Question

Find the probability distribution of the number of boys in families having three children, assuming equal probability for a boy and a girl.

Hint:

The binomial distribution can be used to model the number of successes (e.g., having a boy) in a fixed number of trials (e.g., children).

Answer 1

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: \[ P(X = 0) = \frac{1}{8}, \quad P(X = 1) = \frac{3}{8}, \quad P(X = 2) = \frac{3}{8}, \quad P(X = 3) = \frac{1}{8}. \]