# A friend of mine is giving a dinner party. His current

## Problem 30E Chapter 2

Probability and Statistics for Engineering and the Sciences | 9th Edition

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Problem 30E

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 30 for serving, how many ways are there to do this? c. ?If 6 bottles are randomly selected, how many ways are there to obtain two bottles of each variety? d. ?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?

Step-by-Step Solution:

Answer : Step 1 of 5 : 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) The number of zinfandel bottles be n =8 The number of zinfandel bottles served be r = 3 If he wants to serve 3 bottles of zinfandel and serving order is important, then the number of ways for selecting 3 out of 8 is 8P3 n! Formula for permutation is P(n, r) = (n r)! where , n = 8 and r = 3 8! P(8, 3) = (8 3)! = 40320 120 = 336 336 ways are there to do. Step 2 of 5 : b) 6 bottles of wine are to be randomly selected from the 30 for serving We have n = 30 and r = 6 Formula for combination is C(n, r) = n! r!(n r)! 30! C(30, 6) = 6!(30 6)! = 593775 593775 ways are there to do. Step 3 of 5 : c) 6 bottles are randomly selected, the claim is to find the number of ways to obtain two bottles of each variety. 2 bottles from 8 bottles of zinfandel, 2 bottles from 10 of merlot, and 2 bottles from 12 of cabernet. Use the product rule fork-tuples, here k= 3 is the number of types of wine: 8 10 12 ( )2 )( 2 2 C(8, 2) = 8! 2!(8 2)! = 28 10! C(10, 2) = 2!(10 2)! = 45 12! C(8, 2) = 2!(12 2)! = 66 Therefore, 28*45*66 = 83160 83,160 ways to obtain two bottles of each variety.

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