This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 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 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....
View
Full
Document
This note was uploaded on 04/17/2011 for the course MATH 431 taught by Professor Balazs during the Spring '05 term at Wisconsin.
 Spring '05
 BALAZS
 Conditional Probability, Probability

Click to edit the document details