If a family had 9 children, 4 are girls and 5 are boys, what is the probability of the first three being two girls and one boy?

My understanding is since order doesn't matter, it's just the multiplication rule. So first girl chance is 4/9, second is 3/8. The chance of the boy is then 5/7. And I would multiply them together. Is that the correct way of approaching this question?

  • $\begingroup$ The question is somewhat ambiguous. You (not unreasonably) interpreted it as meaning two girls then a boy. I think there is a good chance it was intended to mean a total of two girls and one boy among the first three kids born. $\endgroup$ Sep 17 '15 at 21:51
  • $\begingroup$ Right, I think I mixed that part up a bit. Assuming it did mean in that particular order, would my solution be correct? If it meant in no particular order like someone mentioned below it would be the naive theory of probability where (chance of 2 girls times chance of 1 boy) / 9 choose 3? $\endgroup$
    – Nick Gong
    Sep 17 '15 at 21:53
  • $\begingroup$ Yes, I am also confused with this. So, I mentioned 'I think' at the last of my answer. $\endgroup$
    – user249332
    Sep 17 '15 at 21:54
  • $\begingroup$ If that specific order, then your solution is correct. If any order is OK, multiply by $3$, the number of possible orders, or (more efficiently) use the approach of Subhadeep Dey. $\endgroup$ Sep 17 '15 at 21:58
  • $\begingroup$ @NickGong Yes, your solution would be correct if it were "two girls then one boy". However, that is only one third of the possibilities for "two girls and one boy". But it's easy enough to give both answers and state why you think the question is ambiguous. $\endgroup$ Sep 17 '15 at 21:58

Now, the denominator will be ${9\choose 3}=84$

The numerator will be ${4\choose 2}\times {5\choose 1}=30$.

So the answer is $\frac {30}{84}=\frac {5}{14}$. I think.


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