40 cars are at an auto dealership: 23 are Fords and 17 are Chevys. Find the probability that in a big sale of?

I would just like to know the answer to this problem and how you do it!40 cars are at an auto dealership: 23 are Fords and 17 are Chevys. Find the probability that in a big sale of?
You need to assume that the choice of a used Ford and a used Chevy are equally likely. (That's a big deal in probability.)

I like to think of these types of problems as marble jar problems. Imagine that you have 40 marbles in a jar. 23 are Ford marbles and 17 are Chevy marbles. You are going to randomly draw 15 marbles out of the jar without replacement. You want to know the probability of choosing 11 Fords and 4 Chevys.

Here's how to solve it with Combinations, nCr.

23C11 = the number of ways to choose 11 Ford marbles from a total of 23 = 1,352,078
17C4 = the number of ways to choose 4 Chevy marbles from a total of 17 = 2,380
40C15 = the number of ways to choose any 15 marbles from the 40 marbles in the jar = 40,225,345,060

P(11 Fords and 4 Chevys) = (23C11)(17C4) / (40C15) = (1,352,078)(2,380)/40,225,345,060 = 0.08

The probability is 0.08 (not very likely.) I used a graphing calculator to find the values for each combination. You can also use this formula to calculate them:

nCr = n! / [(n-r)!r!

For example,

17C4 = 17! / [(17-4)!4! = 17! / [13! 4!]
17C4 = (17)(16)(15)(14)(13!)/[(13!)(4)(3)(2)]
The 13! and the other factors in the denominator cancel out and we are left with 2,380.

Hope this helps!