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.