Question

Construct a probability distribution for a discrete random variable uses the probable pity experiment of a family has three children. Consider the random variable for the number of boys. Find the probability of getting at least one boy in the family

Answer 1

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:

`S=\lbrace GGG,GGB,GBG,BGG,GBB,BGB,BBG,BBB\rbrace`

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:

`P(0)=\frac{number\text{ of favorable outcomes}}{total\text{ number of outcomes}}=(1)/(8)=0.125`

Follow the same procedures for one, two, and three boys to construct the table below:

`P(1)=(3)/(8)=0.375,\quad P(2)=(3)/(8)=0.375,\quad P(3)=(1)/(8)=0.125`

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:

`P(X\geqslant1)=P(1)+P(2)+P(3)`

Substitute the probabilities:

`P(X\geqslant1)=0.375+0.375+0.125=0.875`

The required probability is 0.875