Let $X$ be the random variable representing the number of boys in a family with 3 children. Since the probability of having a boy or a girl is equal (i.e., $\frac{1}{2}$), we have the following probability distribution:
- Probability of having 0 boys (all girls):
\[
P(X = 0) = \left(\frac{1}{2}\right)^3 = \frac{1}{8}.
\]
- Probability of having 1 boy:
\[
P(X = 1) = \binom{3}{1} \left(\frac{1}{2}\right)^3 = 3 \times \frac{1}{8} = \frac{3}{8}.
\]
- Probability of having 2 boys:
\[
P(X = 2) = \binom{3}{2} \left(\frac{1}{2}\right)^3 = 3 \times \frac{1}{8} = \frac{3}{8}.
\]
- Probability of having 3 boys (no girls):
\[
P(X = 3) = \left(\frac{1}{2}\right)^3 = \frac{1}{8}.
\]
Thus, the probability distribution is:
\[
P(X = 0) = \frac{1}{8}, \quad P(X = 1) = \frac{3}{8}, \quad P(X = 2) = \frac{3}{8}, \quad P(X = 3) = \frac{1}{8}.
\]
