Answer :
It is required to construct a probability distribution for a discrete random variable of a family that has three children.
Let the random variable X represent the number of boys.
The sample space is given as:
[tex]S=\lbrace GGG,GGB,GBG,BGG,GBB,BGB,BBG,BBB\rbrace[/tex]Since X represents the number of boys, the probability P(X) for X=0, is the probability that the family has 0 boys, that is, GGG:
[tex]P(0)=\frac{number\text{ of favorable outcomes}}{total\text{ number of outcomes}}=\frac{1}{8}=0.125[/tex]Follow the same procedures for one, two, and three boys to construct the table below:
[tex]P(1)=\frac{3}{8}=0.375,\quad P(2)=\frac{3}{8}=0.375,\quad P(3)=\frac{1}{8}=0.125[/tex]Notice that the sum of P(X)=1. This must be so since it is a probability distribution.
The probability of getting at least one boy is:
[tex]P(X\geqslant1)=P(1)+P(2)+P(3)[/tex]Substitute the probabilities:
[tex]P(X\geqslant1)=0.375+0.375+0.125=0.875[/tex]The required probability is 0.875
