Chapter 3 lecture notes Math 431, Spring 2011 Summary: Chapter 3 lecture notes Math 431, Spring 2011 Instructor: David F. Anderson Chapter 3: Conditional Probability and Independence Section 3.2: Conditional Probabilities Example 1. Consider rolling a die. We know that S = {1, 2, 3, 4, 5, 6}. Let E be the event that the outcome is in {2, 3, 6}. We know that P(E) = 1/2. Now suppose I tell you that the roll was an even number. Given this knowledge, what is the probability that E occurred? Solution: Let F = {2, 4, 6}. Since we know that each of 2, 4, 6 is equally likely, the proba- bility of E now appears to be Cond. Prob. of 2 + Cond. Prob. of 3 + Cond. Prob. of 6 = 1/3 + 0 + 1/3 = 2/3. Example 2. Now consider rolling an unfair die. We know that S = {1, 2, 3, 4, 5, 6}, but assume that P({i}) = .1 for i {1, 2, 3, 4, 5} and P({6}) = .5. Let E be the event that the outcome is in {2, 3, 6}. We know that P(E) = P({2}) + P({3}) + P({6}) = .7. Now suppose I tell you that the roll was an even number. Given this knowledge, what is the probability that E occurred? Solution: Let F = {2, 4, 6}. We have that P(F) = P({2}) + P({4}) + P({6}) = .1 + .1 + .5 = .7. Clearly, the outcome {3} did not occur, so that has zero probability now. What about {2} Collections: Mathematics