Question

Suppose that 60 batteries are shipped to an auto parts store, and that 5 of those are defective. A fleet manager then buys 6 of the batteries from the store. In how many ways can at least 4 defective batteries be included in the purchase

Answer 1

Answer:

In 8910 ways can at least 4 defective batteries be included in the purchase

Step-by-step explanation:

The order in which the batteries are there is not important, which means that the combinations formula is used 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.

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

In this question:

4 defective:

4 defective from a set of 5 and 2 non-defective from a set of 60 - 5 = 55. So

$C_{5,4}C_{60,2} = \frac{5!}{1!4!} \times \frac{60!}{2!58!} = 8850$

5 defective:

5 defective from a set of 5 and 1 non-defective from a set of 60 - 5 = 55. So

$C_{5,5}C_{60,5} = \frac{5!}{0!5!} \times \frac{60!}{1!59!} = 60$

In how many ways can at least 4 defective batteries be included in the purchase?

$T = 8850 + 60 = 8910$

In 8910 ways can at least 4 defective batteries be included in the purchase