Boys and Girls

A company hosts a dinner for working mothers who have at least one son. Ms. Jackson, a mother of two, is invited. What is the probability that both of her children are boys?

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#The sample space

Two children, listed by birth order, give four equally likely cases, $BB$ , $BG$ , $GB$ , $GG$ . The invitation tells me only that Ms. Jackson is not the $GG$ case, since she has at least one son. It does not single out a particular child.

#The count

Three cases survive, $BB$ , $BG$ , $GB$ , each still equally likely. Both boys is one of the three, so

\PP(\text{both boys} \mid \text{at least one son}) = \frac{\PP(BB)}{\PP(\text{at least one son})} = \frac{1/4}{3/4} = \frac{1}{3}. \tag{1}

BBboth boys

BGone of each

GBone of each

GGno son

The four birth orders are equally likely. The dinner excludes two girls, leaving three cases, and both boys is one of them, so the probability is one third.

#The tempting wrong turn

The instinct is to say that the other child is a boy or a girl with equal chance, giving a half. That would be right if the dinner had named a child, for instance if I already knew the elder child was a son. Then only $BB$ and $BG$ remain and the answer is a half. Knowing only that some child is a son keeps $GB$ in play, and that extra case drags the probability down to a third.