# At the whistlestop cafe, 15 of the 19 patrons are mechanics. If a sample of 8 was taken, what is the probability that exactly 5 patrons are mechanics?

*print*Print*list*Cite

In this problem, there are 19 patrons and 15 mechanics, so there are 4 who are not. In the sample of 8 people, we want 5 mechanics and 3 who are not. We can use combinations here. Remember that to calculate a combination, we use C(n,k) combination operation = n!/[k!(n-k)!] where n is the total and k is how many we want to choose.

The probability of choosing 5 mechanics and 3 whp are not =

C(15,5) * C(4,3) / C(19,8)

Because:

C(15,5) represents 5 mechanics chosen out of a possible 15

C(4,3) represents 3 mechanics chosen out of a possible 4

C(19,8) represents and random 8 people chosen out of a possible 19

C(15,5) * C(4,3) / C(19,8) = (3003 * 4)/ 75582 = 0.158926...

**The probability of exactly 5 mechanics is 15.9%.**

At the whistle stop cafe, 15 of the 19 patrons are mechanics. A sample of 8 was taken. The probability that exactly 5 patrons are mechanics has to be determined.

It is possible to create a sample of 8 patrons from the 19 people at the cafe in C(19, 8) ways. It is possible to create samples that have exactly 5 mechanics in C(15, 5)*C(4, 3) ways. The required probability is (C(15, 5)*C(4, 3))/C(19, 8)

=> `(15!*4!*8!*11!)/(10!*5!*3!*19!)`

=> 15.89%

**The probability of the sample of 8 having exactly 5 mechanics is 15.89%**