USDA chicken inspection. In Exercise 3.19 (p. 170), you learned that one in every 100 slaughtered chickens passes USDA inspection with fecal contamination. Consider a random sample of three slaughtered chickens that all pass USDA inspection. Let x equal the number of chickens in the sample that has fecal contamination.

1. Find $\mathbf{p}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$for x = 0, 1, 2, 3.
2. Graph $\mathbf{p}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$.
3. Find $\mathbf{P}\mathbf{(}\mathbf{x}\mathbf{\le }\mathbf{\text{1}}\mathbf{)}$.

Let C and N be denoted as contaminated and uncontaminated chickens, respectively, and accordingly, a list of outcomes of 3 randomly picked slaughtered chickens are shown below

The calculations $\mathbf{p}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ for each of the values of x representing the number of contaminated slaughtered chickens are shown below.

Therefore, for x equal to 0 and 1, the probability is $\frac{\mathbf{\text{1}}}{\mathbf{\text{8}}}$, and for x equal to 1 and 2, it is $\frac{\mathbf{\text{3}}}{\mathbf{\text{8}}}$.

Here the values of x denote the number of contaminated chickens being randomly picked, and accordingly, the probability distribution table is shown below:

In the graph, the vertical axis consists of the probability distribution values, and the horizontal axis consists of the number of contaminated chickens as represented by x.The blue bars represent the probability distributions of the number of contaminated chickens.

In this case, for simplicity, C and N have been chosen to denote contaminated and uncontaminated chickens, respectively, as shown below

$\mathbf{P}\mathbf{(}\mathbf{x}\mathbf{\le }\mathbf{\text{1}}\mathbf{)}$means that the three chickens chosen will contain a maximum of one contaminated chicken, and it can be observed that only four types of combinations can occur.

Therefore, by dividing 4 by 8, it $\mathbf{P}\mathbf{(}\mathbf{x}\mathbf{\le }\mathbf{\text{1}}\mathbf{)}$comes out to be$\frac{\mathbf{1}}{\mathbf{2}}$.