 Solution

Treat the four fixed row totals as four independent samples from binomial distributions. Under independence, each has the same unknown probability $p$ of a third-child boy. The pooled maximum-likelihood estimator is $\hat{p}=100/200=1/2$, so each expected cell count is $25$. The Pearson chi-squared test of homogeneity statistic isThere are four independently varying row probabilities under the alternative, and one estimated common probability under the null; the degrees of freedom are therefore $4-1=3$. Equivalently this is the $(4-1)(2-1)$ contingency-table count. The fact that the pooled estimate happens to equal one half does not make it a specified null probability and does not change the degrees of freedom to four.

Since $10.08>7.8147$, reject independence at the 5% level using the stated chi-squared calibration. All expected counts are sufficiently large for the usual asymptotic test.