SolutionHW2-SP11

SolutionHW2-SP11 - Remark: independence assumption is...

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STAT 427 Home Work Assignment #2 Spring 2011 Dr. Goel Solution REMARK: In part c,   1 2 3 1 2 3 1 2 3 P exactly one ( ' ') ( ' ') ( ' ' ) P A A A P A A A P A A A
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E. P(M| SS)= P(M and SS)/P(SS) =(.08+.07+.12)/( .08+0.07+0.12+0.04+0.02+0.05+0.03+0.07+0.08) =0.27/0.56=0.50 F. P(SS|M) = P(SS and M)/P(M) = (.08+.07+.12)/(.08+.07+.12+0.1+0.05+0.07) = 0.27/0.49 = 0.55 Now, P(LS|M) = 1-P(SS|M )= 1-0.55 = 0.45
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2.76 P(no error in all questions)=[1-P(error in one question)]^10=0.9^10=0.349 P(at least one error)=1-P(no error)=1-0.349=0.651 For general case, P(no error)=(1-p)^n and P(at least one error)=1-(1-p)^n
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Unformatted text preview: Remark: independence assumption is highly important here because it enables us to reduce the probability of intersections of events to multiplication of individual probability. 2.80 2.97 If there is impurity, then the probability that 2 out of 3 experiments are detected is: 3*0.2*0.8^2=0.384 If there is no impurity, the probability that 2 out of 3 experiments are detected is: 3*0.1^2*0.9=0.027 So the posterior probability is: 0.384*0.4/(0.384*0.4+0.027*0.6)=0.9045...
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SolutionHW2-SP11 - Remark: independence assumption is...

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