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Chapter3 - Chapter 3 lecture notes Math 431 Spring 2011...

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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 } and { 6 } ? We know that P ( F ) = . 7 and that P ( { 2 } ) = . 1 , P ( { 4 } ) = . 1 , P ( { 6 } ) = . 5 . Thus, if seems like Cond. Prob. of 2 + Cond. Prob. of 3 + Cond. Prob. of 6 = . 1 /. 7 + 0 + . 5 /. 7 = 6 / 7 . Venn Diagram: 1
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How did we intuitively compute that? For E to occur given that F already has, the outcome must be in EF (so you intuitively ignored the outcome “3” in E ). Because we know that F S has occurred, we make F our new, or reduced sample space. Next, the probability of EF occurring given that F has should be the probability of EF relative to the probability of F . Definition 1. Let E, F S with P ( F ) > 0, then P ( E | F ) = P ( EF ) P ( F ) . In the first example above, EF = { 2 , 6 } and so P ( E | F ) = P ( EF ) P ( F ) = P ( { 2 , 6 } ) P ( { 2 , 4 , 6 } ) = 2 / 6 3 / 6 = 2 3 . The above is taken to be the definition of conditional probability in all cases. Example 3. The probability that a person lives to 80 is .64. The probability that a person lives until 90 is .51. What is the probability that a person will live to 90, given that that person is 80. Solution: Let A be the event of living until 90 and B be the event of living until 80. Note that A AB . Then, P ( A | B ) = P ( AB ) P ( B ) = P ( A ) P ( B ) = . 51 . 64 . 797 . Example 4. A coin is flipped twice. Assuming all points of S = { ( H, H ) , ( T, H ) , ( H, T ) , ( T, T ) } are equally likely, what is the conditional probability that both flips are heads, given that ( a ) the first flip is a head? ( b ) at least one flip is lands on heads. Solution: Let E be the event that both flips are heads. Let A be the event that the first flip is a head. Then P ( E | A ) = P ( EA ) P ( A ) = P ( { ( H, H ) } ) P ( { ( H, T ) , ( H, H ) } ) = 1 / 4 2 / 4 = 1 2 . Next, let B be the event that at least one flip is a head. Then P ( E | B ) = P ( EB ) P ( B ) = P ( { ( H, H ) } ) P ( { ( H, T ) , ( H, H ) , ( T, H ) } ) = 1 / 4 3 / 4 = 1 3 . This seems counter-intuitive as “the other could be heads or tails.” So it seems that answer should be 1/2. However, the information “at least one coin landed on heads” only tells us that ( T, T ) didn’t happen. The remaining three options are equally likely.
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