Question

A friend of mine is giving a dinner party. His current wine supply includes 8 bottles of zinfandel, 10 of merlot, and 12 of cabernet (he only drinks red wine), 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?
b.) If 6 bottles of wine are to be randomly selected from the 30for serving, how many ways are there to do this?
c.) IF 6 bottles are randomly selected, how many ways are thereto obtain two bottles of each variety?
d.) IF 6 bottles are randomly selected, what is the probabilitythat this results in two bottles of each variety being chosen?
e.) If 6 bottles are randomly selected, what is the probabilitythat all of them are the same variety?

Answer 1

a) 336

b) 593775

c) 83160

d) P=0.14

e) P=0.0019

Explanation:

We have wine supply includes 8 bottles of zinfandel, 10 of merlot, and 12 of cabernet.

a) If he wants to serve 3 bottles of zinfandel and serving order is important. We get:

C=8·7·6=336

b) {30}_C_{6}=\frac{30!}{6!(30-6)!}

{30}_C_{6}=593775

c) {8}_C_{2} · {10}_C_{2} · {12}_C_{2}=

=\frac{8!}{2!(8-2)!} · \frac{10!}{2!(10-2)!} · \frac{12!}{2!(12-2)!}

=28 · 45 · 66

=83160

d) We calculate the number of possible combinations:

{30}_C_{6}=593775

We calculate the number of favorable combinations:

{8}_C_{2} · {10}_C_{2} · {12}_C_{2}=83160

The probability that this results in two bottles of each variety being is

P=83160/593775

P=0.14

e) We calculate the number of possible combinations:

{30}_C_{6}=593775

We calculate the number of favorable combinations:

{8}_C_{6} + {10}_C_{6} + {12}_C_{6}= 28+210+924=1162

The probability is

P=1162/593775

P=0.0019