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?
Best Answer
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 $$\frac{\frac12(2^{10}-{10\choose 5})-11}{2^{10}-11}=\frac{375}{1013}\approx 0.37 $$