A shipment of 20 similar laptop computers to a retail outlet contains 3 that are defective. If a school makes a random purchase of 2 of these computers, find the probability mass function for the number of defectives.

Step-by-step explanation:

Total number of laptops, N = 20

Defective laptops, n = 3

School purchase 2 laptops

Case 1:

Both laptops are non defective

Probability to choose 2 laptops which are non defective

Number of non defective laptops = 20 - 3 = 17

Total number of laptops = 20

Probability, P = 17/20 = 0.85

Case 2:

One laptop is defective

Probability to chose 1 defective and one non defective laptop

P' = 3/20 x 17/19 = 0.134

Case 3:

Both laptops are defective

Choosing of 2 laptops out of 3 = 3 C 2 = 3

choosing of 2 laptops out of 20 = 20 C 2 = 190

Probability of choosing two defective laptops = 3 / 190 = 0.0158

P'' =

1. 22

2. 1

3. 7

4. 18

step-by-step explanation

e2020 got it right

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!