A friend of mine is giving a dinner party. His current wine supply includes 8 bo
ID: 3203464 • Letter: A
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. If he wants to serve 3 bottles of zinfandel and serving order is important, how many ways are there to do this? If 6 bottles of wine are to be randomly selected from the 30 for serving, how many ways are there to do this? If 6 bottles are randomly selected, how many ways are there to obtain two bottles of each variety? If 6 bottles are randomly selected, what is the probability that this results in two bottles of each variety being chosen e. If 6 bottles are randomly selected, what is the probability that all of them are the same variety?Explanation / Answer
a) This can be done by selecting 3 bottles out of 8 (=8C3 ways) and then ordering 3 selected bottles in different possible ways (=3!)
So, total number of ways = 8C3*3! = (8!/5!*3!)*3! = 8!/5! = 6*7*8= 336
b) Number of ways of selecting 6 bottles out of 30 = 30C6 = 30!/(6!*24!) = 593775
c) Number of ways = Number ways of selecting 2 zinfandel * 2 of merlot * 2 of cabernet
= (8C2)*(10C2)*(12C2) = 28*45*66 = 83160
d) Probability = Answer to part c/Answer to part b = 83160/593775 = 0.140053
e) Number of ways of selecting 6 bottles of same variety = Number of ways selecting 6 zinfandel + Number of ways of selcting 6 merlot + Number of ways of selecting 6 cabernet
= 8C6 + 10C6 + 12C6 = 28 + 210 +924 = 1162
Probability = 1162/593775 = 0.001957