A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first

Explanation

To solve the problem, we need to find the probability that a red ball drawn came from the first bag. We can use Bayes' theorem for this purpose.

### Step-by-Step Solution:

1. **Define Events**:
- Let \( B_1 \) be the event that the ball is drawn from Bag 1.
- Let \( B_2 \) be the event that the ball is drawn from Bag 2.
- Let \( R \) be the event that the ball drawn is red.

2. **Determine Prior Probabilities**:
- Since one of the two bags is selected at random, we have:
\[ P(B_1) = P(B_2) = \frac{1}{2} \]

3. **Calculate Probability of Drawing a Red Ball from Each Bag**:
- For Bag 1, which contains 4 red and 4 black balls (total 8 balls):
\[ P(R | B_1) = \frac{4}{8} = \frac{1}{2} \]
- For Bag 2, which contains 2 red and 6 black balls (total 8 balls):
\[ P(R | B_2) = \frac{2}{8} = \frac{1}{4} \]

4. **Use Total Probability to Find \( P(R) \)**:
- We can find the total probability of drawing a red ball using the law of total probability:
\[ P(R) = P(R | B_1) P(B_1) + P(R | B_2) P(B_2) \]
- Substituting the values we have:
\[ P(R) = \left(\frac{1}{2} \cdot \frac{1}{2}\right) + \left(\frac{1}{4} \cdot \frac{1}{2}\right) \]
\[ P(R) = \frac{1}{4} + \frac{1}{8} = \frac{2}{8} + \frac{1}{8} = \frac{3}{8} \]

5. **Apply Bayes' Theorem**:
- We want to find \( P(B_1 | R) \):
\[ P(B_1 | R) = \frac{P(R | B_1) P(B_1)}{P(R)} \]
- Substituting the known values:
\[ P(B_1 | R) = \frac{\left(\frac{1}{2}\right) \left(\frac{1}{2}\right)}{\frac{3}{8}} \]
\[ P(B_1 | R) = \frac{\frac{1}{4}}{\frac{3}{8}} = \frac{1}{4} \cdot \frac{8}{3} = \frac{2}{3} \]

### Final Answer:
The probability that the ball drawn is from the first bag is \( \frac{2}{3} \).