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October 1, 2014

October 1, 2014

Posted by **Preet** on Monday, July 23, 2012 at 1:11am.

P(A)=0.8, P(B|A)=0.3, P(B|A^c)=0.8

Find:

1) P(B)

2) P(A|B)

3) P(A^c|B^c)

- Finite Math -
**MathMate**, Monday, July 23, 2012 at 9:07amIn the following, A' will be used in place of A

^{c}for ease of typography.

(1) since A and A' are partitions of A, we can apply the law of total probability

P(B)=P(B|A)*P(A)+P(B|A')*P(A')

(2) Use Bayes Theorem

P(A|B)

=P(A∩B)/P(B) ... def of cond. prob

=P(B∩A)/P(B) ... commutativity

=[P(B|A)*P(A)]/P(B) ... def of cond. prob

(3) We need to use

P(A'∩B')=1-P(A∪B) ... complement

and

P(A∪B)=P(A)+P(B)-P(A∩B) ... prob. of union of sets

P(A'|B')

=P(A'∩B')/P(B') .. def of cond.prob

=(1-P(A∪B))/(1-P(B)) ... compl

=(1-P(B∪A))/(1-P(B)) ... commutativity

=[1-(P(A)+P(B)-P(A∩B)]/(1-P(B)) ... union of sets

=[1-(P(A)+P(B)-P(B∩A)]/(1-P(B)) ... commutativity

=[1-(P(A)+P(B)-P(B|A)*P(A)]/(1-P(B)) ... def. of cond. prob.

Please check and understand every line. Post if you have questions.

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