Probability from a Sample Space. In Exercises 33–36, use the given sample space or construct the required sample space to find the indicated probability.

Three Children Use this sample space listing the eight simple events that are possible when a couple has three children (as in Example 2 on page 135): {bbb, bbg, bgb, bgg, gbb, gbg, ggb, ggg}. Assume that boys and girls are equally likely, so that the eight simple events are equally likely. Find the probability that when a couple has three children, there is exactly one girl.

The possible combinations of children when a couple has three children is given as (bbb, bbg, bgb, bgg, gbb, gbg, ggb, ggg), where b means a boy, and g means a girl.

Theprobability of an event is a value that measures the chance of an event happening. It has the given formula:

$P\left(A\right)=\frac{\text{Number}\text{of}\text{favorable}\text{outcomes}\text{of}\text{A}}{\text{Total}\text{number}\text{of}\text{outcomes}}$

The set of all outcomes of an event is called thesample space.

Let S be the sample space for the genders of three children as shown below:

S =(bbb, bbg, bgb, bgg, gbb, gbg, ggb, ggg), where b represents a boy, and g represents a girl.

The total number of combinations of genders = 8

The number of combinations that contain only one girl is equal to three. They are (bbg, gbb, bgb).

The probability of event E; only one girl out of three children is calculated as shown:

$$\begin{gathered} P\left(E\right)=\frac{3}{8} \\ =0.375 \end{gathered}$$

Therefore, the probability of only one girl out of three children is equal to 0.375.