Draw an appropriate tree diagram, and use the multiplication principle to calculate the probabilities of all the outcomes. HINT [See Example 3.]
Each day, there is a 10% chance that you will sell an automobile. You know that 20% of all the automobiles you sell are two-door models and the rest are four-door models.


To solve this problem, we can use a **tree diagram** and the **multiplication principle** to calculate the probabilities of all outcomes. Here's how the process works:

### 1. Understanding the problem:
- The probability of selling an automobile on any given day is 10% or \( P(\text{Sell}) = 0.10 \).
- The probability of not selling an automobile on a given day is \( P(\text{No Sell}) = 1 - P(\text{Sell}) = 0.90 \).
- If an automobile is sold, 20% are two-door models: \( P(\text{Two-Door}|\text{Sell}) = 0.20 \).
- The remaining 80% are four-door models: \( P(\text{Four-Door}|\text{Sell}) = 0.80 \).

### 2. Drawing the tree diagram:

We'll break the process into steps using branches to represent each decision point:

- First branch: Decide whether you sell an automobile or not.
- Second branch: If you sell, determine whether it's a two-door or a four-door model.

### Tree Diagram Explanation:
- **Start** → Two options: "Sell" or "No Sell"
    - From "Sell" → Two options: "Two-Door" or "Four-Door"
    - From "No Sell" → No further branches (as no car is sold).

### 3. Multiplication Principle:

Now, we calculate the probabilities of all the possible outcomes using the multiplication principle (multiply the probabilities along each branch):

#### Outcomes:
1. **No Sell**:  
   Probability:  
   \( P(\text{No Sell}) = 0.90 \)

2. **Sell and Two-Door**:  
   Probability:  
   \( P(\text{Sell}) \times P(\text{Two-Door}|\text{Sell}) = 0.10 \times 0.20 = 0.02 \)

3. **Sell and Four-Door**:  
   Probability:  
   \( P(\text{Sell}) \times P(\text{Four-Door}|\text{Sell}) = 0.10 \times 0.80 = 0.08 \)

### 4. Final Probabilities of All Outcomes:

- **No Sell**: \( 0.90 \)
- **Sell and Two-Door**: \( 0.02 \)
- **Sell and Four-Door**: \( 0.08 \)

### Conclusion:

The probabilities of all the possible outcomes are:
- Not selling a car: 90%
- Selling a two-door car: 2%
- Selling a four-door car: 8%

Would you like me to illustrate the tree diagram as well?

### 5 Follow-Up Questions:
1. What is the total probability of selling a car?
2. How would the probabilities change if the chance of selling a car was 15%?
3. What happens to the probabilities if 50% of sold cars are two-door models?
4. How could you extend the tree diagram to account for multiple sales on different days?
5. How would the probabilities adjust if a special discount day increased the chance of selling a car to 25%?

**Tip:** The multiplication principle can be applied to any sequence of events where the outcome of one event depends on the previous events!

Draw an appropriate tree diagram, and use the multiplication principle to calculate the probabilities of all the outcomes. HINT [See Example 3.] Each day, there is a 10% chance that you will sell an automobile. You know that 20% of all the automobiles you sell are two-door models and the rest are four-door models.

Learn how to calculate the probabilities of selling two-door or four-door automobiles using a tree diagram and the multiplication principle. This problem demonstrates key concepts in probability theory, perfect for students in Grades 9-11.

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