Question

The computers of six faculty members in a certain department are to be re- placed. two of the faculty members have selected laptop machines and the other four have chosen desktop machines. suppose that only two of the se- tups 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 com- puters are numbered 1,2,...,6, then one outcome consists of computers 1 and 2, another consists of computers 1 and 3, and so on).
(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?

Answer 1

Answer:

A) The probability that both selected setups are for laptop computers is 0.067

B)The probability that both selected setups are desktop machines is 0.4

C)The probability that at least one selected setup is for a desktop computer is 0.933

D)The probability that at least on computer of each type is chosen for setup is 0.533

Step-by-step explanation:

Number of laptops = 2

Number of desktops = 4

Total number of outcomes = 15

a) what is the probability that both selected setups are for laptop computers?

Total number of outcomes = 15

So, the probability that both selected setups are for laptop computers = $\frac{1}{15}=0.067$

b)what is the probability that both selected setups are desktop machines?

Number of desktops = 4

Number of desktops to be chosen = 4

We will use combination

No. of ways to select two desktops =$^4C_2=\frac{4!}{2!(4-2)!}=6$

So,the probability that both selected setups are desktop machines=$\frac{6}{15}=0.4$

(c) what is the probability that at least one selected setup is for a desktop computer?

P(at least 1 desktop)=1-P(No desktop)

P(at least 1 desktop)=1-P(both laptops)

P(at least 1 desktop)=1-0.067=0.933

So,the probability that at least one selected setup is for a desktop computer is 0.933

d) what is the probability that at least on computer of each type is chosen for setup?

No. of ways to select one desktop $=^4C_1=\frac{4!}{1!(4-1)!}=4$

No. of ways to select one laptop =$^2C_1=\frac{2!}{1!(2-1)!}=2$

So, No. of ways to select one laptop and one desktop= $4 \times 2 = 8$

So,the probability that at least on computer of each type is chosen for setup=$\frac{8}{15}=0.533$