One bag contains 3 red and 5 black balls. Another bag contains 6 red and 4 black balls. A ball is transferred from first bag to the second bag and then a ball is drawn from the second bag. Find the probability that the ball drawn is red.

Answer

To solve the problem step by step, we will analyze the situation carefully and calculate the required probabilities.

### Step 1: Identify the initial conditions
- Bag 1 contains 3 red balls and 5 black balls.
- Bag 2 contains 6 red balls and 4 black balls.

### Step 2: Calculate the total number of balls in each bag
- Total balls in Bag 1 = 3 (red) + 5 (black) = 8 balls.
- Total balls in Bag 2 = 6 (red) + 4 (black) = 10 balls.

### Step 3: Determine the probabilities of transferring a ball from Bag 1 to Bag 2
- Probability of transferring a red ball from Bag 1:
\[ P(\text{Red transferred}) = \frac{3}{8} \]

- Probability of transferring a black ball from Bag 1:
\[ P(\text{Black transferred}) = \frac{5}{8} \]

### Step 4: Calculate the probability of drawing a red ball from Bag 2 after transferring a red ball
- If a red ball is transferred, Bag 2 will have:
- Red balls = 6 + 1 = 7
- Black balls = 4
- Total balls = 7 + 4 = 11

- Probability of drawing a red ball from Bag 2 in this case:
\[ P(\text{Red drawn | Red transferred}) = \frac{7}{11} \]

### Step 5: Calculate the probability of drawing a red ball from Bag 2 after transferring a black ball
- If a black ball is transferred, Bag 2 will have:
- Red balls = 6
- Black balls = 4 + 1 = 5
- Total balls = 6 + 5 = 11

- Probability of drawing a red ball from Bag 2 in this case:
\[ P(\text{Red drawn | Black transferred}) = \frac{6}{11} \]

### Step 6: Use the law of total probability to find the overall probability of drawing a red ball
- The overall probability of drawing a red ball from Bag 2 can be calculated as:
\[ P(\text{Red drawn}) = P(\text{Red transferred}) \times P(\text{Red drawn | Red transferred}) + P(\text{Black transferred}) \times P(\text{Red drawn | Black transferred}) \]

Substituting the values:
\[ P(\text{Red drawn}) = \left(\frac{3}{8} \times \frac{7}{11}\right) + \left(\frac{5}{8} \times \frac{6}{11}\right) \]

### Step 7: Calculate the individual probabilities
1. For the first term:
\[ \frac{3}{8} \times \frac{7}{11} = \frac{21}{88} \]

2. For the second term:
\[ \frac{5}{8} \times \frac{6}{11} = \frac{30}{88} \]

### Step 8: Add the probabilities
\[ P(\text{Red drawn}) = \frac{21}{88} + \frac{30}{88} = \frac{51}{88} \]

### Final Answer
The probability that the ball drawn is red is:
\[ \frac{51}{88} \]

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