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Lec05 Total prob and independence

# Lec05 Total prob and independence - 1 Total Probability...

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Total Probability Rule and Independence 1

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An old friend, the weighted average 2 Weighted Average: There are two suppliers A and B. A supplies 80% of parts with fraction defective .01. B supplies 20% of parts with fraction defective .07. Guess the fraction of defective parts. Greater than 0.01? Greater than 0.07? Closer to 0.01 or 0.07?
2 Suppliers with Notation 3 Given: event D = part is defective event A= part comes from A P(A) =.8 P(D/A)=.01 P(A’) =.2 P(D/A’)=.07 Solution: Events A and A’ are exhaustive and mutually exclusive P(D)=P(D/A)P(A)+P(D/A’)P(A’)=

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Statement of Total Probability Rule 4 Consider events D S and A S ( )= ( / ) ( )+ ( / ’) ( ’) P D P D A P A P D A P A : Proof ( )= ( )+ ( ’) P D P D⋂A P D⋂A & A and A are exhaustive mutually exclusive Apply mult rule to get ( )= ( )+ ( ’)= P D P D⋂A P D⋂A ( / ) ( )+ ( / ’) ( ’) P D A P A P D A P A
Suppose you have many suppliers, A1, A2, …, A n 5

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Example with 4 Suppliers 6 Fin d fra c tio n d e fe c tiv e u n its w ith t h e s e s u p p lie rs An s : 0 .0 2 8 Supplier % of supply fraction def.
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