Question

A man is giving a dinner party. His current wine supply includes 8 bottles of zinfandel, 10 of merlot, and 12 of cabernet, all from different wineries. a) If he wants to serve 3 bottles of zinfandel and serving order is important, how many ways are there to do this? (2 points) b) If 6 bottles are randomly selected, what is the probability that this results in two bottles of each variety being chosen? (4 points) c) If 6 bottles are randomly selected, what is the probability that all of them are the same variety?

Answer 1

a. 336

b. 14.01%

c. 0.2%

Explanation:

a. We have that the number of zinfandel bottles is 8 and that the number of zinfandel served is 3, therefore:

n = 8 and r = 3

we can calculate it by means of permutation:

nPr = n! / (n-r)!

replacing:

8P3 = 8! / (8-3)!

8P3 = 336

Which means there are 336 ways.

b. First we must calculate the ways to choose 2 bottles of each variety, through combinations:

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

We know that there are 8 bottles zinfandel, 10 of merlot, and 12 of cabernet, and we must choose 2 of each, therefore it would be:

8C2 * 10C2 * 12C2

8! / (2! * (8-2)! * 10! / (2! * (10-2)! * 12! / (2! * (12-2)!

28 * 45 * 66 = 83160

Now we must calculate the total number of ways, that is, choose 6 bottles of the 30 total (8 + 10 + 12)

30C6 = 30! / (6! * (30-6)! = 593775

Thus:

83160/593775 = 0.1401

In other words, the probability is 14.01%

c. In this case, we must calculate the number of ways of 8 bottles zinfandel, 10 of merlot, and 12 of cabernet choose 6, that is to say that they are all of the same variety, therefore:

8C6 + 10C6 + 12C6

8! / (6! * (8-6)! + 10! / (6! * (10-6)! + 12! / (6! * (12-6)!

28 + 210 + 924 = 1162

And that divide it by the total amount that we calculated the previous point, 30C6 = 593775

Thus:

1162/593775 = 0.002

In other words, the probability 0.2%