A couple plans to have children until they get a boy,but they agree that they will not have more than four children even if all are girls. Create a probability model for the number of children they will have.Assume that boys and girls are equally likely.

A) $$\begin{array}{lcccc}\text{ Children }& 1& 2& 3& 4\\ \text{ P(Children) }& 0.5& 0.25& 0.125& 0.125\end{array}$$
B) $$\begin{array}{lccccc}\text{ Children }& 1& 2& 3& 4& 5\\ \text{ P(Children) }& 0.5& 0.25& 0.125& 0.0625& 0.0625\end{array}$$
C) $$\begin{array}{lccc}\text{ Children }& 1& 2& 3\\ \mathrm{P}\text{ (Children) }& 0.5& 0.25& 0.25\end{array}$$
D) $$\begin{array}{lcccc}\text{ Children }& 1& 2& 3& 4\\ \mathrm{P}\text{ (Children) }& 0.5& 0.25& 0.125& 0.0625\end{array}$$
E) $$\begin{array}{lcccc}\text{ Children }& 1& 2& 3& 4\\ \text{ P(Children) }& 0.25& 0.25& 0.25& 0.25\end{array}$$

The probability of having a boy (and thus stopping) is 0.5 for the first child. If the first is a girl, the probability of having a boy as the second child is 0.5, but we must also account for the probability of the first being a girl (0.5), making it 0.5 * 0.5 = 0.25 for the second child. This pattern continues, with the probability halving each time. However, if they have three girls in a row (probability 0.125), they will stop at four children regardless, making the probability of having four children 0.125, not halving again because they stop regardless of the fourth child's gender.