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Suppose that there are six prospective jurors, four men and two women, who might be impaneled to sit on the jury in a criminal case. Two jurors are randomly selected from these six to fill the two remaining jury seats.

a. List the simple events in the experiment
b. What is the probability that both impaneled jurors are women?

Exercise:

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?

Respuesta :

Answer:

(1)

(a) Shown below.

(b) The probability that both impaneled jurors are women is 0.0667.

(2)

(a) Sampling 3 people and noting their gender.

(b) Shown below.

(c) The probability of each simple event is 0.125.

(d) The probability of selecting only one male is 0.375.

(e) The probability of selecting all 3 females is 0.125.

Step-by-step explanation:

(1)

Let the 4 men be denoted as: M₁, M₂, M₃ and M₄.

And the 2 women be denoted as: W₁ and W₂.

(a)

A jury of two is to be selected.

The simple events in this experiment are:

(M₁, M₂), (M₁, M₃), (M₁, M₄), (M₁, W₁), (M₁, W₂)

(M₂, M₃), (M₂, M₄), (M₂, W₁), (M₂, W₂)

(M₃, M₄), (M₃, W₁), (M₃, W₂)

(M₄, W₁), (M₄, W₂)

(W₁, W₂)

(b)

The total possible number of jury selections is, N = 15.

The possible combination such that both the jurors are woman is, n = 1.

Compute the probability of selecting  two women jurors as follows:

[tex]P(2\ juror\ are\ women)=\frac{n}{N} =\frac{1}{15}=0.0667[/tex]

Thus, the probability that both impaneled jurors are women is 0.0667.

(2)

(a)

The experiment consists of sampling 3 people from the voter registration and driving records and noting the gender of each person.

(b)

The simple events are:

S = {(M, M, M), (M, M, F), (M, F, M), (F, M, M), (M, F, F), (F, F, M), (F, M, F), (F, F, F)}

Total number of simple events = 8.

(c)

If the probability of selecting a male is same as the probability of selecting a female, i.e. P (M) = P (F) = [tex]\frac{1}{2}[/tex] then,

The probability of each simple event is:

[tex]\frac{1}{2} \times\frac{1}{2} \times\frac{1}{2}=\frac{1}{8}=0.125[/tex]

(d)

The number of simple events with only 1 male is n = 3.

Compute the probability that only one of the three is a man as follows:

[tex]P(1\ male)=\frac{n}{N} =\frac{3}{8} =0.375[/tex]

Thus, the probability of selecting only one male is 0.375.

(e)

The number of simple events with all 3 females is n = 1.

Compute the probability that all 3 females are selected as follows:

[tex]P(3\ female)=\frac{n}{N} =\frac{1}{8} =0.125[/tex]

Thus, the probability of selecting all 3 females is 0.125.

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