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Three pairs of siblings, each pair consisting of one girl
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05 Aug 2012, 03:32
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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? \((A) \, \frac{1}{2}\) \((B) \, \frac{1}{4}\) \((C) \, \frac{1}{6}\) \((D) \, \frac{1}{8}\) \((E) \, \frac{1}{16}\)
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Re: Three pairs of siblings, each pair consisting of one girl
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05 Aug 2012, 04:08
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?
\((A) \, \frac{1}{2}\) \((B) \, \frac{1}{4}\) \((C) \, \frac{1}{6}\) \((D) \, \frac{1}{8}\) \((E) \, \frac{1}{16}\) 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.
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Re: Three pairs of siblings, each pair consisting of one girl
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05 Aug 2012, 23:23
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?
\((A) \, \frac{1}{2}\) \((B) \, \frac{1}{4}\) \((C) \, \frac{1}{6}\) \((D) \, \frac{1}{8}\) \((E) \, \frac{1}{16}\) 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.
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Re: Three pairs of siblings, each pair consisting of one girl
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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?
\((A) \, \frac{1}{2}\) \((B) \, \frac{1}{4}\) \((C) \, \frac{1}{6}\) \((D) \, \frac{1}{8}\) \((E) \, \frac{1}{16}\) 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.
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Re: Three pairs of siblings, each pair consisting of one girl
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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!!!
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Re: Three pairs of siblings, each pair consisting of one girl
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07 Nov 2012, 02:51
EvaJager wrote: EvaJager wrote: 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?



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Re: Three pairs of siblings, each pair consisting of one girl
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Re: Three pairs of siblings, each pair consisting of one girl
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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?
\((A) \, \frac{1}{2}\) \((B) \, \frac{1}{4}\) \((C) \, \frac{1}{6}\) \((D) \, \frac{1}{8}\) \((E) \, \frac{1}{16}\) 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, couldnt understand why to devide by 6! ??



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Re: Three pairs of siblings, each pair consisting of one girl
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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?
\((A) \, \frac{1}{2}\) \((B) \, \frac{1}{4}\) \((C) \, \frac{1}{6}\) \((D) \, \frac{1}{8}\) \((E) \, \frac{1}{16}\) 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 ? thanks



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Re: Three pairs of siblings, each pair consisting of one girl
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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?
\((A) \, \frac{1}{2}\) \((B) \, \frac{1}{4}\) \((C) \, \frac{1}{6}\) \((D) \, \frac{1}{8}\) \((E) \, \frac{1}{16}\) 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



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Three pairs of siblings, each pair consisting of one girl
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05 Aug 2015, 04:07
vipulgoel wrote: 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?
\((A) \, \frac{1}{2}\) \((B) \, \frac{1}{4}\) \((C) \, \frac{1}{6}\) \((D) \, \frac{1}{8}\) \((E) \, \frac{1}{16}\) 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*5Let 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.



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Re: Three pairs of siblings, each pair consisting of one girl
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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
36+24+6/6!



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Re: Three pairs of siblings, each pair consisting of one girl
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06 Aug 2015, 04:51
Hi Bunuel, 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. 6!*1/8 ==> 90 (favourable outcome) /6! ==> 90/720 ==> 1/8. 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.



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Re: Three pairs of siblings, each pair consisting of one girl
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20 Jun 2017, 17:55
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?
\((A) \, \frac{1}{2}\) \((B) \, \frac{1}{4}\) \((C) \, \frac{1}{6}\) \((D) \, \frac{1}{8}\) \((E) \, \frac{1}{16}\) Let’s call one pair of siblings Abby and John. Since two people can be arranged in 2 ways, the probability is 1/2 that Abby sits to the left of John and 1/2 that she sits to the right of John. Using that same logic, we see there is a 1/2 x 1/2 x 1/2 = 1/8 probability that all 3 girls are seated to the left of their male siblings. Answer: D
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Re: Three pairs of siblings, each pair consisting of one girl
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28 Jun 2018, 20:13
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?
\((A) \, \frac{1}{2}\) \((B) \, \frac{1}{4}\) \((C) \, \frac{1}{6}\) \((D) \, \frac{1}{8}\) \((E) \, \frac{1}{16}\) I think Bunuel just did a beautiful job. However, my solution is a little bit quantitative. Let's call L the left position of the girl, and R the right one. The question is: what is the probability that LLL happens ? The following cases would happen: LLL : 1 case RRR : 1 case LRR : 3 cases (choose 2 out 3 pairs that have the girl sitting on the right) LLR: 3 cases (choose 2 out 3 pairs that have the girl sitting on the left) Total cases: 8 The probability = 1/8.




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