The Flotterons plan to purchase paint for the walls and for the trim in their living room. They select one color for the walls and one color for the trim at random from following colors. Walls: Java, Chocolate, Mahogany Trim: Alabaster, Vanilla, Oyster Complete parts a) through e). Use the fundamental counting principle to determine the answer to part a). Assume that each event is equally likely to occur. List the sample space, Choose the correct answer below. A. J, C, M, A, V, O B. JA, JV, JO, CA, CV, CO, MA, MV, MO O C. JA, JV, JO, CA, CV, CO, MA, MV, MO, JC, JM, CM, AV, AO, VO O D. JC, JM, CM, AV, AO, VO c) Determine the probability that they select Vanilla. The probability is $$ \square $$. (Simplify your answer.) d) Determine the probability that they select Vanilla and Chocolate. The probability is (Simplify your answer.) e) Determine the probability that they select a paint for the trim other than Vanilla. -.

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### Part a: List the Sample Space

The Flotterons have 3 colors for the walls and 3 colors for the trim. We can use the fundamental counting principle to determine the total number of combinations.

1. **Colors for Walls**: Java (J), Chocolate (C), Mahogany (M)
2. **Colors for Trim**: Alabaster (A), Vanilla (V), Oyster (O)

Using the fundamental counting principle, the total number of combinations is given by multiplying the number of choices for walls by the number of choices for trim:

$$
\text{Total combinations} = \text{Number of wall colors} \times \text{Number of trim colors} = 3 \times 3 = 9
$$

Now, let's list the sample space:

- JA (Java and Alabaster)
- JV (Java and Vanilla)
- JO (Java and Oyster)
- CA (Chocolate and Alabaster)
- CV (Chocolate and Vanilla)
- CO (Chocolate and Oyster)
- MA (Mahogany and Alabaster)
- MV (Mahogany and Vanilla)
- MO (Mahogany and Oyster)

Thus, the sample space is: 
$$
\{ JA, JV, JO, CA, CV, CO, MA, MV, MO \}
$$

The correct answer is **B**: JA, JV, JO, CA, CV, CO, MA, MV, MO.

### Part c: Probability of Selecting Vanilla

To find the probability of selecting Vanilla, we need to count how many combinations include Vanilla and divide by the total number of combinations.

The combinations that include Vanilla are:
- JV (Java and Vanilla)
- CV (Chocolate and Vanilla)
- MV (Mahogany and Vanilla)

There are 3 combinations that include Vanilla. The total number of combinations is 9.

Thus, the probability $$ P(V) $$ is:

$$
P(V) = \frac{\text{Number of combinations with Vanilla}}{\text{Total combinations}} = \frac{3}{9} = \frac{1}{3}
$$

### Part d: Probability of Selecting Vanilla and Chocolate

The only combination that includes both Vanilla and Chocolate is:
- CV (Chocolate and Vanilla)

Thus, the probability $$ P(CV) $$ is:

$$
P(CV) = \frac{\text{Number of combinations with Vanilla and Chocolate}}{\text{Total combinations}} = \frac{1}{9}
$$

### Part e: Probability of Selecting a Trim Other Than Vanilla

The combinations that do not include Vanilla are:
- JA (Java and Alabaster)
- JO (Java and Oyster)
- CA (Chocolate and Alabaster)
- CO (Chocolate and Oyster)
- MA (Mahogany and Alabaster)
- MO (Mahogany and Oyster)

There are 6 combinations that do not include Vanilla. The total number of combinations is 9.

Thus, the probability $$ P(\text{not V}) $$ is:

$$
P(\text{not V}) = \frac{\text{Number of combinations without Vanilla}}{\text{Total combinations}} = \frac{6}{9} = \frac{2}{3}
$$

### Summary of Answers
- Part a: **B**
- Part c: $$ \frac{1}{3} $$
- Part d: $$ \frac{1}{9} $$
- Part e: $$ \frac{2}{3} $$
- **Part a:** B - **Part c:** $$ \frac{1}{3} $$ - **Part d:** $$ \frac{1}{9} $$ - **Part e:** $$ \frac{2}{3} $$

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