Question

A warehouse contains ten printing machines, two of which are defective. A company selects seven of the machines at random, thinking all are in working condition. What is the probability that all seven machines are nondefective?

Answer 1

Answer:

0.0667 = 6.67% probability that all seven machines are nondefective.

Step-by-step explanation:

The machines are chosen from the sample without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes 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 successes.

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:

10 machines means that $n = 10$

2 defective, so 10 - 2 = 8 work correctly, which means that $k = 8$

Seven are selected, which means that $n = 7$

What is the probability that all seven machines are nondefective?

This is P(X = 7). So

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

$P(X = 7) = h(7,10,7,8) = \frac{C_{8,7}*C_{2,0}}{C_{10,7}} = 0.0667$

0.0667 = 6.67% probability that all seven machines are nondefective.