A man has a bag containing 10 red balls and 20 blue balls. He randomly takes out one ball at a time without looking and stops when he gets two balls of the same color. What is the probability that he stops after taking out exactly three balls?
Let's analyze this problem step-by-step:
1. The man wants to stop when he gets two balls of the same color.
2. If he stops after taking out exactly three balls, it means that the first two balls must be of different colors, and the third ball must match one of the first two.
3. There are two possible scenarios for this situation:
   a. First ball is red, second ball is blue, and the third ball is red.
   b. First ball is blue, second ball is red, and the third ball is blue.
4. Let's calculate the probability for each scenario:
   a. P(Red-Blue-Red) = (10/30) × (20/29) × (9/28) = 1/29
   b. P(Blue-Red-Blue) = (20/30) × (10/29) × (19/28) = 19/87
5. Since these two scenarios are mutually exclusive, we can add their probabilities to find the overall probability: P(Stopping after three balls) = 1/29 + 19/87 = 36/203.

Final answer: The probability that the man stops after taking out exactly three balls is 36/203.