Question

Taco Bell has decided to keep their doors open all night. There are 25 employees that have volunteered for the night shift. a) If a night crew consists of 8 employees, how many different crews are possible? b) If the employees are ranked 1,2,...,25 in order of competence, how many of these crews would not have the best employee? c) How many of the crews would have at least 1 of the 10 best employees? d) If one of these crews is selected at random to work on a particular night, what is the probability that the best employee will not work that night?

Answer 1

Answer:

a) 1,081,575

b) 735,471

c) 1,075,140

d) 0.3159

Step-by-step explanation:

a)

This is a combination of 25 elements taken 8 at a time

$\binom{25}{8}=\frac{25!}{8!(25-8)!}=\frac{25!}{8!17!}=1081575$

b)

Supposing number 1 is the best, if we leave him/her out we will now have 24 employees, so now it is a combination of 24 elements taken 8 at a time

$\binom{24}{8}=\frac{24!}{8!(24-8)!}=\frac{24!}{8!16!}=735471$

c)

Let's compute the complement of this set and then subtract.

That is to say, compute how many crews do not have any of the 10 best.

This would be a combination of 25-10 = 15 taken 8 at a time

$\binom{15}{8}=\frac{15!}{8!(15-8)!}=\frac{15!}{8!7!}=6435$

If we subtract this figure from the total of possible crews, we get 1,081,575 - 6,435 = 1,075,140 possible crews with at least one of the 10 best.

d)

From the figure obtained in c) we subtract the figure obtained in b) 1,075,140 - 735,471 = 339,669 and this is the number of crews that do not have number 1, so the probability that in a crew with at least one of the 10 best number 1 is not there is

339,669/1,075,140 = 0.3159