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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PhD in Applied Mathematics
Love GMAT Quant questions and running.

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.
_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

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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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.
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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

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.

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.

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.