We are given two boxes:

• Box A: Contains 4 green and 5 black coins (no blue coins mentioned, so there are no blue coins in Box A).
• Box B: Contains 6 green and 3 black coins.

Objective:

We need to find the conditional probability that a green coin was transferred from Box A to Box B, given that the coin drawn from Box B is green.

Problem Setup:

Let:

• $G_A$ be the event that a green coin was transferred from Box A to Box B.
• $G_B$ be the event that a green coin is drawn from Box B.

We are tasked with finding $P(G_A \mid G_B)$, the probability that a green coin was transferred from Box A to Box B given that the coin drawn from Box B is green.

Using Bayes' Theorem:

$P(G_A \mid G_B) = \frac{P(G_B \mid G_A) \cdot P(G_A)}{P(G_B)}$

Step 1: Calculate $P(G_A)$

$P(G_A)$ is the probability that a green coin was transferred from Box A to Box B. Since Box A contains 4 green and 5 black coins, the probability of transferring a green coin from Box A is:

$P(G_A) = \frac{4}{9}$

Step 2: Calculate $P(G_B \mid G_A)$

$P(G_B \mid G_A)$ is the probability that a green coin is drawn from Box B, given that a green coin was transferred from Box A. After transferring a green coin, Box B will have 7 green coins and 3 black coins. The probability of drawing a green coin from Box B in this case is:

$P(G_B \mid G_A) = \frac{7}{10}$

Step 3: Calculate $P(G_B \mid \neg G_A)$

Next, we calculate $P(G_B \mid \neg G_A)$, the probability of drawing a green coin from Box B given that a black coin was transferred from Box A. In this case, Box B will still have 6 green coins and will now have 4 black coins. The probability of drawing a green coin is:

$P(G_B \mid \neg G_A) = \frac{6}{10}$

Step 4: Calculate $P(G_B)$

The total probability of drawing a green coin from Box B, $P(G_B)$, can be computed using the law of total probability:

$P(G_B) = P(G_B \mid G_A) \cdot P(G_A) + P(G_B \mid \neg G_A) \cdot P(\neg G_A)$

We know that $P(\neg G_A)$ (the probability that a black coin was transferred) is:

$P(\neg G_A) = \frac{5}{9}$

Thus, we have:

$P(G_B) = \left( \frac{7}{10} \cdot \frac{4}{9} \right) + \left( \frac{6}{10} \cdot \frac{5}{9} \right)$

$P(G_B) = \frac{28}{90} + \frac{30}{90} = \frac{58}{90} = \frac{29}{45}$

Step 5: Calculate $P(G_A \mid G_B)$

Finally, we apply Bayes' Theorem to find $P(G_A \mid G_B)$:

$P(G_A \mid G_B) = \frac{P(G_B \mid G_A) \cdot P(G_A)}{P(G_B)}$

Substituting the known values:

$P(G_A \mid G_B) = \frac{\frac{7}{10} \cdot \frac{4}{9}}{\frac{29}{45}} = \frac{\frac{28}{90}}{\frac{29}{45}} = \frac{28}{90} \cdot \frac{45}{29} = \frac{28}{58} = \frac{14}{29}$

Final Answer:

The probability that a green coin was transferred from Box A to Box B given that the coin drawn from Box B is green is:

$\boxed{\frac{14}{29}}$

---

Would you like any further clarification or details on this solution?

Here are some related questions you might find useful:

1. What if the number of coins in each box changes? How does that affect the probability?
2. How would the problem change if there were blue coins in Box A?
3. What if we drew a coin from Box A directly, without transferring? How would that affect the probabilities?
4. How would the calculations change if Box B had a different number of black coins?
5. Can you explain Bayes' Theorem in more detail with another example?

Tip: Whenever dealing with conditional probabilities, remember that Bayes' Theorem is a powerful tool to reverse conditionality, especially in complex, multi-step problems.