Question

Pinworm Infestation. Use Procedure 5.1 on page 236 to solve part (g) of Exercise 5.165.

Short answer

The probabilities are:

x	0	1	2	3
$P\left(X=x\right)$	0.001	0.027	0.243	0.729

Step-by-step solution

Step 1. Given information.

The given statement from Exercise 5.165 is:

Pinworm infestation, which is commonly found in children, can be treated with the drug pyrantel pamoate.

According to the Merck Manual, the treatment is effective in 90% of cases. Suppose that three children with pinworm infestation are given pyrantel pamoate.

Step 2. Obtain the probability distribution of the random variable X, the number of children out of three who are cured.

First, identify the success:

Treatment with pyrantel pamoate cures a person with pinworm.

Now determine the probability of success (p):

In 90% of situations, the likelihood of a person being healed by treatment is high.

Therefore, $p=0.9$.

Determine the number of trials (n).

Given that pyrantel pamoate is given to three infants with pinworm infestation. Therefore,$n=3$.

Step 3. Use the binomial probability formula.

The formula is:

$P\left(X=x\right)=\left(\begin{array}{c}n\\ x\end{array}\right)P^x{\left(1-p\right)}^{n-x}$

Insert the values in the formula,

$P\left(X=x\right)=\left(\begin{array}{c}3\\ x\end{array}\right){\left(0.9\right)}^x{\left(1-0.9\right)}^{3-x}$

We can see that X is a binomial random variable with parameters and the binomial distribution $p=0.9\mathrm{and}n=3$.

We must now calculate the probabilities of all Xvalues. That is, Xis a random variable with values ranging from 0 to 3.

Step 4. Use the binomial probability formula.

$$\begin{gathered} P\left(X=0\right)=\left(\begin{array}{c}3\\ 0\end{array}\right){\left(0.9\right)}^0{\left(0.1\right)}^3 \\ ={\left(0.1\right)}^3 \\ =0.001 \\ P\left(X=1\right)=\left(\begin{array}{c}3\\ 1\end{array}\right){\left(0.9\right)}^1{\left(0.1\right)}^2 \\ =3\times 0.9\times {\left(0.1\right)}^2 \\ =0.0270 \\ P\left(X=2\right)=\left(\begin{array}{c}3\\ 2\end{array}\right){\left(0.9\right)}^2{\left(0.1\right)}^1 \\ =3\times {\left(0.9\right)}^2\times 0.1 \\ =0.243 \\ P\left(X=3\right)=\left(\begin{array}{c}3\\ 3\end{array}\right){\left(0.9\right)}^0{\left(0.1\right)}^3 \\ =1\times {\left(0.9\right)}^3 \\ =0.7290 \end{gathered}$$