Questions: A family has three children. If the genders of these children are listed in the order they are born, there are eight possible outcomes: BBB, BBG, BGB, BGG, GBB, GBG, GGB, and GGG. Assume these outcomes are equally likely. Let x represent the number of children that are girls. Find the probability distribution of x. Part 1 of 2 (a) Find the number of possible values for the random variable X. There are possible values for the random variable X.

Solution Steps

The random variable \( X \) represents the number of girls in a family with three children. The possible outcomes for the genders of the children are:

• \( BBB \) (0 girls)
• \( BBG \) (1 girl)
• \( BGB \) (1 girl)
• \( BGG \) (2 girls)
• \( GBB \) (1 girl)
• \( GBG \) (2 girls)
• \( GGB \) (2 girls)
• \( GGG \) (3 girls)

Counting the unique values of \( X \), we find that there are 4 possible values: \( 0, 1, 2, 3 \).

The mean \( \mu \) of the distribution is calculated as follows:

\[ \mu = E(X) = 0 \times 0.125 + 1 \times 0.125 + 1 \times 0.125 + 2 \times 0.125 + 1 \times 0.125 + 2 \times 0.125 + 2 \times 0.125 + 3 \times 0.125 = 1.5 \]

The variance \( \sigma^2 \) is calculated using the formula:

\[ \sigma^2 = E(X^2) - (E(X))^2 \]

Calculating \( E(X^2) \):

\[ E(X^2) = (0 - 1.5)^2 \times 0.125 + (1 - 1.5)^2 \times 0.125 + (1 - 1.5)^2 \times 0.125 + (2 - 1.5)^2 \times 0.125 + (1 - 1.5)^2 \times 0.125 + (2 - 1.5)^2 \times 0.125 + (2 - 1.5)^2 \times 0.125 + (3 - 1.5)^2 \times 0.125 = 0.75 \]

Thus, the variance is:

\[ \sigma^2 = 0.75 \]

The standard deviation \( \sigma \) is the square root of the variance:

\[ \sigma = \sqrt{\sigma^2} = \sqrt{0.75} \approx 0.866 \]

Final Answer