1. **State the problem:** We have two bags: - Bag A: 7 blue, 4 red, 1 yellow discs (total 12 discs). - Bag B: 3 blue, 6 red discs (total 9 discs). A disc is drawn at random from Bag A and placed into Bag B. Then a disc is drawn at random from Bag B. We want to find the probability that the disc drawn from Bag B is red. 2. **Formula and approach:** We use the law of total probability considering the color of the disc transferred from Bag A to Bag B. Let $R_B$ be the event "disc drawn from Bag B is red". Then: $$P(R_B) = P(R_B|\text{blue transferred})P(\text{blue transferred}) + P(R_B|\text{red transferred})P(\text{red transferred}) + P(R_B|\text{yellow transferred})P(\text{yellow transferred})$$ 3. **Calculate probabilities of transferring each color from Bag A:** - $P(\text{blue transferred}) = \frac{7}{12}$ - $P(\text{red transferred}) = \frac{4}{12} = \frac{1}{3}$ - $P(\text{yellow transferred}) = \frac{1}{12}$ 4. **Calculate $P(R_B|\text{color transferred})$ for each case:** - If blue transferred, Bag B has 4 blue, 6 red discs (total 10). So: $$P(R_B|\text{blue transferred}) = \frac{6}{10} = \frac{3}{5}$$ - If red transferred, Bag B has 3 blue, 7 red discs (total 10). So: $$P(R_B|\text{red transferred}) = \frac{7}{10}$$ - If yellow transferred, Bag B has 3 blue, 6 red, 1 yellow discs (total 10). So: $$P(R_B|\text{yellow transferred}) = \frac{6}{10} = \frac{3}{5}$$ 5. **Calculate total probability:** $$P(R_B) = \frac{7}{12} \times \frac{3}{5} + \frac{4}{12} \times \frac{7}{10} + \frac{1}{12} \times \frac{3}{5}$$ 6. **Simplify each term:** $$\frac{7}{12} \times \frac{3}{5} = \frac{21}{60}$$ $$\frac{4}{12} \times \frac{7}{10} = \frac{28}{120} = \frac{14}{60}$$ $$\frac{1}{12} \times \frac{3}{5} = \frac{3}{60}$$ 7. **Sum all terms:** $$P(R_B) = \frac{21}{60} + \frac{14}{60} + \frac{3}{60} = \frac{38}{60}$$ 8. **Simplify fraction:** $$\frac{38}{60} = \frac{\cancel{2} \times 19}{\cancel{2} \times 30} = \frac{19}{30}$$ **Final answer:** $$\boxed{\frac{19}{30}}$$ This means the probability that the disc drawn from Bag B is red after the transfer is $\frac{19}{30}$.