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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}

  

Source: Anderson, David F. - Department of Mathematics, University of Wisconsin at Madison

 

Collections: Mathematics