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'' =

85.560%

step-by-step explanation:

conditional probability is defined as:

p(a|b) = p(a∩b) / p(b)

in other words, the probability that a occurs, given that b has occurred = the probability that both a and b occur / the probability that b occurs.

in this example:

p(turns 64 | turns 44) = p(turns 64 & turns 44) / p(turns 44).

if someone turns 64, they must have already turned 44, so:

p(turns 64 | turns 44) = p(turns 64) / p(turns 44)

on page 42, we see that of the original 100,000, 94944 turn 44, and 81234 turn 64.   therefore:

p(turns 64 | turns 44) = 81234 / 94944

p(turns 64 | turns 44) = 0.85560