Let's go through each part of the problem step by step to ensure clarity and provide solutions for each case.

(b) If 6 bottles of wine are to be randomly selected from the 31 for serving, how many ways are there to do this?

We are selecting 6 bottles from the 31 (11 Zinfandel, 13 Merlot, and 7 Cabernet). Since the order does not matter, we use combinations. The number of ways to choose 6 bottles from 31 is given by:

$\binom{31}{6} = \frac{31!}{6!(31-6)!} = \frac{31!}{6! \cdot 25!}$

Let’s calculate this.

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

We need to select exactly 2 bottles of Zinfandel, 2 bottles of Merlot, and 2 bottles of Cabernet. The number of ways to do this is:

$\binom{11}{2} \times \binom{13}{2} \times \binom{7}{2}$

where:

• $\binom{11}{2}$ is the number of ways to select 2 Zinfandel bottles,
• $\binom{13}{2}$ is the number of ways to select 2 Merlot bottles,
• $\binom{7}{2}$ is the number of ways to select 2 Cabernet bottles.

Let’s calculate this as well.

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

The probability is the ratio of favorable outcomes (choosing two bottles of each variety) to the total possible outcomes (choosing 6 bottles from 31):

$P = \frac{\text{ways to choose two bottles of each variety}}{\text{ways to choose 6 bottles from 31}} = \frac{\binom{11}{2} \times \binom{13}{2} \times \binom{7}{2}}{\binom{31}{6}}$

Let’s compute this probability and round to three decimal places.

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

We need to calculate the probability that all 6 bottles come from one of the three varieties. There are 3 possibilities here: selecting 6 Zinfandel, 6 Merlot, or 6 Cabernet bottles. The total number of favorable outcomes is:

• $\binom{11}{6}$ ways to choose 6 Zinfandel bottles,
• $\binom{13}{6}$ ways to choose 6 Merlot bottles,
• $\binom{7}{6}$ ways to choose 6 Cabernet bottles (note that $\binom{7}{6} = 7$).

Thus, the probability is:

$P = \frac{\binom{11}{6} + \binom{13}{6} + \binom{7}{6}}{\binom{31}{6}}$

Let me compute all the values for parts (b), (c), (d), and (e).Here are the results for each part:

(b) The total number of ways to randomly select 6 bottles from the 31 is:

$736,281$

(c) The number of ways to obtain exactly two bottles of each variety is:

$90,090$

(d) The probability that the selection results in two bottles of each variety is:

$0.122$

(e) The probability that all 6 bottles are of the same variety is:

$0.003$

Would you like more details or have any further questions?

Here are 5 related questions for further exploration:

1. How would the answer change if there were 15 bottles of Zinfandel instead of 11?
2. What if the total number of selected bottles was increased to 7? How would the probabilities change?
3. How can combinations be used to model other real-world selections, like choosing items for a menu?
4. If the wines were divided by year, how would that affect the selection process?
5. How would adding a fourth wine variety change the calculation of probabilities?

Tip: In probability, always ensure whether the problem requires combinations (order doesn't matter) or permutations (order matters) before solving!