Question

Jury Duty Three people are randomly selected from voter registration and driving records to report for jury duty. The gender of each person is noted by the county clerk.
a. Define the experiment.
b. List the simple events in S.
c. If each person is just as likely to be a man as a woman, what probability do you assign to each simple event?
d. What is the probability that only one of the three is a man?
e. What is the probability that all three are women?

Answer 1

Answer:

(a) The experiment defined here is a random variable that includes the selecting of 3 people from the set of voter registration and driving records.

(b) The simple events in sample space, S = (M, M, M), (M, F, M), (M, M, F), (F, M, M), (F, M, F), (F, F, M), (M, F, F), and (F, F, F).

(c) If each person is just as likely to be a man as a woman, then the probability for each of the simple event can be assigned as $0.5 \times 0.5 \times 0.5 = 0.125$.

(d) The probability that only one of the three is a man is 0.375.

(e) The probability that all three are women is 0.125.

Step-by-step explanation:

We are given that three people are randomly selected from voter registration and driving records to report for jury duty. The gender of each person is noted by the county clerk.

(a) The experiment defined here is a random variable that includes the selecting of 3 people from the set of voter registration and driving records.

(b) As we know that the gender of each person is noted by the county clerk, which means one is male and another female.

So, the simple events in sample space, S = (M, M, M), (M, F, M), (M, M, F), (F, M, M), (F, M, F), (F, F, M), (M, F, F), and (F, F, F).

Here, M is denoted for male and F for female.

(c) If each person is just as likely to be a man as a woman, then the probability for each of the simple event can be assigned as $0.5 \times 0.5 \times 0.5 = 0.125$.

Because there is 50-50 chance of selecting males or females.

(d) The probability that only one of the three is a man is given by;

The total cases in the sample space = 8

Number of cases of only one man out of three = 3

So, the required probability =  $\frac{3}{8}$ = 0.375.

(e) The probability that all three are women is given by;

The total cases in the sample space = 8

Number of cases of all three are women = 1

So, the required probability =  $\frac{1}{8}$ = 0.125.