# Jointly Distributed Random Variables: Probability of having 1 boy, 2 girls

Suppose that $15\%$ of families in a community have no children, $20\%$ have 1, $35\%$ have 2, and $30\%$ have 3 children. Suppose that it is equally likely and independently to have a boy or a girl. If a family is chosen at random from the community, then B, the number of boys, and G, the number of girls will have the joint probability mass function values:

How did my textbook arrive at the probability $0.1125$ for having $2$ boys and $1$ girl (or $2$ girls and $1$ boy)?

I've calculated the values for the other possibilities:

$P\{B=0,G=0\}=0.15$ (given)

$P\{B=1,G=0\}=P\{B=0,G=1\}=0.2\times0.5=0.1$

$P\{B=1,G=1\}=0.35\times0.5=0.175$

$P\{B=2,G=0\}=P\{B=0,G=2\}=0.35\times0.5^2=0.875$

$P\{B=3,G=0\}=P\{B=0,G=3\}=0.30\times0.5^3=0.0375$

It seems that $P\{B=2,G=1\}=P\{B=1,G=2\}=0.30\times0.5^2=0.075$, the probability of having $2$ boys (OR $2$ girls) AND having 3 children. Shouldn't that be the same as the probability of having $1$ girl (OR $1$ boy) AND having 3 children? But $0.30\times0.5^2=0.075$ does not equal $0.3\times0.5=0.15$...

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Given that $B+G=3$, the probability of having $B=2, G=1$ is given by the binomial coefficient: $$P\{B=2,G=1|B+G=3\}=2^{-3}\left(\begin{array}{c}3\\2\end{array}\right)=2^{-3} 3=0.375$$
Therefore, $$P\{B=2,G=1\}=P\{B=2,G=1|B+G=3\}P\{B+G=3\}=0.375\times0.3=0.1125$$