Question

A friend of mine is giving a dinner party. His current wine supply includes 8 bottles of zinfandel, \({\rm{10}}\) of merlot, and \({\rm{12}}\) of cabernet (he only drinks red wine), all from different wineries.

a. If he wants to serve \({\rm{3}}\) bottles of zinfandel and serving order is important, how many ways are there to do this?

b. If \({\rm{6}}\) bottles of wine are to be randomly selected from the \({\rm{30}}\) for serving, how many ways are there to do this?

c. If \({\rm{6}}\) bottles are randomly selected, how many ways are there to obtain two bottles of each variety?

d. If \({\rm{6}}\) bottles are randomly selected, what is the probability that this results in two bottles of each variety being chosen?

e. If \({\rm{6}}\) bottles are randomly selected, what is the probability that all of them are the same variety?

Short answer

1. \({{\rm{P}}_{{\rm{3,8}}}}{\rm{ = 336}}\)
2. \({C_{6,30}} = 593775;\)
3. \({\rm{83160 }}\) ways
4. \({\rm{0}}{\rm{.14}}\)
5. \({\rm{0}}{\rm{.002}}\)

Step-by-step solution

describing the factorial method use to explanation

There are 30 bottles of wine in all, with 8 bottles of zinfandel, 10 bottles of merlot, and 12 bottles of cabernet. Note: For factorial, we use the idea "! \({\rm{n! = n}}\)\({\rm{(n - 1)* \ldots *2*1}}\)

If he wants to serve \({\rm{3}}\) bottles of zinfandel and serving order is important 

A permutation is a subset that is ordered. The number of permutations of size \({\rm{k}}\)for \({\rm{n}}\) persons in a group is represented as \({{\rm{P}}_{{\rm{k,n}}}}\).The proposal is this:

\({{\rm{P}}_{{\rm{k,n}}}}{\rm{ = }}\frac{{{\rm{n!}}}}{{{\rm{(n - k)!}}}}\)

We utilize permutations because the order is critical. There are \({\rm{n = 8}}\)bottles of zinfandel in this case, and we wish to serve \({\rm{k = 3}}\)bottles, thus there are

\(\begin{aligned} {{P}_{k,n}}&={{P}_{3,8}}=\frac{8!}{(8-3)!}=\frac{8!}{5!}=\frac{8*7*6*5!}{5!} \\&=8*7*6=336 \\\end{aligned}\)

If \({\rm{6}}\) bottles of wine are to be randomly selected from the \({\rm{30}}\) for serving, how many ways are there to do this

-Because order is irrelevant in this case, we employ combinations.

An unordered subset is referred to as a combination. The number of combinations of size \({\rm{k}}\) for \({\rm{n}}\)persons in a group is indicated as \({{\rm{C}}_{{\rm{k,n}}}}\)

\(\left( {\begin{aligned}{\rm{n}}\\{\rm{k}}\end{aligned}} \right)\)

which we interpret as \({\rm{k}}\)is chosen from \({\rm{n}}\)items. The proposal is this:

\({{\rm{C}}_{{\rm{k,n}}}}{\rm{ = }}\left( {\begin{aligned}{\rm{n}}\\{\rm{k}}\end{aligned}} \right){\rm{ = }}\frac{{{{\rm{P}}_{{\rm{k,n}}}}}}{{{\rm{k!}}}}{\rm{ = }}\frac{{{\rm{n!}}}}{{{\rm{k!(n - k)!}}}}\)

When \({\rm{n = 30}}\)and \({\rm{k = 6}}\)are used, we get

\({{\rm{C}}_{{\rm{k,n}}}}{\rm{ = }}{{\rm{C}}_{{\rm{6,30}}}}{\rm{ = }}\left( {\begin{aligned}{{\rm{30}}}\\{\rm{6}}\end{aligned}} \right){\rm{ = }}\frac{{{\rm{30!}}}}{{{\rm{6!(30 - 6)!}}}}{\rm{ = }}\frac{{{\rm{30!}}}}{{{\rm{6!24!}}}}{\rm{ = 593775}}\)

If \(\left( {\rm{b}} \right){\rm{.}}\) bottles are randomly selected, collected ways are there to obtain two bottles of each variety

There must be two of each of the six bottles chosen. We'd want to use:

For k-Tuples, there is a Product Rule.

