In Exercises 25–28, find the probabilities and answer the questions.

Whitus v. Georgia In the classic legal case of Whitus v. Georgia, a jury pool of 90 people was supposed to be randomly selected from a population in which 27% were minorities. Among the 90 people selected, 7 were minorities. Find the probability of getting 7 or fewer minorities if the jury pool was randomly selected. Is the result of 7 minorities significantly low? What does the result suggest about the jury selection process?

A sample of 90 people is selected. It is given that 7 of them were minorities.

27% of the people were minorities.

Let X denote the minorities.

Let success be defined as selecting a minority out of the given sample.

The number of trials (n) is given to be equal to 90.

The probability of success is given as follows:

$$\begin{gathered} p=27\% \\ =\frac{27}{100} \\ =0.27 \end{gathered}$$

The probability of failure is given as follows:

$$\begin{gathered} q=1-p \\ =1-0.27 \\ =0.73 \end{gathered}$$

The number of successes required in 90 trials should be less than or equal to 7.

The binomial probability formula is as follows:

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

Using the binomial probability formula, the probability of getting less than or equal to 7 minorities is computed below:

$P\left(X⩽7\right)=P\left(X=0\right)+P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right)+P\left(X=4\right)+P\left(X=5\right)+P\left(X=6\right)+P\left(X=7\right)$

The probabilities of the individual terms can be computed in the following manner:

$$\begin{gathered} P\left(X=0\right)={}^{90}{C}_0{\left(0.27\right)}^0{\left(0.73\right)}^{90} \\ =0.0000000000005 \\ P\left(X=1\right)={}^{90}{C}_1{\left(0.27\right)}^1{\left(0.73\right)}^{89} \\ =0.00000000001665 \\ . \\ . \\ . \\ P\left(X=7\right)={}^{90}{C}_7{\left(0.27\right)}^7{\left(0.73\right)}^{83} \\ =0.0000035 \end{gathered}$$

By adding up the individual probabilities, the following value is obtained:

$$\begin{gathered} P\left(X⩽7\right)=P\left(X=0\right)+P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right)+P\left(X=4\right)+P\left(X=5\right)+P\left(X=6\right)+P\left(X=7\right) \\ =0.0000045 \end{gathered}$$

Therefore, the probability of selecting less than or equal to 7 minorities is equal to 0.0000045.

The number of successes (x) of a binomial probability value is said to be significantly low if$P\left(x\text{or}\text{fewer}\right)⩽0.05$.

Here, the number of successes (minorities) is equal to 7.

$$\begin{gathered} P\left(7\text{or}\text{fewer}\right)=0.0000045 \\ ⩽0.05 \end{gathered}$$

Thus, it can be said that 7 minorities out of 90 people can be considered a significantly low number of successes.

Since a total of 7 minorities out of 90 people is a significantly low value, it can be said that it is highly unlikely that the process will select fewer than or equal to 7 minorities in a panel of 90 people.