Respuesta :
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:
[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 question:
4 defective:
4 defective from a set of 5 and 2 non-defective from a set of 60 - 5 = 55. So
[tex]C_{5,4}C_{60,2} = \frac{5!}{1!4!} \times \frac{60!}{2!58!} = 8850[/tex]
5 defective:
5 defective from a set of 5 and 1 non-defective from a set of 60 - 5 = 55. So
[tex]C_{5,5}C_{60,5} = \frac{5!}{0!5!} \times \frac{60!}{1!59!} = 60[/tex]
In how many ways can at least 4 defective batteries be included in the purchase?
[tex]T = 8850 + 60 = 8910[/tex]
In 8910 ways can at least 4 defective batteries be included in the purchase