Assume that there are \({{\rm{n}}_{\rm{1}}}\)methods to pick the first element of an ordered pair, and that there are \({{\rm{n}}_{\rm{2}}}\)ways to select the second member for each selected element. The number of pairings is \({{\rm{n}}_{\rm{1}}}{{\rm{n}}_{\rm{2}}}\)in this case.

Similarly, there are \({{\rm{n}}_{\rm{1}}}{{\rm{n}}_{\rm{2}}}{\rm{* \ldots *}}{{\rm{n}}_{\rm{k}}}\)potential \({\rm{k}}\)-tuples given an ordered collection of \({\rm{k}}\)elements, where the \({{\rm{k}}^{{\rm{th}}}}\)can be selected in \({{\rm{n}}_{\rm{k}}}\)ways.

Where \({n_1}\)represents the number of elements from which we can choose two bottles out of eight for the first element in the ordered group, \({{\rm{n}}_{\rm{2}}}\)represents the number of elements from which we can choose two bottles out of ten for the second element in the ordered group, and \({{\rm{n}}_{\rm{3}}}\)represents the number of elements from which we can choose two bottles out of twelve for the third element in the ordered group. We utilize combinations because we don't care which two bottles we take (unordered). Therefore

\(\begin{aligned}{{n}_{1}}&={{C}_{2,8}}=\left( \begin{aligned}8 \\2 \\\end{aligned} \right)=\frac{8!}{2!(8-2)!}=\frac{8!}{2!6!}=\frac{40320}{1440}=28 \\{{n}_{2}}&={{C}_{2,10}}=\left( \begin{aligned}10 \\2 \\\end{aligned} \right)=\frac{10!}{10!(10-2)!}=\frac{10!}{2!8!}=\frac{3628800}{80640}=45 \\{{n}_{3}}&={{C}_{2,12}}=\left( \begin{aligned}12 \\2 \\\end{aligned} \right)=\frac{12!}{12!(12-2)!}=\frac{12!}{2!10!}=\frac{479001600}{7257600}=66 \\\end{aligned}\)

We now have the final findings as a result of this.

\({{\rm{n}}_{\rm{1}}}{{\rm{n}}_{\rm{2}}}{{\rm{n}}_{\rm{3}}}{\rm{ = 28*45*66 = 83160}}\)

If \({\rm{6}}\) bottles are randomly selected, the probability that this results in two bottles of each variety being chosen are

There's a good chance that two bottles of each flavor will be picked from the six bottles available.

\({\rm{P(D) = }}\frac{{{\rm{83160}}}}{{{\rm{593775}}}}{\rm{ = 0}}{\rm{.14}}\)

Where we used,

\({\rm{P(A) = }}\frac{{{\rm{\# \;favorable outcomes in\;A}}}}{{{\rm{\# \;outcomes in the sample space\;}}}}\)

\(\left( {\rm{c}} \right)\)gives us the number of positive outcomes in A, and \(\left( {\rm{d}} \right)\)gives us the number of outcomes in the sample space \(\left( {\rm{b}} \right){\rm{.}}\)

If \({\rm{6}}\) bottles are randomly selected, the probability that all of them are the same variety are as

We want to use

\({\rm{P(A) = }}\frac{{{\rm{\# \;favorable outcomes in\;A}}}}{{{\rm{\# \;outcomes in the sample space\;}}}}\)

favorable outcomes in the sample space \({\rm{A}}\) ,event that all vines are the same

\(\begin{aligned}{{\rm{P(\{ \;All vines bottles are same\;\} ) = P(\{ \;All are zindfandel\;\} ) + P(\{ \;All are merlot\;\} )}}}\\{{\rm{ + P(\{ \;All are cabernet\;\} )}}}\end{aligned}\)

\({\rm{ + P(\{ }}\)They're all cabernet \({\rm{\} )}}\)

\(\mathop {\rm{ = }}\limits^{{\rm{(1)}}} \left( {\begin{aligned}{\rm{8}}\\{\rm{6}}\end{aligned}} \right){\rm{ + }}\left( {\begin{aligned}{{\rm{10}}}\\{\rm{6}}\end{aligned}} \right){\rm{ + }}\left( {\begin{aligned}{{\rm{12}}}\\{\rm{6}}\end{aligned}} \right)\)

\({\rm{ = 28 + 210 + 924 = 1162}}\)

As previously stated, we employ combinations.

As a result, the outcomes in the sample space are presented in \({\rm{(b)}}\)

\({\rm{P(E) = }}\frac{{{\rm{1162}}}}{{{\rm{593775}}}}{\rm{ = 0}}{\rm{.002}}\)