siblings are seated that way is 1/2*1/2*1/2=1/8.

Answer: D.

By far the fastest and most elegant solution!

Those who want to play with combinatorics are invited to provide an alternate solution.
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: Three pairs of siblings, each pair consisting of one girl [#permalink]

Show Tags

05 Aug 2012, 23:23

3

This post received KUDOS

1

This post was BOOKMARKED

EvaJager wrote:

Three pairs of siblings, each pair consisting of one girl and one boy, are randomly seated at a table. What is the probability that all three girls are seated on the left of their boy siblings?

Place the first pair of siblings - we have 6 possibilities for one of them, and 5 for the other one; this gives 6*5, but we have to divide by 2, as only in half of them, the girls will sit on the left of her brother. So, 6*5/2=15 possibilities. To place the second pair of siblings - similarly, we have 4*3/2=6 possibilities. Finally, for the last and third pair - 2*1/2 = 1 possibility.

Therefore, the requested probability is 15*6/6!= 3*5*6/(2*3*4*5*6)=1/(2*4) = 1/8.

Answer D.
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: Three pairs of siblings, each pair consisting of one girl [#permalink]

Show Tags

06 Aug 2012, 21:45

Thanks Bunuel, you once again showed that in GMAT in most cases it is more logical thinking than doing quants. I have tried this one with different approaches but still could not come up with solution, but after your explanation it seems so easy and i wonder how i could not come up myself.

Thanks!!!
_________________

If you found my post useful and/or interesting - you are welcome to give kudos!

Therefore, the requested probability is 15*6/6!= 3*5*6/(2*3*4*5*6)=1/(2*4) = 1/8.

Answer D.

Aren't there only 5! total arrangements around a table for 6 people?

We are not told that these 6 are seated around a table, so we don't have circular arrangement. The question implies that they are seated like in a row.
_________________

Re: Three pairs of siblings, each pair consisting of one girl [#permalink]

Show Tags

08 Nov 2012, 10:06

EvaJager wrote:

EvaJager wrote:

Three pairs of siblings, each pair consisting of one girl and one boy, are randomly seated at a table. What is the probability that all three girls are seated on the left of their boy siblings?

Place the first pair of siblings - we have 6 possibilities for one of them, and 5 for the other one; this gives 6*5, but we have to divide by 2, as only in half of them, the girls will sit on the left of her brother. So, 6*5/2=15 possibilities. To place the second pair of siblings - similarly, we have 4*3/2=6 possibilities. Finally, for the last and third pair - 2*1/2 = 1 possibility.

Therefore, the requested probability is 15*6/6!= 3*5*6/(2*3*4*5*6)=1/(2*4) = 1/8.

Re: Three pairs of siblings, each pair consisting of one girl [#permalink]

Show Tags

12 Feb 2014, 03:09

Bunuel wrote:

EvaJager wrote:

Three pairs of siblings, each pair consisting of one girl and one boy, are randomly seated at a table. What is the probability that all three girls are seated on the left of their boy siblings?

Notice that we need a girl to be to the left of her sibling, but not necessarily right to the left of him (meaning that if B and G are siblings, then GB arrangement as well as for example G*B arrangement is possible).

Now, the probability that one particular sibling is seated that way is 1/2 (a girl can be either to the left of her sibling or to the right), the probability that two siblings are seated that way is 1/2*1/2 and the probability that all three siblings are seated that way is 1/2*1/2*1/2=1/8.

Answer: D.

Hi Bunuel,

first sibling can be seated in 1/2 ways. but how do we come about the second sibling probability of 1/2 ? I am bit confused here, can you explain please ?

Re: Three pairs of siblings, each pair consisting of one girl [#permalink]

Show Tags

05 Aug 2015, 03:21

EvaJager wrote:

EvaJager wrote:

Three pairs of siblings, each pair consisting of one girl and one boy, are randomly seated at a table. What is the probability that all three girls are seated on the left of their boy siblings?

