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

Answer:

0.2384 = 23.84% probability that 3 were laser printers.

Step-by-step 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:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

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.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

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).

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 3) = h(3,20,5,8) = \frac{C_{8,3}*C_{12,2}}{C_{20,5}} = 0.2384$

0.2384 = 23.84% probability that 3 were laser printers.