LEC04 Conditional prob

LEC04 Conditional prob - Conditional Probability example S...

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Conditional Probability example: S = all Rutgers students E = engineers W = women consider these different probabilities P(W) = .5 P(E) = .05 P(W / E) = .2 P (E / W ) = .02 P(E W) = .01
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Definition of Conditional Probability given 2 events E S and F S P E F P E F P F P F ( / ) ( ) ( ) ( ) = 0 2
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Example: Related Defects in a Product A product can have 2 types of defects, A and B, with probabilities given below. Defect B yes no Defect yes .02 .05 A no .03 .90 P(A) = .07 P(B) = .05 P(A B) = .02 Find P(A/B) A B P A B P A B P B ( / ) ( ) ( ) . . . = = = 02 05 40 3
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Example Continued Note: P(A)=.07 while P(A/B) = .4 Find P(B/A) and compare to P(B). Find P(B’/A) and compare to P(B/A). Compare P(B/A) and P(A B) = .02 (note commutative law) Defect B yes no Defect yes .02 .05 A no .03 .90 4
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Multiplication Rule for P(E F) Definition of conditional probability: 2 events E S and F S P E F P E F P F P F ( / ) ( ) ( ) ( ) = 0 Multiplication Rule: P E F P E F P F ( ) ( / ) ( ) = 5
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Solving a Problem Using the Multiplication Rule A lot consists of 10 units - 7 good and 3 bad.
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