Here is a solution using combinatorics: Place the first pair of siblings - we have 6 possibilities for one of them, and 5 for the other one; this gives 6*5, but we have to divide by 2, as only in half of them, the girls will sit on the left of her brother. So, 6*5/2=15 possibilities. To place the second pair of siblings - similarly, we have 4*3/2=6 possibilities. Finally, for the last and third pair - 2*1/2 = 1 possibility.

Therefore, the requested probability is 15*6/6!= 3*5*6/(2*3*4*5*6)=1/(2*4) = 1/8.

Answer D.

Hi Eva I dont get eather of the solution , by Bunual or by you, please explain Place the first pair of siblings - we have 6 possibilities for one of them, and 5 for the other one; this gives 6*5

Three pairs of siblings, each pair consisting of one girl and one boy, are randomly seated at a table. What is the probability that all three girls are seated on the left of their boy siblings?

Here is a solution using combinatorics: Place the first pair of siblings - we have 6 possibilities for one of them, and 5 for the other one; this gives 6*5, but we have to divide by 2, as only in half of them, the girls will sit on the left of her brother. So, 6*5/2=15 possibilities. To place the second pair of siblings - similarly, we have 4*3/2=6 possibilities. Finally, for the last and third pair - 2*1/2 = 1 possibility.

Therefore, the requested probability is 15*6/6!= 3*5*6/(2*3*4*5*6)=1/(2*4) = 1/8.

Answer D.

Hi Eva I dont get eather of the solution , by Bunual or by you, please explain Place the first pair of siblings - we have 6 possibilities for one of them, and 5 for the other one; this gives 6*5

Let the siblings be : in the order Girl Boy A B C D E F

Let the arrangement be _ _ _ _ _ _

So if you place lets say A on the first dash (=6 ways you can place A), you only have 5 places to let B go to. Thus you get 6*5. Eva has divided this and other possible arrangements by 2 to account for the fact 50% of the combinations will have AB_ _ _ _ while 50% will be BA_ _ _ _ . Only cases with AB_ _ _ _ type of combinations are allowed. We can safley assume 50% for either cases as there is no case for a 'bias' in these arrangements.

Bunuel has done the same , albeit in a slightly different manner. Probability of any girl sibling sitting to the right of the boy sibling = 50% or 1/2 (same as above)

Final probability = probability of 1st sibling girl to the left of the boy sibling * probability of 2nd sibling girl to the left of the boy sibling *probability of 3rd sibling girl to the left of the boy sibling = 1/2 * 1/2 * 1/2 = 1/8

Probability can be calculated in 2 ways:

Probability = total favorable cases / total cases (which is what Eva has done) or

Probability = probability of case 1* probability of case 2*probability of case 3 etc .... (which is what Bunuel has done).

You can choose whichever method suits you.
_________________

Re: Three pairs of siblings, each pair consisting of one girl [#permalink]

Show Tags

05 Aug 2015, 04:27

Thanks Now I got it.little more help, where i am wrong in this ...

Let the siblings be : in the order Girl Boy A B C D E F

Only three cases are available

ABCDEF ....(3!) no of ways three siblings can be arranges like CDABEF(one of the case out of 6) = 6 + ACEBDF ...3! *3! ( no of ways ACE and BDF can be arranged them self) = 36 + ABCEDF OR CDAEBF OR EFACBD ( no of ways ne sibling comes extrem left , then remaining two girls then remaining two boys)

3c1(ne subling out of three) *2! (ways two girls arranged among themself ) * 2! (ways 2 boys arranged among themself ) =24

Please let me know if my approach to the problem is correct. I have seen your other solutions where we multiply by 1/2 whenever we have a condition of sitting/ standing only on left or right.

So instead of multiplying 1/2 for each pair, i took (1/2)^3, 3 is the number of pairs. I want to know if this approach is correct in case i have to apply to similar questions.

gmatclubot

Re: Three pairs of siblings, each pair consisting of one girl
[#permalink]
06 Aug 2015, 04:51

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Post-MBA I became very intrigued by how senior leaders navigated their career progression. It was also at this time that I realized I learned nothing about this during my...