Question

9. Suppose a friend of yours is hosting a wine tasting. His wine supply includes three different types of deluxe wine. All bottles come from the same winery and all wines were harvested in 2017. He currently owns 8 bottles of zinfandel, 10 bottles of merlot, and 12 bottles of cabernet. Throughout the wine tasting, your friend will only open a new bottle of wine when there are no other bottles of wine open at the time.

   1. State any and all event definitions.

   2. Suppose the wine tasting will consist of two bottles chosen at random. How many different

      sequences of offerings are there?

   3. Create a tree diagram of the sample space and label all events and their corresponding

      probabilities.

   4. What is the probability that all bottles in the wine tasting are of the same type of wine?

Answer 1

Let M show the Merlot, Z shows Zinfandel and C shows Cabernet.

Total number of bottles = 8 + 10 + 12 = 30

Since first wine can be selected in 3 ways and second also can be selected in 3 ways so possible number of sequences are:

S = { ZZ, ZM, ZC, MZ, MM, MC, CZ, CM, CC}

Out of 30 bottles,  8 bottles are of zinfandel so probability of getting of first bottle of zinfandel is

P(Z) = 8/30

After that 7 bottles of  zinfandel remaining out of 29 remaining bottles so

P(Z |Z) = 7/29

Therefore,

P(ZZ) = P(Z|Z)P(Z) = (7/29) * (8/30) = 0.0644

Likewise:

P(ZM) = (8/30) * (10/29) = 0.0920

P(ZC) = (8/30) * (12/29) = 0.1103

P(MZ) =  (10/30) * (8/29) = 0.0920

P(MM) = (10/30) * (9/29) = 0.1034

P(MC) = (10/30) * (12/29) = 0.1379

P(CZ) = (12/30) * (8/29) = 0.1103

P(CM) = (12/30) * (10/29) = 0.1379

P(CC) = (12/30) * (11/29) = 0.1517

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Following is the tree diagram:

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The  probability that all bottles in the wine tasting are of the same type of wine is

P(ZZ) + P(MM) + P(CC) = 0.0644 + 0.1034 + 0.1517 = 0.3195