Solution to Exercise 3.69.

  Let X be the number of male children (``failures'') before there are two female childres (``successes''). Then X is Negative Binomial with parameters r = 2 and p = 0.5.

(a) Applying the Proposition onf page 131,

$$P[ X = x ] \; = \; \left( \begin{array} {c} x+1 \\ 1 \end{array} \right) (.5)^2 (1-.5)^x \; = \; (x+1)(.5)^{x+2} .$$

(b) If the family has 4 children, then 2 are male and 2 are female, so we are to find

$$P[ X = 2 ] \; = \; 3*(.5)^4 \; = \; 0.1875 .$$

(c) ``At most 4 children'' is the same as ``at most 2 males.'' First computing the probabilities of and 1 males before the requisite 2 females:

$$P[ X = 0 ] \; = \; (.5)^2 = 0.25 , \quad P[ X = 1 ] \; = \; 2*(.5)^3 = 0.25 .$$

So the probability of ``At most 4 children'' is

$$P[ X \le 2 ] \; = \; 0.1875 + 0.25 + 0.25 \; = \; 0.6875 .$$

(d)

$$E[X] \; = \; \frac{2*(1-.5)}{.5} \; = \; 2 .$$

The total number of children will be X+2, so the expected number of children will be

$$E[X+2] \; = \; E[X] + 2 \; = \; 4 .$$