Need help with a probability problem?

a. Exactly of the chosen cars are foreign

You failed to ask for a specific number. But we can handle this, anyhow. There are

C(16,4) = 16! / (12! 4!)

= 16 * 15 * 14 * 13 / (4 * 3 * 2 * 1)

= 4 * 5 * 7 * 13

= 1,820 different combinations of 4 cars, all equally probable.

So we can just compute how many of these combinations include any possible number of foreign cars.

[I am using a formula for combinations, which I hope you know. But if you need any help with permutations or combinations formulas, their derivation, or their use, I recommend the online tutorial:

http://www.mathsisfun.com/combinatorics/combinatio...

for both clarity and complete coverage of the topic.]

0 foreign cars: all 4 must be American.

C(6,4) = 6! / (4! 2!) = 6 * 5 / 2 = 15

so the probability is 15/1,820 = 3/364

1 foreign car: 3 must be American.

C(10,1) C(6,3) = [10! / (9! 1!)] [6! / (3! 3!)]

= 10 * 6 * 5 * 4 / (3 * 2 * 1)

= 10 * 5 * 4 = 200

so the probability is 200/1,820 = 10/91

2 foreign cars: the other two are American.

C(10,2) C(6,2) = [10! / (8! 2!)] [6! / (4! 2!]

= 10 * 9 * 6 * 5 / (2 * 2)

= 5 * 9 * 3 * 5 = 675

so the probability is 675/1,820 = 135/364

3 foreign cars: the fourth is American.

C(10,3) C(6,1) = [10! / (7! 3!)] [6! / (5! 1!)]

= 10 * 9 * 8 * 6 / (3 * 2 * 1)

= 10 * 9 * 8 = 720

so the probability is 720/1,820 = 36/91

4 foreign cars (none American):

C(10,4) = 10! / (6! 4!)

= 10 * 9 * 8 * 7 / (4 * 3 * 2 * 1)

= 10 * 3 * 7 = 210

so the probability is 210/1,820 = 3/26

We note that the total of these five cases should add up to our calculated total:

15 + 200 + 675 + 720 + 210 = 1,820

b. all the chosen cars are foreign

This was our last case, above. We had 210 possible ways to choose 4 foreign cars,

for a probability of 210/1,820 = 3/26

c. at most 3 chosen cars are foreign

This only excludes the case where 4 foreign cars are selected.

So let's use the answer to part b:

1 - 3/26 = 23/26