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Unformatted text preview: Lecture Stat 302 Introduction to Probability  Slides 5 AD Jan. 2010 AD () Jan. 2010 1 / 20 Conditional Probabilities Conditional Probability . Consider an experiment with sample space S . Let E and F be two events, then the conditional probability of E given F is denoted by P ( E j F ) and satis&es if P ( F ) > P ( E j F ) = P ( E \ F ) P ( F ) Intuition : If F has occured, then, in order for E to occur, it is necessary that the occurence be both in E and F , hence it must be in E \ F . Once F has occured, F is the new sample space. Equally likely outcomes . In this case, we have P ( E j F ) = # outcomes in E \ F # outcomes in F = # outcomes in E \ F # outcomes in S  {z } P ( E \ F ) / & # outcomes in F # outcomes in S ¡  {z } P ( F ) . AD () Jan. 2010 2 / 20 Example Mr and Mrs. Smith have 2 children, boys or girls, the 4 con&gurations being equally likely. What is the probability that the two children are two girls if 1) you are not given any additional information, 2) the elder child is a girl, 3) one of the two children is a girl. Answer to 1) : By assumption, we have P ( f G , G g ) = P ( f G , B g ) = P ( f B , G g ) = P ( f B , B g ) = 1 4 so if you don¡t have any additional information P ( f G , G g ) = 1. Answer to 2) : Now we know that the elder child is a girl so the event F = ( f G , G g , f G , B g ) occured. We are interested in P ( f G , G gj F ) = P ( F \ f G , G g ) P ( F ) = P ( f G , G g ) P ( F ) = 1 4 2 4 = 1 2 . Answer to 3): Now we know that one of the two children is a girl so the event F = ( f G , G g , f G , B g , f B , G g ) occured. We have P ( f G , G gj F ) = P ( F \ f G , G g ) P ( F ) = P ( f G , G g ) P ( F ) = 1 4 3 4 = 1 3 . AD () Jan. 2010 3 / 20 Examples Example : A math teacher gave her class two tests. 25% of the class passed both tests and 42% of the class passed the &rst test. What percent of those who passed the &rst test also passed the second test? Answer : Let E i = ¡Event passed test i ¢. We have P ( E 1 ) = . 42 and P ( E 1 \ E 2 ) = . 25 so P ( E 2 j E 1 ) = P ( E 1 \ E 2 ) P ( E 1 ) = . 25 . 42 = . 60 Example : The probability that it is Friday and that a student is absent is 0.03. Since there are 5 school days in a week, the probability that a school day is Friday is 0.2. What is the probability that a student is absent given that today is Friday? Answer : We have P ( Absent j Friday ) = P ( Absent on Friday ) P ( Friday ) = . 03 . 2 = . 15 . AD () Jan. 2010 4 / 20 The Multiplication Rule Let E 1 , E 2 , . . . , E n be a sequence of events, then we have P ( E 1 \ E 2 \ &&& \ E n ) = P ( E 1 ) P ( E 2 j E 1 ) ¡ ¡ P ( E 3 j E 1 \ E 2 ) &&& P ( E n j E 1 \ &&& \ E n ¢ 1 ) Proof follows from P ( E 1 \ &&& \ E n ) = P ( E 1 \&&&\ E n ) P ( E 1 \&&&\ E n ¢ 1 ) P ( E 1 \ E 2 \ &&& \ E n ¢ 1 ) = P ( E n j E 1 \ &&& \ E n ¢ 1 ) P ( E 1 \&&&\ E n ¢ 1 ) P ( E 1 \&&&\ E n ¢ 2 ) P ( E 1 \ &&& \ E n ¢ 2 ) i.e. apply repeatedly P ( A j B ) = P ( A \ B ) / P ( B ) where A = E 1 \ &&& \ E i and B = E...
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 Spring '10
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 Conditional Probability, Probability, Sudden infant death syndrome, HIV test

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