A friend of mine is giving a dinner party. His current wine supply includes 12 b

Question

A friend of mine is giving a dinner party. His current wine supply includes 12 bottles of zinfandel, 8 of merlot, and 9 of cabernet (he only dinks 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? 1320 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 (e) If 6 bottles are randomly selected, how many ways are there to obtain two bottles of each variety? ways (d) If 6 bottles were 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 at them are the same variety?

Explanation / Answer

Answer:

1. The order is important means we want a permutation. Use the permutation rule with n= 12 and k=3

= 12*11*10= 1320.

2. if the order is unimportant then we want a combination of 6 = 29! / 6! (29-6)! = 47,5020

if order is important then we want a permutation = 29*28*27*26*25*24 = 3,42,014,400

3.if they order two bottles of each variety= (12! / 2!(12-10)!) * (8! (8-2)! 2!) * (9! / 2! (9-2)!) = 66,528

4.probability that results in two bottles of each variety being choosen is

=[(12! / 2!(12-10)!) * (8! (8-2)! 2!) * (9! / 2! (9-2)!)] / [29! / (29-6)! 6!) = 66258/ 475020 = 0.14005305

5.Probability = [(12! / 2!(12-10)!) + (8! (8-2)! 2!) + (9! / 2! (9-2)!)] / [29! / (29-6)! 6!) = 130 / 475020 = 2.73672e-4.

please let me know if you have any problem with my solution

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