Three imported wines are to be ranked from lowest to highest by a purported wine expert. That
is, one wine will be identified as best, another as second best, and the remaining wine as worst.    a Describe one sample point for this experiment.
b List the sample space.
c Assume that the “expert” really knows nothing about wine and randomly assigns ranks to
the three wines. One of the wines is of much better quality than the others. What is the
probability that the expert ranks the best wine no worse than second best?

Let's break down the problem step by step.

### (a) Describe one sample point for this experiment.

A sample point in this experiment represents one possible ranking of the three wines, which are ranked from **lowest to highest** (i.e., worst to best). Since we have three wines, a sample point could be any permutation of the three wines. We can label the wines as \( W_1 \), \( W_2 \), and \( W_3 \).

For example, one sample point could be:

- \( W_2 \) is ranked the best,
- \( W_1 \) is ranked second,
- \( W_3 \) is ranked the worst.

This can be represented as: \( (W_2, W_1, W_3) \).

### (b) List the sample space.

The sample space for this experiment consists of all possible rankings (permutations) of the three wines. There are 3 wines, and we want to rank all of them, so the number of possible rankings is the number of permutations of 3 objects, which is \( 3! = 6 \).

The sample space \( S \) is:
\[
S = \{ (W_1, W_2, W_3), (W_1, W_3, W_2), (W_2, W_1, W_3), (W_2, W_3, W_1), (W_3, W_1, W_2), (W_3, W_2, W_1) \}
\]
Each ordered triple represents a different ranking of the three wines.

### (c) What is the probability that the expert ranks the best wine no worse than second best?

We assume the expert is randomly assigning ranks and knows nothing about wine. Let's say \( W_1 \) is the best wine.

We are interested in the probability that \( W_1 \) is ranked **no worse than second best**. In other words, \( W_1 \) should be either in the first or second position in the ranking.

#### Step 1: Count the favorable outcomes.

The total number of favorable outcomes where \( W_1 \) is either first or second can be listed:

1. \( W_1 \) is first: \( (W_1, W_2, W_3), (W_1, W_3, W_2) \)
2. \( W_1 \) is second: \( (W_2, W_1, W_3), (W_3, W_1, W_2) \)

These give us 4 favorable outcomes.

#### Step 2: Calculate the probability.

The total number of possible outcomes (sample space size) is 6.

Thus, the probability \( P \) that the best wine is ranked no worse than second is:
\[
P = \frac{4}{6} = \frac{2}{3}
\]

### Summary of Answers:
- (a) A sample point is a possible ranking, e.g., \( (W_2, W_1, W_3) \).
- (b) The sample space is \( \{ (W_1, W_2, W_3), (W_1, W_3, W_2), (W_2, W_1, W_3), (W_2, W_3, W_1), (W_3, W_1, W_2), (W_3, W_2, W_1) \} \).
- (c) The probability that the best wine is ranked no worse than second is \( \frac{2}{3} \).

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Would you like further details or explanations? Here are five related questions to expand on the topic:

1. How would the sample space change if there were four wines instead of three?
2. How is the calculation affected if two wines have equal quality and are indistinguishable?
3. Can you use combinatorics to find the probability if the expert has a bias for one wine?
4. How does the probability change if we allow ties in the ranking?
5. What is the probability that the expert ranks the worst wine in the lowest position?

**Tip:** When calculating probabilities involving multiple outcomes, listing all possible outcomes first can make it easier to identify favorable events.

Three imported wines are to be ranked from lowest to highest by a purported wine expert. That is, one wine will be identified as best, another as second best, and the remaining wine as worst. a) Describe one sample point for this experiment. b) List the sample space. c) Assume that the 'expert' really knows nothing about wine and randomly assigns ranks to the three wines. One of the wines is of much better quality than the others. What is the probability that the expert ranks the best wine no worse than second best?

Explore a probability problem involving ranking three wines from best to worst using permutations. This detailed solution covers key combinatorics and probability concepts, including how to calculate the chance of ranking the best wine no worse than second. Suitable for Grades 9-12.

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