Respuesta :
Answer:
0.000004 = 0.0004% probability of getting 7 or fewer minorities if the jury pool was randomly selected.
Step-by-step explanation:
For each person, there are only two possible outcomes. Either it was a minority, or it was not. The probability of a person being a minority is independent of any other person. This means that the binomial probability distribution is used 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.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
Jury of 90 people:
This means that [tex]n = 90[/tex]
27% were minorities.
This means that [tex]p = 0.27[/tex]
Find the probability of getting 7 or fewer minorities if the jury pool was randomly selected.
This is:
[tex]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)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{90,0}.(0.27)^{0}.(0.73)^{90} \approx 0[/tex]
[tex]P(X = 1) = C_{90,1}.(0.27)^{1}.(0.73)^{89} \approx 0[/tex]
[tex]P(X = 2) = C_{90,2}.(0.27)^{2}.(0.73)^{88} \approx 0[/tex]
[tex]P(X = 3) = C_{90,3}.(0.27)^{3}.(0.73)^{87} \approx 0[/tex]
[tex]P(X = 4) = C_{90,4}.(0.27)^{4}.(0.73)^{86} \approx 0[/tex]
[tex]P(X = 5) = C_{90,5}.(0.27)^{5}.(0.73)^{85} \approx 0[/tex]
[tex]P(X = 6) = C_{90,6}.(0.27)^{6}.(0.73)^{84} \approx 0[/tex]
[tex]P(X = 7) = C_{90,7}.(0.27)^{7}.(0.73)^{83} = 0.000004[/tex]
So
[tex]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) = 7*0 + 0.000004[/tex]
0.000004 = 0.0004% probability of getting 7 or fewer minorities if the jury pool was randomly selected.
