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

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05 Aug 2012, 04:32

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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]

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05 Aug 2012, 05:35

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]

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06 Aug 2012, 00:23

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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]

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06 Aug 2012, 22:45

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]

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08 Nov 2012, 11:06

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]

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12 Feb 2014, 04:09

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]

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05 Aug 2015, 04:21

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]

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05 Aug 2015, 05:27

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
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06 Aug 2015, 05:51

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