Question

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.

a. State any and all event definitions.

b. Suppose the wine tasting will consist of two bottles chosen at random. How many different sequences of offerings are there?

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

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

Answer 1

Let 'Z' denote Zinfandel, 'M' denote Merlot and 'C' denote Cabernet.

a. 3 bottles are chosen at random.

Let 'A' be the event that bottles chosen are different.

Let 'B' be the event that no Zinfandel bottles are chosen.

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b. If two bottles are chosen then there are possibilities that the bottles are of same wine or they can be of different wines.

If same wines then { ZZ, MM, CC}

If different {ZM, ZC, MZ, MC, CZ, CM}

ZM and MZ are not same since the order matters. ZM means the 1st chosen was Zinfandel and in MZ 2nd chosen was Zinfandel.

Joining the two we have sample space

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

c. The probabilities are

Event	Probability
P(ZZ)	0.064
P(ZM)	0.092
P(ZC)	0.110
P(MZ)	0.092
P(MM)	0.103
P(MC)	0.138
P(CZ)	0.110
P(CM)	0.138
P(CC)	0.152
Sum	1

The probabilities are explained below.

For every wine to be chosen 1st the probability will be = (no. of that wine bottle)/ Total no. of bottle.

Eg. there are 10 Merlot and 30 in all

For choosing Merlot = 10/30

For every next bottle, the probability will be = P (1st chosen) *(no. of the same remaining bottle)/ (total remaining bottle)

For the underline part :If the 1st and 2nd bottle are same then the numerator will reduce by 1 for that wine but if the bottles are different then the numerator will not change .

Eg. Assume 1st is Merlot, now there are 29 in all remaining and 9 of Merlot . The rest have unchanged.

2nd is Merlot therefore probabiltiy = (10/30) * (9/29) = 0.103

If 2nd was Zinfandel the probability = (10/30) * (8/29)= 0.092

We multiply the probabilities since the event is taking place together.

d. The probabiltiy that the bottles are of same wine = P(ZZ) + P(MM) + P(CC)

We add these since we dont know which event it will be.

From the above table

Probability = 0.064 + 0.103 + 0.152

= 0.319

The probability of choosing same wine is .