Question

In the class 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 se- lected, 7 were minorities. The baseline hypothesis, or null, was that the jury selection process was random. The test hypothesis, or alternative, was the process was not random (suggesting bias).
Find the probability of getting 7 or fewer minorities in a world where the jury pool selection pro- cess was supposed to be random.
a. .35400000
b. .00000354
C..00000451
d. 00000000

Answer 1

Answer:

C..00000451

Step-by-step explanation:

For each person in the jury, there are only two possible outcomes. Either they were minorities, or they were not. The probability of a person being a minority is independent of any other person. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Population in which 27% were minorities.

This means that $p = 0.27$

90 people

This means that $n = 90$

Find the probability of getting 7 or fewer minorities in a world where the jury pool selection pro- cess was supposed to be random.

This is

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

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{90,0}.(0.27)^{0}.(0.73)^{90} \approx 0$

$P(X = 1) = C_{90,1}.(0.27)^{1}.(0.73)^{89} \approx 0$

$P(X = 2) = C_{90,2}.(0.27)^{2}.(0.73)^{88} \approx 0$

$P(X = 3) = C_{90,3}.(0.27)^{3}.(0.73)^{87} = \approx 0$

$P(X = 4) = C_{90,4}.(0.27)^{4}.(0.73)^{86} = 0.00000002$

$P(X = 5) = C_{90,5}.(0.27)^{5}.(0.73)^{85} = 0.00000015$

$P(X = 6) = C_{90,6}.(0.27)^{6}.(0.73)^{84} = 0.00000080$

$P(X = 7) = C_{90,7}.(0.27)^{7}.(0.73)^{83} = 0.00000354$

So

$P(X \leq 7) = 0.00000002 + 0.00000015 + 0.00000080 + 0.00000354 = 0.00000451$

So the correct answer is given by option C.