Question

Axline Computers manufactures personal computers at two plants, one in Texas and the other in Hawaii. The Texas plant has 40 employees; the Hawaii plant has 20. A random sample of 10 employees is to be asked to fill out a benefits questionnaire.

a. What is the probability that none of the employees in the sample work at the plant in Hawaii?
b. What is the probability that one of the employees in the sample works at the plant in Hawaii?
c. What is the probability that two or more of the employees in the sample work at the plant in Hawaii?
d. What is the probability that nine of the employees in the sample work at the plant in Texas?

Answer 1

The probability that no employee in the sample work at the plant in Hawaii is 0.0279, the probability that one of the employees in the sample works at the plant in Hawaii is 0.2656, the probability that two or more of the employees in the sample work at the plant in Hawaii is 0.7065, and probability that nine of the employees in the sample work at the plant in Texas is 0.06.

Step-by-step explanation:

To find the probability that none of the employees in the sample work at the plant in Hawaii, we need to find the probability of selecting 0 employees from the Hawaii plant out of a total sample size of 10 employees. Since there are 20 employees at the Hawaii plant and 60 employees in total, the probability is:

P(selecting 0 employees from Hawaii) = (20C0 * 40C10) / (60C10) = 0.0279

To find the probability that one of the employees in the sample works at the plant in Hawaii, we need to find the probability of selecting 1 employee from the Hawaii plant out of a total sample size of 10 employees. The probability is:

P(selecting 1 employee from Hawaii) = (20C1 * 40C9) / (60C10) = 0.2656

To find the probability that two or more of the employees in the sample work at the plant in Hawaii, we need to find the probability of selecting 2 or more employees from the Hawaii plant out of a total sample size of 10 employees. The probability is:

P(selecting 2 or more employees from Hawaii) = 1 - P(selecting 0 employees from Hawaii) - P(selecting 1 employee from Hawaii) = 1 - 0.0279 - 0.2656 = 0.7065

To find the probability that nine of the employees in the sample work at the plant in Texas, we need to find the probability of selecting 9 employees from the Texas plant out of a total sample size of 10 employees. The probability is:

P(selecting 9 employees from Texas) = (40C9 * 20C1) / (60C10) = 0.06

Answer 2

Answer: a) 0.01822, b) 0.0897, c) 0.892082, d) 0.01822

Step-by-step explanation:

Since we have given that

Number of employees at Texas plant = 40

Number of employees at Hawaii plant = 20

Probability of employees at Texas plant =
`(40)/(60)=0.67`

Probability of employees at Hawaii plant =
`(20)/(60)=0.33`

a. What is the probability that none of the employees in the sample work at the plant in Hawaii?

Here, n = 10

We will use "Binomial distribution":

`P(X=0)=^(10)C_0(0.67)^(10)(0.33)^0=0.01822`

b. What is the probability that one of the employees in the sample works at the plant in Hawaii?

`P(X=1)=^(10)C_1(0.67)^9(0.33)=0.0897`

c. What is the probability that two or more of the employees in the sample work at the plant in Hawaii?

`P(X\geq 2)= 1-P(X=0)-P(X=1)\\\\=1-0.01822-0.0897\\\\=0.892082`

d. What is the probability that nine of the employees in the sample work at the plant in Texas?

It means there would be 1 employee at the plant in Hawaii.

So, it will be

`P(X=1)=^(10)C_1(0.67)^9(0.33)=0.0897`

Hence, a) 0.01822, b) 0.0897, c) 0.892082, d) 0.01822