Where does P(A|Bbar) = 1/5 and P(Abar|Bbar) = 4/5 come from? I know that if you have one, you can figure out the other. But is one of them just a given?

College Algebra
7th Edition
ISBN:9781305115545
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter1: Equations And Graphs
Section: Chapter Questions
Problem 10T: Olympic Pole Vault The graph in Figure 7 indicates that in recent years the winning Olympic men’s...
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Where does P(A|Bbar) = 1/5 and P(Abar|Bbar) = 4/5 come from? I know that if you have one, you can figure out the other. But is one of them just a given? 

Example 4.1: Assessing Uncertainty at Bender Company
(slide 2 of 2)
O The uncertain situation is depicted
graphically in the form of a
probability tree.
P(AIB) = 3/4
P(A and B) = (3/4)(2/3)
%3D
A
P(B) = 2/3
P(AIB) = 1/4
P(A and B) = (1/4)(2/3)
B
P(AIB) = 1/5
P(A and B) = (1/5)(1/3)
P(B) = 1/3
A
O The addition rule for mutually exclusive
events implies that:
P(ĀIB) = 4/5
P(Ā and B) = (4/5)(1/3)
%3D
P(A) = P(A and B) + P(A and B) = 1/2 + 1/15 = 17/30 = 0.5667
© Cengage Learming
Transcribed Image Text:Example 4.1: Assessing Uncertainty at Bender Company (slide 2 of 2) O The uncertain situation is depicted graphically in the form of a probability tree. P(AIB) = 3/4 P(A and B) = (3/4)(2/3) %3D A P(B) = 2/3 P(AIB) = 1/4 P(A and B) = (1/4)(2/3) B P(AIB) = 1/5 P(A and B) = (1/5)(1/3) P(B) = 1/3 A O The addition rule for mutually exclusive events implies that: P(ĀIB) = 4/5 P(Ā and B) = (4/5)(1/3) %3D P(A) = P(A and B) + P(A and B) = 1/2 + 1/15 = 17/30 = 0.5667 © Cengage Learming
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Cengage Learning