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05 Introduction to Probability Part 1

A survey was recently conducted among 200 sports

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A survey was recently conducted among 200 sports journalists about the outcomes of the Phillies’ upcoming games against Reds and Cardinals. 50 thought Phillies would beat Reds (R) 110 thought Phillies would beat Cardinals (C) 60 thought Phillies would lose both P(R) = 50/200 = 0.25, P(C) = 0.55, P(R’∩ C’) = 0.30
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33 Example: Phillies baseball! Venn Diagram Representation for Phillies baseball 33 Sample space R C P(R and C)=0.10 R: Phillies beating Reds C: Phillies beating Cardinals P(R) = 0.25 P(C) = 0.55 P(R or C)=0.70 P(R’ ∩ C’) = P(R U C)’ = 0.30 30 journalists 20 journalists 90 journalists 60 journalists R C
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a) What is the probability that a respondent said that Phillies would win both games? Need P(R ∩ C) First, compute P(R U C) = 1 – P(R’ ∩ C’) = 1 – 0.30 = 0.70 Next, use P(R U C) = P(R) + P(C) – P(R∩C) 0.70 = 0.25 + 0.55 - P(R∩C) P(R∩C) = 0.80 – 0.70 = 0.10 Example: Phillies baseball !
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b) What is the probability that a respondent said that Phillies would beat Cardinals but lose to Reds? Need P(C∩R’) Recall: P(C) = P(C∩R) + P(C∩R’) 0.55 = 0.10 + P(C∩R’) P(C∩R’) = 0.45 Example: Phillies baseball !
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36 Mutually exclusive events 36 Mutually Exclusive = no overlap between events For mutually exclusive events, the intersection is always zero: A C A: the sum is even C: the sum is 7 or 11 2 4 6 8 10 12 7 11 0 C) P(A = 0 C) P(A = 0 C) P(A =
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37 Compound Probability for Mutually Exclusive Events A collection of events, is said to be mutually exclusive if for all pairs For two events For a collection of mutually exclusive events 37 = 0 ) P(E ) P(E ) E P(E ) P(E ) P(E ) E P(E 2 1 2 1 2 1 2 1 + = - + = 0 E E j i = k 2 1 E , , E , E ) P(E ) P(E ) P(E ) E E P(E k 2 1 k 2 1 + + + =
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A survey was recently conducted among 200 sports...

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