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21 day ago

15

The computers of six faculty members in a certain department are to be replaced. Two of the faculty members have selected laptop

machines and the other four have chosen desktop machines. Suppose that only two of the setups can be done on a particular day, and the two computers to be set up are randomly selected from the six (implying 15 equally likely outcomes; if the computers are numbered 1, 2, . . . , 6, then one outcome consists of computers 1 and 2, another consists of computers 1 and 3, and so on).
Required:
a. What is the probability that both selected setups are for laptop computers?
b. What is the probability that both selected setups are desktop machines?
c. What is the probability that at least one selected setup is for a desktop computer?
d. What is the probability that at least on computer of each type is chosen for setup?

1 answer:

lawyer [9.2K]21 day ago

8 0

Answer:

a. $\frac{1}{15}$

b. $\frac{2}{5}$

c. $\frac{14}{15}$

d. $\frac{8}{15}$

Step-by-step explanation:

There are four desktop computers and two laptops.

On a specific day, we will set up 2 of these computers.

To find:

a. What is the probability that both selected computers are laptops?

b. What is the probability that both computers are desktops?

c. What is the probability that at least one computer is a desktop?

d. What is the probability that at least one of each type of computer is included?

Solution:

Using the probability formula for event E:

$P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}$

a. The number of successful outcomes for both computers being laptops = $_2C_2$ = 1

Total possible outcomes = 15

The needed probability is $\frac{1}{15}$ .

b. The successful outcomes for both being desktop computers = $_4C_2=6$

Total possible outcomes = 15

The required probability is $\frac{6}{15} = \frac{2}{5}$ .

c. For at least one desktop:

Two scenarios exist:

1. 1 desktop and 1 laptop:

Successful outcomes = $_2C_1\times _4C_1 = 8$

2. Both are desktops:

Successful outcomes = $_4C_2=6$

Total successful outcomes = 8 + 6 = 14

The needed probability is $\frac{14}{15}$ .

d. 1 desktop and 1 laptop:

Successful outcomes = $_2C_1\times _4C_1 = 8$

Total outcomes = 15

The required probability is $\frac{8}{15}$ .

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