Question

A friend of mine is giving a dinner party. His current wine supply includes 10 bottles of zinfandel, 8 of merlot, and 11 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? ways (b) If 6 bottles of wine are to be randomly selected from the 29 for serving, how many ways are there to do this? ways (c) If 6 bottles are randomly selected, how many ways are there to obtain two bottles of each variety? ways (d) If 6 bottles are randomly selected, what is the probability that this results in two bottles of each variety being chosen? (Round your answer to three decimal places.) (e) If 6 bottles are randomly selected, what is the probability that all of them are the same variety? (Round your answer to three decimal places.)

Answer 1

Answer:

a) 720, b) 475020, c) 69300, d) 0.146, e) 0.001

Step-by-step explanation:

It is given that my friend has 10 bottles of zinfandel, 8 of merlot, and 11 of cabernet.

a)

If he wants to serve 3 bottles of zinfandel and serving order is important, then the total number of ways is

$10\times 9\times 8=720$

Therefore if he wants to serve 3 bottles of zinfandel and serving order is important, then the total number of ways are 720.

b)

The total number of bottles is

$10+8+11=29$

Combination is defined as

$^nC_r=\frac{n!}{r!(n-r)!}$

where, n is total possible outcomes and r is selected outcomes.

we have to select 6 bottles out of 29. so,

$^{29}C_{6}=475020$

Therefore ff 6 bottles of wine are to be randomly selected from the 29 for serving, then the total number of ways are 475020.

c)

If we want to select 2 bottles of each variety, then total number of ways are

$^{10}C_{2}\times ^{8}C_{2}\times ^{11}C_{2}=69300$

Therefore if 6 bottles are randomly selected with two bottles of each variety, then the total possible ways are 69300.

d)

Probability is defined as

$P=\frac{\text{Total outcomes}}{\text{Favorable outcomes}}$

$\frac{^{10}C_{2}\times ^{8}C_{2}\times ^{11}C_{2}}{^{29}C_{6}}=\frac{69300}{475020}\approx 0.146$

Therefore the probability that two bottles of each variety being chosen is 0.146.

e)

If 6 bottles are randomly selected, then the probability that all of them are the same variety is

$\frac{^{10}C_{6}+^{8}C_{6}+^{11}C_{6}}{^{29}C_{6}}=\frac{700}{475020}\approx 0.001$

Therefore if 6 bottles are randomly selected, then the probability that all of them are the same variety is 0.001.