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Let's analyze the problem presented in the image:

There are three bags:
- Bag 1 contains three red balls and one yellow ball.
- Bag 2 contains one red ball and three yellow balls.
- Bag 3 contains two red balls and two yellow balls.

A person randomly selects one of the bags and randomly chooses one ball from the selected bag.

The problem asks: What is the probability that the person selected Bag 1?

### Solution:
1. **Determine the total number of bags**: There are 3 bags.
2. **Determine the probability of selecting each bag**: Since the selection is random, the probability of selecting any one bag is \(\frac{1}{3}\).
3. **Determine the probability of selecting a ball from each bag**:
   - **Bag 1**: There are 4 balls in total (3 red and 1 yellow). The probability of selecting any ball from Bag 1 is \(\frac{1}{4}\).
   - **Bag 2**: There are 4 balls in total (1 red and 3 yellow). The probability of selecting any ball from Bag 2 is \(\frac{1}{4}\).
   - **Bag 3**: There are 4 balls in total (2 red and 2 yellow). The probability of selecting any ball from Bag 3 is \(\frac{1}{4}\).

4. **Calculate the probability of selecting a ball from each bag given that a specific color is chosen**:
   - For example, if we consider selecting a red ball, the probabilities would be:
     - **Bag 1**: Probability of selecting a red ball is \(\frac{3}{4}\).
     - **Bag 2**: Probability of selecting a red ball is \(\frac{1}{4}\).
     - **Bag 3**: Probability of selecting a red ball is \(\frac{1}{2}\).

5. **Calculate the overall probability of selecting a red ball**:
   - Overall probability = (Probability of selecting Bag 1 * Probability of selecting a red ball from Bag 1) + (Probability of selecting Bag 2 * Probability of selecting a red ball from Bag 2) + (Probability of selecting Bag 3 * Probability of selecting a red ball from Bag 3)
   - \( P(\text{red ball}) = \left(\frac{1}{3} \times \frac{3}{4}\right) + \left(\frac{1}{3} \times \frac{1}{4}\right) + \left(\frac{1}{3} \times \frac{1}{2}\right) = \frac{1}{3} \left(\frac{3}{4} + \frac{1}{4} + \frac{2}{4}\right) = \frac{1}{3} \left(\frac{6}{4}\right) = \frac{1}{3} \left(\frac{3}{2}\right) = \frac{1}{2} \)

6. **Using Bayes' Theorem to find the probability of selecting Bag 1 given that a red ball was chosen**:
   - \( P(\text{Bag 1} | \text{red ball}) = \frac{P(\text{red ball} | \text{Bag 1}) \times P(\text{Bag 1})}{P(\text{red ball})} \)
   - \( P(\text{Bag 1} | \text{red ball}) = \frac{\left(\frac{3}{4}\right) \times \left(\frac{1}{3}\right)}{\frac{1}{2}} = \frac{\frac{3}{12}}{\frac{1}{2}} = \frac{\frac{3}{12}}{\frac{6}{12}} = \frac{3}{6} = \frac{1}{2} \)

Therefore, the probability that the person selected Bag 1 given that a red ball was chosen is \(\frac{1}{2}\).

Would you like further details or have any questions?

### Relative Questions:
1. What is the probability of selecting Bag 2 given that a red ball was chosen?
2. What is the probability of selecting Bag 3 given that a red ball was chosen?
3. How would the probabilities change if the number of balls in each bag were different?
4. What is the probability of selecting a yellow ball from any of the bags?
5. How would you solve the problem if there were four bags instead of three?

### Tip:
Understanding the application of Bayes' Theorem is crucial for solving conditional probability problems effectively.

There are three bags. Bag 1 contains three red balls and one yellow ball. Bag 2 contains one red ball and three yellow balls. Bag 3 contains two red balls and two yellow balls. A person randomly selects one of the bags and randomly chooses one ball from the selected bag. What is the probability that the person selected Bag 1 given that they picked a red ball?

Solve a conditional probability problem involving three bags of red and yellow balls. The problem explores the application of Bayes' Theorem to find the probability of selecting Bag 1 given that a red ball was chosen. Suitable for Grades 9-12.

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