Chapter 8, Problem 1RE

In Exercises 1–6, find the probability distribution for the given random variable and draw a histogram.

A couple has two children; $X=$ the number of boys. (Assume an equal likelihood of a child being a boy or a girl.)

To determine

To calculate: The probability distribution for the random variable X, if there is a couple that has two children and X is a number of boys, assuming that the chances of a child being a boy or a girl are same.

Answer to Problem 1RE

Solution:

The probability distribution for a number of boys is:

$x$	$0$	$1$	$2$
$P\left(X=x\right)$	$0.25$	$0.5$	$0.25$

Explanation of Solution

Given Information:

There is a couple that has two children and X is a number of boys, assuming that the chances of a child being a boy or a girl are the same.

Formula Used:

If n is the number of trials and X is the number of successes, then the probability for a particular value of X is given by:

$P\left(X=x\right)=C\left(n,x\right)p^x{q}^{n-x}$

Calculation:

Consider the number of children $n=2$,

Evaluate the probability of success which is the probability of a boy.

$\begin{array}{c}p=\frac{1}{2}\\ =0.5\end{array}$

Evaluate the probability of failure which is the probability of a girl.

$\begin{array}{c}q=1-\frac{1}{2}\\ =0.5\end{array}$

Possible values of X being a boy child are $0,1,2$.

Now estimate the probability of each value of n by applying the formula $P\left(x\right)=C\left(n,x\right)p^x{q}^{n-x}$.

Evaluate the probability $P\left(X=0\right)$,

$\begin{array}{c}P\left(X=0\right)=C\left(2,0\right)p^0{q}^{2-0}\\ =\frac{2!}{0!\left(2-0\right)!}\times {\left(0.5\right)}^0\times {\left(0.5\right)}^2\\ =1\times 1\times {\left(0.5\right)}^2\\ =0.25\end{array}$

Evaluate the probability $P\left(X=1\right)$,

$\begin{array}{c}P\left(X=0\right)=C\left(2,1\right)p^1{q}^{2-1}\\ =\frac{2!}{1!\left(2-1\right)!}\times {\left(0.5\right)}^1\times {\left(0.5\right)}^1\\ =2\times \left(0.5\right)\times \left(0.5\right)\\ =0.5\end{array}$

Evaluate the probability $P\left(X=2\right)$,

$\begin{array}{c}P\left(X=0\right)=C\left(2,2\right)p^2{q}^{2-2}\\ =\frac{2!}{2!\left(2-2\right)!}\times {\left(0.5\right)}^2\times {\left(0.5\right)}^0\\ =1\times {\left(0.5\right)}^2\times 1\\ =0.25\end{array}$

The probability distribution for the number of boys is,

$x$	$0$	$1$	$2$
$P\left(X=x\right)$	$0.25$	$0.5$	$0.25$

Graph:

The steps to draw the histogram of the binomial distribution is $n=2,p=0.5$ are shown below:

Step 1: Place x values on the horizontal axis.

Step2. Place the value of probability $P\left(x\right)$ on the vertical axis.

Step3. Construct a bar over each value of x extending from $x-0.5$ to $x+0.5$ equal to the height of the corresponding $P\left(x\right)$.

Draw the graph of the binomial distribution as follows:

Interpretation:

The histogram of the probability distribution of a boy child out of two children where the probability of selecting a girl or boy child is equal.