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

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  = $\dfrac{40}{60}=0.67$

Probability of employees at Hawaii plant  = $\dfrac{20}{60}=0.33$

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