-1
$\begingroup$

A family has 3 children. Assume that each child is as likely to be a boy as it is to be a girl. Find the probability that the family has 3 girls if it is known the family has at least one girl.

Outcomes: GGG, BBB, GGB, GBG, GBB, BBG, BGB, BGG (8 in total)

When I first approached this problem, I simply did (1)(1/2)(1/2) = 1/4 because I thought there was a 100% certainty that one of the children is a girl; hence a probability of 1. And then the remaining children have a 50% chance of being either a girl or boy. 1/4 is the correct answer, but I now do not believe this way of thinking was correct.

My second approach was to view this problem as a conditional probability where P(A|B) = P(A and B)/P(B) where event A is the family having 3 girls and event B is family having at least one girl.

Therefore, P(A|B) = (1/8) / (7/8) where 1/8 is GGG and 7/8 has at least one girl using the complement. But this is obviously incorrect because my online homework does not accept this as an answer. It accepts 1/4.

I saw an online solution with similar question except for boys and it showed the solution using conditional probability as: (1/8) / (1/8 + 3/8) = 1/4. I don't understand how P(B) = 1/8 + 3/8. Here is the link: https://www.quora.com/A-family-has-three-children-What-is-the-probability-that-they-are-all-boys-given-that-at-least-one-of-them-is-a-boy

Any help in plain, simple language is appreciated!

$\endgroup$

closed as unclear what you're asking by Did, Adam Hughes, C. Falcon, Namaste, астон вілла олоф мэллбэрг Jan 12 '17 at 3:00

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ It seems the "online solution with similar question" really is not so similar after all. We can't explain how $P(B) = 1/8 + 3/8$ (at least, we can't be sure we're correctly explaining it) if you don't edit your text above to show the exact wording of that question (or at least show a link to the online solution). In the absence of a clear statement for that problem, I would make the same guess as Brian Scott. $\endgroup$ – David K Jan 11 '17 at 22:19
  • $\begingroup$ Some of the on-line schools (you did not mention which one is involved here) have pretty poor math problems/answers. The second approach is the only reasonable interpretation of the problem as described. $\endgroup$ – hardmath Jan 12 '17 at 1:43
6
$\begingroup$

The answer that you dismiss as ‘obviously incorrect’ is correct. The fact that the family has at least one girl rules out exactly one of the $8$ otherwise equally likely possibilities, namely the BBB possibility. That means that the actual sample space has $7$ equally likely possibilities, only one of which is $3$ girls, so the probability that the family has $3$ girls is $\frac17$.

The online solution that you mention is either wrong or for a problem that differs in more than just changing girls to boys. It would be correct, for example, for the probability that a family with with three kids, at least two of whom are boys, has three boys.

$\endgroup$
  • $\begingroup$ My online homework accepted 1/4 as an input but claims that 1/7 is incorrect. Not sure why.. $\endgroup$ – J.K Jan 11 '17 at 22:58
  • 1
    $\begingroup$ @J.K: $\frac14$ would be the correct answer if we knew that a specific one of the children was a girl. For instance, if we know that the eldest child is a girl, then the probability that all three are girls is $\frac14$, since the sample space now has just the four elements BB, BG, GB, and GG (second child, third chile). $\endgroup$ – Brian M. Scott Jan 11 '17 at 23:01
  • $\begingroup$ I guess the wording of the problem was a bit ambiguous. Thank you! $\endgroup$ – J.K Jan 11 '17 at 23:08
  • $\begingroup$ @J.K: You’re welcome! $\endgroup$ – Brian M. Scott Jan 11 '17 at 23:09

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