Question 1201611
To estimate the probabilities based on the data provided, we will use the following information:

- **Atlantic Coast Conference (ACC)**: 10 appearances in the championship game over 20 years.
- **Southeastern Conference (SEC)**: 8 appearances in the championship game over 20 years.
- **Both conferences had a team in the championship game together only once** (1994).

### Step 1: Calculate the Probabilities

**Total Years**: 20

#### a) Probability the ACC will have a team in the championship game

The probability $ P(\text{ACC}) $ is calculated as:

$$
P(\text{ACC}) = \frac{\text{Number of times ACC had a team}}{\text{Total years}} = \frac{10}{20} = 0.5
$$

#### b) Probability the SEC will have a team in the championship game

The probability $ P(\text{SEC}) $ is calculated as:

$$
P(\text{SEC}) = \frac{\text{Number of times SEC had a team}}{\text{Total years}} = \frac{8}{20} = 0.4
$$

#### c) Probability the ACC and SEC will both have teams in the championship game

Since both conferences had a team in the championship game together only once, the probability $ P(\text{ACC} \cap \text{SEC}) $ is:

$$
P(\text{ACC} \cap \text{SEC}) = \frac{1}{20} = 0.05
$$

#### d) Probability at least one team from these two conferences will be in the championship game

To find the probability that at least one team from the ACC or SEC will be in the championship game, we can use the formula for the union of two events:

$$
P(\text{ACC} \cup \text{SEC}) = P(\text{ACC}) + P(\text{SEC}) - P(\text{ACC} \cap \text{SEC})
$$

Substituting the values we calculated:

$$
P(\text{ACC} \cup \text{SEC}) = 0.5 + 0.4 - 0.05 = 0.85
$$

#### e) Probability that the championship game will not have a team from one of these two conferences

The probability that neither the ACC nor the SEC has a team in the championship game is the complement of the probability that at least one of them does:

$$
P(\text{Neither ACC nor SEC}) = 1 - P(\text{ACC} \cup \text{SEC}) = 1 - 0.85 = 0.15
$$

### Summary of Probabilities

- a) $ P(\text{ACC}) = 0.5 $
- b) $ P(\text{SEC}) = 0.4 $
- c) $ P(\text{ACC} \cap \text{SEC}) = 0.05 $
- d) $ P(\text{ACC} \cup \text{SEC}) = 0.85 $
- e) $ P(\text{Neither ACC nor SEC}) = 0.15 $ 

These probabilities provide a clear picture of the likelihood of teams from the ACC and SEC participating in the NCAA basketball championship game based on historical data.