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A couple is planning to have 3 children. Assuming that having a boy and having a girl are equally likely, and that the gender of one child has no influence on (or, is independent of) the gender of another, what is the probability that the couple will have exactly 2 girls?

Now here is my sample space $$S=\{(BBB),(BBG),(BGG),(GGG)\}$$ which would lead me to believe that the probability of $A$ the couple having exactly two girls is $$P(A)=\frac{1}{4}$$ which turns out to be incorrect.

Now they give the sample space as $$S=\{ (BBB), (BBG), (BGB), (GBB), (GGB), (GBG), (BGG), (GGG) \}.$$ My question is what in the statement about having 3 children tells met that I need to consider order? Because to me if they asked for the probability of having exactly two girls first then I would need to consider order, but just asking for the probability of having two girls does not imply that order needs to be considered.

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    $\begingroup$ From your sample space, the probability is $\frac14$, IF the four events are equally likely. What are your reasons for believing this is the case? $\endgroup$
    – David
    Feb 20, 2017 at 23:08
  • $\begingroup$ @David now $P(G)=P(B)=\frac{1}{2}$ which implies $P((BBB))=\frac{1}{8}$ but $P((BBG))=P(BGB)=P(GBB)=P(B)P(B)P(G)=\frac{1}{8}$ therefore my sample space is incorrect because the possible outcomes have to be equally likely. $\endgroup$
    – Andrew
    Feb 20, 2017 at 23:24
  • $\begingroup$ I think that you have more or less answered your own question with that comment, yes? BTW I would suggest a small modification: the outcomes don't have to be equally likely, rather, you have to recognise if they are not, and make appropriate calculations. (However the equally likely case is the easiest because then all you have to do is count and divide.) $\endgroup$
    – David
    Feb 20, 2017 at 23:35

2 Answers 2

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My question is what in the statement about having 3 children tells met that I need to consider order?

Because, you have three distinct children being borne with each birth equally likely to be either sex.

This clearly generates an ordered sequence of three independent choices with two options; often phrased as "Selection with repetition".

Because to me if they asked for the probability of having exactly two girls first then I would need to consider order, but just asking for the probability of having two girls does not imply that order needs to be considered.

That actually should clue you in that order affects the measure.   Because the probability of having two girls and a boy in no particular order must equal the sum of probabilities of having two girls and a boy in each of the three particular orders.

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The reasoning would be that there is more than one way of having exactly two girls. Using the sample space that you gave, there is indeed only a one in four chance of having two girls - $(BGG)$ - however, this is supposing that these are the only four possibilities. Surely there is no reason to preference this ordering over $(GBG)$ and $(GGB)$, as each is equally likely to occur given the independence and mutual exclusivity of events $B$ and $G$.

Furthermore, $(GGG)$ and $(BBB)$ are the only possible ways of having exactly three girls or exactly three boys, so their probability of occurring is surely not the same as that of having exactly two girls as this can be done in three ways. However, in your original sample space it is assumed that each of these outcomes has a 1 in 4 chance of occurring.

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