Three pairs of siblings, each pair consisting of one girl

Question

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}

Bunuel · Accepted Answer

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.

EvaJager · Answer

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.

EvaJager · Answer

By far the fastest and most elegant solution!

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

ziko · Answer

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!!!

Thanks!!!

BN1989 · Answer

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

Bunuel · Answer

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.

bhavinshah5685 · Answer

Hi, couldnt understand why to devide by 6! ??

vishalrastogi · Answer

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

vipulgoel · Answer

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

ENGRTOMBA2018 · Answer

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.

vipulgoel · Answer

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!

Swaroopdev · Answer

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.

JeffTargetTestPrep · Answer

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

elainetianfong · Answer

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 L-L-L happens ?
The following cases would happen:
L-L-L : 1 case
R-R-R : 1 case
L-R-R : 3 cases (choose 2 out 3 pairs that have the girl sitting on the right)
L-L-R: 3 cases (choose 2 out 3 pairs that have the girl sitting on the left)
Total cases: 8
The probability = 1/8.

abhi0612 · Answer

Probability of being on the right is equal to probability of being on the left so for each pair the probablity is 1/2. Therefore, the ans would be : (1/2)^3 = 1/8

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Regor60 · Answer

Pick the first male, 3 choices.

Only 1 choice for his sibling 🤪.

One choice for which side to place the sibling, so a total of:

3*1*1=3 choices

Place 2nd male, two choices.

1 choice for sibling, 1 choice, left, for seat.

Total choices:

2*1*1= 2

3rd sibling pair only 1 choice

Total choices meeting requirements:

3*2*1= 6

Total available choices follow logic above, but recognizing that female sibling has 2 choices of left or right for each pair,so

3*2=6
2*2=4
1*2=2

Or 6*4*2= 48 potential arrangements.

Probability: 6/48= 1/8

Note that there is no reduction in available seats to the left or right just because a prior pair has been seated, since there is no limitation in the question to the number of seats, so one can assume an equal availability of seats to the left or right or 1/2 probability of being seated to the left or right for all sibling pairs. This is based on an infinite number of seats point of view.

The other point of view is that even with a limited set of seats, every pair has the same opportunity to be properly seated since no specific pair has been identified as occupying a specific set of seats, thereby not constraining the remaining pairs choices.

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Three pairs of siblings, each pair consisting of one girl

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Three pairs of siblings, each pair consisting of one girl

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