Suppose that 40 batteries are shipped to an auto parts store, and that 4 of those are defective. A fleet manager then buys 8 of the batteries from the store. In how many ways can at least 3 defective batteries be included in the purchase?

From the given above, we will include the case in which 3 and all 4 defectives are included in the purchase.

For 3:
(36C5) x (4C3) = 1507968
For 4:
(36C4) x (4C4) = 58905
Adding these numbers will give us an answer of 1566873.

Answer Link

joaobezerra joaobezerra · Answer 2 · 2021-10-26T20:11:20+02:00

Using the combination formula, it is found that at least 3 defective batteries can be included in the purchase in 1,566,873 ways

In this problem, the order in which the batteries are selected is not important, which means that the combination formula is used to solve it.

Combination formula:

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

In this problem, two outcomes are desired:

3 defective from a set of 4 and 5 non-defective from a set of 36.
4 defective from a set of 4 and 4 non-defective from a set of 36.

Then:

[tex]T = C_{4,3}C_{36,5} + C_{4,4}C_{36,4} = \frac{4!}{3!1!}\frac{36!}{5!31!} + \frac{4!}{4!0!}\frac{36!}{4!32!} = 1,566,873[/tex]

There are 1,566,873 ways in which at least 3 defective batteries can be included.

A similar problem is given at https://brainly.com/question/24278448