0
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – DylanSp Apr 7 '16 at 17:15
  • $\begingroup$ Sorry? If exactly two are boys then exactly eight are girls. What am I missing? $\endgroup$ – lulu Apr 7 '16 at 17:15
  • 1
    $\begingroup$ Is it possible you meant: "at least two are boys"? $\endgroup$ – lulu Apr 7 '16 at 17:16
  • $\begingroup$ 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. $\endgroup$ – Shababb Karim Apr 7 '16 at 17:24
  • 2
    $\begingroup$ 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"). $\endgroup$ – lulu Apr 7 '16 at 17:31
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Thank you. How did you calculate there are 11 cases with one or no boys? $\endgroup$ – Shababb Karim Apr 7 '16 at 18:16
  • 1
    $\begingroup$ Okay got it. C(10, 1) + C(10, 0) right? Thanks! $\endgroup$ – Shababb Karim Apr 7 '16 at 18:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.