Question

A friend of mine is giving a dinner party. His current wine supply includes 10 bottles of zinfandel, 8 of merlot, and 11 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? 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 (c) If 6 bottles are randomly selected, how many ways are there to obtain two bottles of each variety? ways (d) If 6 bottles are randomly selected, what is the probability that this results in two bottles of each variety being chosen? (Round your answer to three decimal places.) (e) If 6 bottles are randomly selected, what is the probability that all of them are the same variety? (Round your answer to three decimal places.)

Answer 1

The probability of a selection of drinks is given by the number of required

ways divided by the number of possible outcome.

(a) The number of ways he can serve 3 bottles out  of the 10 bottles of

zinfandel is 720 ways.

(b) The number of ways to select 6 bottles out of the 29 bottles is 475020 ways.

(c) The number of ways of selecting 2 bottles of each variety is 69,300 ways.

(d) The probability of selecting two bottles of each variety, when six bottles are selected is approximately 0.146

(e) The probability that six bottles are the same variety is approximately 0.001.

Reasons:

Number of bottles of zinfandel = 10

Number of bottles of merlot = 8

Number of bottles of cabernet = 11

(a) Number of bottles of zinfandel he wants to serve = 3 bottles

Number of ways of serving 3 bottles out of 10, with the order of serving being important is 10 permutation 3

$^nP_r = \mathbf{\dfrac{n!}{(n - r)!}}$

Which gives;

$^{10}P_3 = \mathbf{\dfrac{10!}{(10 - 3)!}} = 720$

• The number of ways he can serve 3 bottles out  of the 10 bottles of zinfandel = 720 ways

(b) The number of ways to select 6 bottles out of the 29 bottles is given by  combination, $^{n}C_r$ as  follows;

$^{29}C_6$ = 475020

The number of ways to select 6 bottles out of the 29 bottles is 475020 ways.

(c) The number of ways 2 bottles can be selected from each variety is given as follows;

Number of ways = $^{10}C_2$ × $^{8}C_2$ × $^{11}C_2$ = 69,300

The number of ways of selecting 2 bottles of each variety = 69,300 ways

(d) The probability is given by the ratio of the number of ways of selecting 2 bottles from each variety to the number of ways of selecting 6 bottles from 29, as follows;

$\displaystyle Probability = \frac{^{10}C_2 \times ^{8}C_2 \times ^{11}C_2}{^{29}C_6} = \frac{69,300}{475020} =\frac{55}{377} \approx 0.146$

The probability of selecting two bottles of each variety, when six bottles are selected P(2 each) ≈ 0.146

(e) The probability that six bottles are the same variety, is given by the sum

of the ways of selecting 6 bottles of a given variety, divided by the ways of

selecting six bottles from the 29 bottles as follows;

$\displaystyle P(6 \ of \ same \ variety) = \frac{^{10}C_6 + ^{8}C_6 + ^{11}C_6}{^{29}C_6} = \frac{700}{475020} = \frac{5}{3393} \approx 1.474 \times 10^{-3}$

P(6 of same variety) ≈ 1.474 × 10⁻³ ≈ 0.001

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