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Here's my exercise:

A family has three children. What's the probability of event $A \cup C$ where:

$A$- Family has children of both sexes.

$C$- Family has at most one girl.

Well I see two ways to look at this conundrum:

The first one would be to differentiate possibilities by triples, $B$ meaning boy, $D$ meaning girl. So we have eight possibilities: $BBB,BBD,BDB,DBB,BDD,DDB,DBD,DDD$. So the answer would be $P(A \cup C)=\frac{7}{8}$.

And this is correct according to solutions in my textbook. Yet my first solution was this, since they ask about if the family has two boys, and not about whether the first and third child are boys, then $BDB$ and $BBD$ are indistinguishable, then we have only four possibilities: one girl and two boys, three boys, three girls, one boy and two girls. Then $P(A \cup C)=\frac{3}{4}$.

Why was my first intuition wrong?

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2 Answers 2

up vote 2 down vote accepted

The only order that fails $A \cup C$ is $DDD$, one out of eight, so $\frac 78$ is correct. Your other approach misses the fact that (unordered) $BDD$ and $BBD$ are three time more likely than $BBB$ and $DDD$

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You distinguish between four possibilities: $BBB, BBD, BDD, DDD$. In three out of four cases the event $A \cup C$ occurs. However this only implies $P(A \cup C) = \frac{3}{4}$ when each of the four cases are equally likely, which is not the case here.

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Ah yes, of course :) Thanks for removing my doubts :) –  Arek Krawczyk Oct 25 '13 at 16:11
    
This is not correct as BBD and BDD are three times as likely as the other two. –  Ross Millikan Oct 25 '13 at 16:14
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