Question 12 of 12

Question 12

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Each birth is independently a boy or girl with probability $\tfrac12$ each; a family stops having children immediately after its first boy.

Find. E[children per family], E[girls per family], E[boys per family], and whether the long-run population sex ratio is imbalanced.

Approach. Model the number of children as a Geometric($p=\tfrac12$) random variable counting trials until the first "success" (a boy); use the standard geometric-distribution mean formula and its "number of failures before success" companion, then reason about aggregate expectations across many independent families.

1. Part (a) — expected number of children. The number of children N in a family is Geometric($p=\tfrac12$): $P(N=k)=(1-p)^{k-1}p$ for $k=1,2,3,\ldots$ (k-1 girls followed by 1 boy). Its mean is $$\boxed{E[N] = \frac1p = \frac{1}{1/2}=2 \text{ children per family}}$$
2. Part (b) — expected number of girls. Every family has exactly $N-1$ girls (all children before the terminal boy), so $$\boxed{E[\text{girls}] = E[N]-1 = 2-1 = 1 \text{ girl per family}}$$ (equivalently, the number of girls follows a Geometric distribution counting "failures before the first success," with known mean $(1-p)/p = (1/2)/(1/2)=1$).
3. Part (c) — expected number of boys. By construction every family has exactly one boy (the family stops the instant a boy is born, never before, never after): $$\boxed{E[\text{boys}] = 1 \text{ boy per family, always exactly 1}}$$
4. Part (d) — long-run population balance. Across a large number of independent families, the expected total girls equals (number of families) × 1, and the expected total boys equals (number of families) × 1 — the two expected totals are equal. So despite the stopping rule appearing to "favor" ending on a boy at the family level, it produces no long-run imbalance: the underlying coin flip for every single birth (boy/girl) remains an independent fair $\tfrac12$/$\tfrac12$ event regardless of when the family chooses to stop, and aggregating independent fair coin flips across the whole population preserves the 1:1 ratio. $$\boxed{\text{No imbalance --- the population-wide boy:girl ratio stays } 1:1 \text{ in the long run}}$$