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Answer :
Answer:
Explained below.
Step-by-step explanation:
The complete question is:
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?
Solution:
In mathematics, the procedure to select k items from n distinct items, without replacement, is known as combinations.
The formula to compute the combinations of k items from n is given by the formula:
[tex]{n\choose k}=\frac{n!}{k!(n-k)!}[/tex]
Permutation is the number of ways to select k items from n distinct items in a specific order.
The formula to compute the permutation or arrangement of k items is:
[tex]^{n}P_{k}=\frac{n!}{(n-k)!}[/tex]
(a)
The number of ways to serve 3 bottles of zinfandel, with a specific order is:
[tex]^{8}P_{3}=\frac{8!}{(8-3)!}=\frac{8\times7\times6\times5!}{5!}=336[/tex]
(b)
The number of ways to select 6 bottles from the 30 is:
[tex]{30\choose 6}=\frac{30!}{6!(30-6)!}=\frac{30!}{6!\times 24!}=593775[/tex]
(c)
The number of ways to select two bottles of each variety is:
[tex]{8\choose 2}\times {10\choose 2}\times {12\choose 2}=\frac{8!}{2!\times6!}\times \frac{10!}{2!\times8!}\times \frac{12!}{2!\times10!}[/tex]
[tex]=\frac{12!}{(2!)^{3}\times 6!}\\\\=83160[/tex]
(d)
Compute the probability of selecting two bottles of each variety if 6 bottles are selected:
[tex]P(\text{2 bottles of each})=\frac{83160}{593775}=0.14[/tex]
(e)
Compute the probability of selecting the same variety of bottles, if 6 bottles are selected:
[tex]P(\text{Same Variety})=\frac{{8\choose 6}+{10\choose 6}+{12\choose 6}}{{30\choose 6}}[/tex]
[tex]=\frac{28+210+924}{593775}\\\\=0.0019570\\\\\approx 0.002[/tex]
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