# A family has $10$ children. What is the probability that they have more girls than boys, given that at least two of the children are boys?

A family has $10$ children. What is the probability that they have more girls than boys, given that at least two of the children are boys?

If it was the case for 3 children then I could have simply find all the combinations and then from that it becomes very easy but for such a big case how am I supposed to find the answer?

• If exactly $2$ of the children are boys, and we assume every child is either a girl or a boy, then we have $10 - 2 = 8$ girls, so the probability is $1$. Either this is a trick question, or we're missing something. – DylanSp Apr 7 '16 at 17:15
• Sorry? If exactly two are boys then exactly eight are girls. What am I missing? – lulu Apr 7 '16 at 17:15
• Is it possible you meant: "at least two are boys"? – lulu Apr 7 '16 at 17:16
• I think you guys are having trouble understanding the question. Given that 2 of 10 are boys. Now there are 8 places left. Of these 8 places 8 of them can be girls and 7 of them can be girls and 1 boy and 6 girls 2 boys. Now we have to find the probability of the cases happening. – Shababb Karim Apr 7 '16 at 17:24
• Right. So it's "at least" two boys, not "exactly" two boys as you wrote. In that case, the only winning scenarios are exactly $6,7,8$ girls. Just compute the probability of each of those cases (and I do mean "exactly", not "at least"). – lulu Apr 7 '16 at 17:31

Out of the $2^{10}=1024$ possible gender assignments, the restriction rules out those $11$ with one or no boy. These removed cases happen to be "favorable" cases, i.e., with more girls than boys. Moreover, there are ${10\choose 5}$ possible cases of a tie, and usually half of the rest would be favorable. Hence the desired probability is $$\frac{\frac12(2^{10}-{10\choose 5})-11}{2^{10}-11}=\frac{375}{1013}\approx 0.37$$