# Tree Diagram Probability [duplicate]

I have a question for practice:

Imagine that you know that your new neighbours have two children, but you don't know whether they are boys or girls or a boy and a girl. Then the mother says, in your hearing, "They were running a special promotion at the store for families with two boys, but we don't have two boys so we couldn't participate." What is the probability the two children are a boy and a girl?

Shouldn't the probability that they have a boy and a girl be 50% or.5? The question says to use a tree diagram but the tree diagram only leads to boy/girl or girl/girl, so 50% chance for either?

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## marked as duplicate by Henry, Austin Mohr, Gerry Myerson, Chris Eagle, Asaf KaragilaOct 12 '12 at 9:09

What about girl/boy? In that order. –  Henry Oct 12 '12 at 1:27
–  Henry Oct 12 '12 at 1:28

Diagrammatically, the original situation is

Each node has a probability of $0.5$ occurring, as you mention. Since you know that they don't have two girls, the other probabilities adjust accordingly. Before each of the four possibilities had equal chances of occurring, i.e. $0.5\cdot 0.5=0.25$. Now that one is eliminated, the other three have equal chances of occurring, i.e. $0.\bar 3$. Since the possibilites are:

• first boy, second boy

• first boy, second girl

• first girl, second boy

And since the last two work for you, your probability is $0.\bar3 +0.\bar 3 = 0.\bar 6 = \frac23$.

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