Part D - Comparing Conditional Probability (possible 24 points) Activity: You learned about Conditional 0.99 P D D = person has the disease Probability in Lesson 8. Use your 0.02 knowledge to answer the following 0.01 H = person is healthy N P = person tests positive questions using the given tree diagram. For example to find P(P | D), 0.03 P N = person tests negative which is the probability that a person 0.98 with a disease will test positive for it H 0.97 N will be P(P | D) = 0.99. Example: Find P(H | P). Since P(H | P) = P ( H and P) P ( P find P(H and P) and P(P). P(H and P) = 0.98 . 0.03 P(P) = P(D and P) or P(H and P) = 0.0294 = 0.02 . 0.99 + 0.98 . 0.03 = 0.0492 So P(H | P) = P ( H and P) 0.0294 Substitute. P (P) 0.0492 = 0.598 Simplify. About 60% of the people who test positive do not actually have the disease. Use the tree diagram in the Example above to find each probability for 1-4. (2 pts each) 1. P(N) 2. P(H and N) 3. P(HIN) 4. P(D | N)