Respuesta :
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 = [tex]\frac{1}{15}=0.067[/tex]
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 =[tex]^4C_2=\frac{4!}{2!(4-2)!}=6[/tex]
So,the probability that both selected setups are desktop machines=[tex]\frac{6}{15}=0.4[/tex]
(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 [tex]=^4C_1=\frac{4!}{1!(4-1)!}=4[/tex]
No. of ways to select one laptop =[tex]^2C_1=\frac{2!}{1!(2-1)!}=2[/tex]
So, No. of ways to select one laptop and one desktop= [tex]4 \times 2 = 8[/tex]
So,the probability that at least on computer of each type is chosen for setup=[tex]\frac{8}{15}=0.533[/tex]
