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!