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Chapter 3: Q76E (page 130)

A family decides to have children until it has three children of the same gender. Assuming P(B) = P(G) =\({\rm{.5}}\), what is the pmf of X = the number of children in the family?

Short Answer

Expert verified

The pmf is obtained as: \(\begin{aligned}{\rm{P(X = 3)}}\\&{\rm{ = }}\frac{{\rm{1}}}{{\rm{4}}}\\&{\rm{ = 0}}{\rm{.25}}\\{\rm{P(X = 4)}}\\&{\rm{ = }}\frac{{\rm{3}}}{{\rm{8}}}\\&{\rm{ = 0}}{\rm{.375}}\\{\rm{P(X = 5)}}\\&{\rm{ = }}\frac{{\rm{3}}}{{\rm{8}}}\\&{\rm{ = 0}}{\rm{.375}}\end{aligned}\).

Step by step solution

01

Define Discrete random variables

A discrete random variable is one that can only take on a finite number of different values.

02

Step 2: Evaluating the pmf

It is given that:

\(\begin{aligned}{\rm{p = P(B)}}\\&{\rm{ = P(G)}}\\&{\rm{ = 0}}{\rm{.5}}\end{aligned}\)

If the family intends to have further children until it has three children of the same gender, the family must have at least three children and a maximum of five children.

Until the family has three children of the same gender, all feasible combinations of boys and girls are:

\(\begin{array}{c}{\rm{\{ BBB,GGG,BBGB,BGBB,GBBB,GGBG,GBGG,BGGG,}}\\{\rm{BBGGB,BGBGB,BGGBB,GBGBB,GBBGB,GGBBB,}}\\{\rm{BBGGG,BGBGG,BGGBG,GGBBG,GBGBG,GBBGG\} }}\end{array}\)

Then we see that there are two combinations that result in three children, six combinations that result in four children, and twelve combinations that result in five children.

For independent events, use the following multiplication rule:

\({\rm{P(A and B) = P(A) \times P(B)}}\)

We may apply the multiplication formula for separate events if each child is born independently of the other children:

\(\begin{aligned}{\rm{P(BBB) = P(GGG)}}\\&{\rm{ = }}{{\rm{p}}^{\rm{3}}}\\&{\rm{ = 0}}{\rm{.}}{{\rm{5}}^{\rm{3}}}\\&{\rm{ = 0}}{\rm{.125}}\\{\rm{P(BBGB) = \ldots }}\\&{\rm{ = P(BGGG)}}\\&{\rm{ = }}{{\rm{p}}^{\rm{4}}}\\&{\rm{ = 0}}{\rm{.}}{{\rm{5}}^{\rm{4}}}\\&{\rm{ = 0}}{\rm{.0625}}\\{\rm{P(BBGGB) = \ldots }}\\&{\rm{ = P(GBBGG)}}\\&{\rm{ = }}{{\rm{p}}^{\rm{5}}}\\&{\rm{ = 0}}{\rm{.}}{{\rm{5}}^{\rm{5}}}\\&{\rm{ = 0}}{\rm{.03125}}\end{aligned}\)

Let X be the family's total number of children. Add up the probability for each conceivable combination:

\(\begin{aligned}{\rm{P(X = 3)}}\\&{\rm{ = P(BBB) + P(GGG)}}\\&{\rm{ = 0}}{\rm{.125 + 0}}{\rm{.125}}\\&{\rm{ = 0}}{\rm{.25}}\\&{\rm{ = }}\frac{{\rm{1}}}{{\rm{4}}}\\{\rm{P(X = 4)}}\\&{\rm{ = P(BBGB) + \ldots + P(BGGG)}}\\&{\rm{ = 6P(BBGB)}}\\&{\rm{ = 6(0}}{\rm{.0625)}}\\&{\rm{ = 0}}{\rm{.375}}\\&{\rm{ = }}\frac{{\rm{3}}}{{\rm{8}}}\end{aligned}\)

\(\begin{aligned}{\rm{P(X = 5)}}\\&{\rm{ = P(BBGGB) + \ldots + P(GBBGG)}}\\&{\rm{ = 12P(BBGGB)}}\\&{\rm{ = 12(0}}{\rm{.03125)}}\\&{\rm{ = 0}}{\rm{.375}}\\&{\rm{ = }}\frac{{\rm{3}}}{{\rm{8}}}\end{aligned}\)

Therefore, the values are:

\(\begin{aligned}{\rm{P(X = 3)}}\\&{\rm{ = }}\frac{{\rm{1}}}{{\rm{4}}}\\&{\rm{ = 0}}{\rm{.25}}\\{\rm{P(X = 4)}}\\&{\rm{ = }}\frac{{\rm{3}}}{{\rm{8}}}\\&{\rm{ = 0}}{\rm{.375}}\\{\rm{P(X = 5)}}\\&{\rm{ = }}\frac{{\rm{3}}}{{\rm{8}}}\\&{\rm{ = 0}}{\rm{.375}}\end{aligned}\)

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