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 $X$ denote the random variable representing the count of boys in a family with three children. Assuming the probability of having a boy or a girl is $\frac{1}{2}$, the probability distribution is as follows: - The probability of having 0 boys (i.e., all girls) is: \[ P(X = 0) = \left(\frac{1}{2}\right)^3 = \frac{1}{8}. \] - The probability of having 1 boy is: \[ P(X = 1) = \binom{3}{1} \left(\frac{1}{2}\right)^3 = 3 \times \frac{1}{8} = \frac{3}{8}. \] - The probability of having 2 boys is: \[ P(X = 2) = \binom{3}{2} \left(\frac{1}{2}\right)^3 = 3 \times \frac{1}{8} = \frac{3}{8}. \] - The probability of having 3 boys (i.e., no girls) is: \[ P(X = 3) = \left(\frac{1}{2}\right)^3 = \frac{1}{8}. \] The complete probability distribution is: \[ P(X = 0) = \frac{1}{8}, \quad P(X = 1) = \frac{3}{8}, \quad P(X = 2) = \frac{3}{8}, \quad P(X = 3) = \frac{1}{8}. \]