Question

during a particular period a university's information technology office received 20 service orders for problems with printers, of which 8 were laser printers and the rest are inkjet printers a sample of 5 of these service orders is to be selected for inclusion in a customer satisfaction survey what is the probability that 3 were laser printers

Answer 1

0.2384 = 23.84% probability that 3 were laser printers.

Explanation:

The printers will be chosen without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

`P(X = x) = h(x,N,n,k) = (C_(k,x)*C_(N-k,n-x))/(C_(N,n))`

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

In this question:

20 orders means that
`N = 8`

8 were laser printers, which means that
`k = 8`

Sample of 5 means that
`n = 5`

What is the probability that 3 were laser printers?

This is P(X = 3).

`P(X = x) = h(x,N,n,k) = (C_(k,x)*C_(N-k,n-x))/(C_(N,n))`

`P(X = 3) = h(3,20,5,8) = (C_(8,3)*C_(12,2))/(C_(20,5)) = 0.2384`

0.2384 = 23.84% probability that 3 were laser printers.