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5 December, 09:16

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 30for serving, how many ways are there to do this?

c.) IF 6 bottles are randomly selected, how many ways are thereto obtain two bottles of each variety?

d.) IF 6 bottles are randomly selected, what is the probabilitythat this results in two bottles of each variety being chosen?

e.) If 6 bottles are randomly selected, what is the probabilitythat all of them are the same variety?

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Answers (1)
  1. 5 December, 09:28
    0
    a) 336

    b) 593775

    c) 83160

    d) P=0.14

    e) P=0.0019

    Step-by-step explanation:

    We have wine supply includes 8 bottles of zinfandel, 10 of merlot, and 12 of cabernet.

    a) If he wants to serve 3 bottles of zinfandel and serving order is important. We get:

    C=8·7·6=336

    b) {30}_C_{6}=/frac{30!}{6! (30-6) !}

    {30}_C_{6}=593775

    c) {8}_C_{2} · {10}_C_{2} · {12}_C_{2}=

    =/frac{8!}{2! (8-2) !} · / frac{10!}{2! (10-2) !} · / frac{12!}{2! (12-2) !}

    =28 · 45 · 66

    =83160

    d) We calculate the number of possible combinations:

    {30}_C_{6}=593775

    We calculate the number of favorable combinations:

    {8}_C_{2} · {10}_C_{2} · {12}_C_{2}=83160

    The probability that this results in two bottles of each variety being is

    P=83160/593775

    P=0.14

    e) We calculate the number of possible combinations:

    {30}_C_{6}=593775

    We calculate the number of favorable combinations:

    {8}_C_{6} + {10}_C_{6} + {12}_C_{6} = 28+210+924=1162

    The probability is

    P=1162/593775

    P=0.0019
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