# Let the random variable X denote the number of girls in a five-child family. If the probability of a female birth is 0.5, what are the probabilities of having 0, 1, 2, 3, 4, and 5 girls?

See below:

#### Explanation:

The probability of having a girl is given as $0.5$. This means that the probability of not having a girl is also $0.5$.

The last thing to look at is the number of ways we can achieve any given result. For instance, there is only 1 way to have 0 girls (and the same goes for 5 girls), but there are 5 different ways we can have 1 girl (and the same goes for 4 girls). This is a combinations question, with:

C_(n,k)=((n),(k))=(n!)/((k!)(n-k)!) with $n = \text{population", k="picks}$

Boiling it all down, we end up with:

0 girls $= \left(\begin{matrix}5 \\ 0\end{matrix}\right) {\left(\frac{1}{2}\right)}^{0} {\left(\frac{1}{2}\right)}^{5} = 1 \times 1 \times \frac{1}{32} = \frac{1}{32}$

1 girl $= \left(\begin{matrix}5 \\ 1\end{matrix}\right) {\left(\frac{1}{2}\right)}^{1} {\left(\frac{1}{2}\right)}^{4} = 5 \times \frac{1}{2} \times \frac{1}{16} = \frac{5}{32}$

2 girls $= \left(\begin{matrix}5 \\ 2\end{matrix}\right) {\left(\frac{1}{2}\right)}^{2} {\left(\frac{1}{2}\right)}^{3} = 10 \times \frac{1}{4} \times \frac{1}{8} = \frac{10}{32} = \frac{5}{16}$

3 girls $= \left(\begin{matrix}5 \\ 3\end{matrix}\right) {\left(\frac{1}{2}\right)}^{3} {\left(\frac{1}{2}\right)}^{2} = 10 \times \frac{1}{8} \times \frac{1}{4} = \frac{10}{32} = \frac{5}{16}$

4 girls $= \left(\begin{matrix}5 \\ 4\end{matrix}\right) {\left(\frac{1}{2}\right)}^{4} {\left(\frac{1}{2}\right)}^{1} = 5 \times \frac{1}{16} \times \frac{1}{2} = \frac{5}{32}$

5 girls $= \left(\begin{matrix}5 \\ 5\end{matrix}\right) {\left(\frac{1}{2}\right)}^{5} {\left(\frac{1}{2}\right)}^{0} = 1 \times \frac{1}{32} \times 1 = \frac{1}{32}$

Notice that if we add all of these up, we get 1 (or the sum total of all probabilities):

$\frac{1}{32} + \frac{5}{32} + \frac{10}{32} + \frac{10}{32} + \frac{5}{32} + \frac{1}{32} = \frac{32}{32} = 1$