Question

2. A company is to hire two new employees. They have prepared a

candidates, all of whom are equally qualified. Of these eight candidates, five are women.
Suppose the company decides to select two persons randomly from these eight candidates.
a. What is the probability that:
1. Both candidate selected are women?
2. At least one candidate selected is a woman?
3. Second candidate is a woman.
3
4. First candidate is a woman given that the second candidate is a woman.

b. Let X denote the number of women in this sample.
1. Write the probability distribution of X.
2. Find the standard deviation of X.​

Answer 1

Explained below

Explanation:

X = number of women candidates selected

Of the 8 candidates, 5 are women.

The probability of selecting a women candidate is, p = 5/8.

(a)

(1)

Compute the probability that both candidate selected are women as follows:

P (Both women) = p2 = (5/8)*(4/7) = 0.3571

(2)

Compute the probability that at least one candidate selected is a woman as follows:

P (At least 1 women) = 1 – P (No women)

= 1 – P (Both men)

= 1 – [(3/8)*(2/7)]

= 0.8929

(3)

Compute the probability that the second candidate is a woman as follows:

P (second candidate is a woman) = P (Both Women) + P (1st Man and 2nd Woman)

= (5/8)*(4/7) + (3/8)*(5/7)

= 0.625

(4)

Compute the probability that first candidate is a woman given that the second candidate is a woman as follows:

P (1st Women|2nd Women) = P (Both Women)/P(2nd Women)

= 0.3571/0.625

= 0.2232

(b)

X is defined as the number of women in this sample.

(1)

The sample candidates selected is, n = 8. And the probability of selecting a women is, p = 5/8.

The event of the selecting a female candidate is independent of others.

The random variable X thus follows a binomial distribution with parameters n = 8 and p = 5/8.

(2)

Compute the standard deviation of X as follows:

SD (X) = √(npq) = √(8*(5/8)*(3/8))=1.369