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Let's go over the problem step-by-step to solve it.

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### Problem Analysis

1. **Bag A** contains:
   - 6 red cards
   - 2 black cards
   - Total cards in Bag A = 8

2. **Bag B** contains:
   - 7 red cards
   - 3 black cards
   - Total cards in Bag B = 10

3. The process of drawing cards:
   - First draw from **Bag A**:
     - If a red card is drawn, it is placed into **Bag B**.
     - If a black card is drawn, it is also placed into **Bag B**.
   - Another draw from **Bag A** follows (regardless of the first draw).
   - After these two draws from Bag A, a final card is drawn from **Bag B**.

4. **Goal**: To find the probability of drawing two red cards from **Bag A**, followed by one black card from **Bag B**.

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### Step-by-Step Solution

1. **First Draw from Bag A**:
   - Probability of drawing a red card from Bag A = \( \frac{6}{8} = \frac{3}{4} \).
   - Probability of drawing a black card from Bag A = \( \frac{2}{8} = \frac{1}{4} \).

2. **Second Draw from Bag A**:
   - Since we're only interested in scenarios where both draws from Bag A are red cards, let's focus on this path:
     - Probability of drawing a red card on the first draw = \( \frac{3}{4} \).
     - After drawing one red card, Bag A has 5 red and 2 black cards left.
     - Probability of drawing a red card on the second draw = \( \frac{5}{7} \).

   - Combined probability of drawing two red cards from Bag A:
     \[
     P(\text{two red cards from Bag A}) = \frac{3}{4} \times \frac{5}{7} = \frac{15}{28}
     \]

3. **Final Draw from Bag B**:
   - After transferring two red cards to Bag B, Bag B now has:
     - 9 red cards
     - 3 black cards
     - Total cards in Bag B = 12

   - Probability of drawing a black card from Bag B = \( \frac{3}{12} = \frac{1}{4} \).

4. **Overall Probability**:
   - The overall probability of drawing two red cards from Bag A followed by one black card from Bag B:
     \[
     P(\text{two red cards from A and one black from B}) = \frac{15}{28} \times \frac{1}{4} = \frac{15}{112}
     \]

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### Answer
The probability of drawing two red cards from Bag A followed by one black card from Bag B is \( \frac{15}{112} \).

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Would you like more details on any step, or have any questions?

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#### Related Questions
1. How would the probability change if we needed at least one black card drawn from Bag A?
2. What is the probability of drawing two black cards from Bag A?
3. How does the probability change if a card drawn from Bag A is not replaced in Bag B?
4. How would the answer be affected if Bag B started with 5 red and 5 black cards?
5. What if we wanted the probability of drawing two cards of the same color from both bags?

#### Tip
In probability problems with multiple stages, break down each event separately and multiply the probabilities for the desired sequence.

Bag A contains 6 red cards and 2 black cards. Bag B contains 7 red cards and 3 black cards. Suppose that a card is randomly taken out of Bag A. If a red card is taken out of Bag A, then that red card is put into Bag B. If a black card is taken out of Bag A, then that card is put back into Bag A. Then a second card is taken out of Bag A and put back into Bag A, irrespective of color. A card is then randomly taken out of Bag B. What is the probability of choosing two red cards from Bag A followed by one black card from Bag B?

This problem involves calculating the probability of drawing two red cards from Bag A followed by one black card from Bag B. Using concepts of conditional probability and the multiplication rule for independent events, this solution provides a step-by-step approach suitable for Grades 11-12.

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