Question

Suppose that four microchips in a production run of fifty are defective. A sample of six is to be selected to be checked for defects. (a) How many different samples can be chosen? (b) How many samples will contain at least one defective chip? (c) What is the probability (as a percent) that a randomly chosen sample of six contains at least one defective chip? (Round your answer to one decimal place.) %

Answer 1

Answer:

a) 15,890,700 different samples can be chosen

b) 6,523,881 will contain at least one defective chip

c) 41.1%

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The order in which the microchips are chosen is not important. So we use the combinations formula to solve this question.

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.

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

(a) How many different samples can be chosen?

Samples of 6 from a set of 50.

Then

$C_{50,6} = \frac{50!}{6!(50-6)!} = 15890700$

15,890,700 different samples can be chosen

(b) How many samples will contain at least one defective chip?

Either a sample contains no defective chip, or it contains at least one. From a), the sum of them is 15890700.

Then

No defective chips:

Four are defective.

So 50-4 = 46 are not.

This is samples of 6 from a set of 46.

$C_{46,6} = \frac{46!}{6!(46-6)!} = 9366819$

9,366,819 samples contain no defective chips.

At least one:

9366819 + n = 15890700

n = 6523881

6,523,881 will contain at least one defective chip

(c) What is the probability (as a percent) that a randomly chosen sample of six contains at least one defective chip

6,523,881 out of 15,890,700

6,523,881/15,890,700 = 0.411

As a percent

41.1%