1. **Problem Statement:** We have two jars: Jar A with 6 green and 4 red balls, Jar B with 2 green and 8 red balls. One ball is drawn from each jar. Given that the balls drawn have different colors, find the probability that the ball from Jar B is green. 2. **Define Events:** Let: - $G_A$ = ball from Jar A is green - $R_A$ = ball from Jar A is red - $G_B$ = ball from Jar B is green - $R_B$ = ball from Jar B is red 3. **Calculate individual probabilities:** - $P(G_A) = \frac{6}{10} = 0.6$ - $P(R_A) = \frac{4}{10} = 0.4$ - $P(G_B) = \frac{2}{10} = 0.2$ - $P(R_B) = \frac{8}{10} = 0.8$ 4. **Event of different colors:** This can happen in two ways: - Jar A green and Jar B red: $P(G_A \cap R_B) = P(G_A) \times P(R_B) = 0.6 \times 0.8 = 0.48$ - Jar A red and Jar B green: $P(R_A \cap G_B) = P(R_A) \times P(G_B) = 0.4 \times 0.2 = 0.08$ 5. **Total probability of different colors:** $$P(\text{different colors}) = 0.48 + 0.08 = 0.56$$ 6. **Find the conditional probability that the ball from Jar B is green given different colors:** We want $P(G_B | \text{different colors})$. By definition of conditional probability: $$P(G_B | \text{different colors}) = \frac{P(G_B \cap \text{different colors})}{P(\text{different colors})}$$ Since $G_B$ and different colors means the case where Jar A is red and Jar B is green: $$P(G_B \cap \text{different colors}) = P(R_A \cap G_B) = 0.08$$ Therefore: $$P(G_B | \text{different colors}) = \frac{0.08}{0.56} = \frac{8}{56} = \frac{1}{7}$$ 7. **Final answer:** The probability that the ball drawn from Jar B is green given the balls have different colors is $\boxed{\frac{1}{7}}$. This corresponds to option D.