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Law of Total Probability and Bayes’ Rule “Event-composition method” • Understand the experiment and sample points. • Using set notation, express the event of interest in terms of events for which the probability is known. • Applying probability rules, combine the known probabilities to determine the probability of the specified event. Problem 2.86 • In a factory, 40% of items produced come from Line 1 and others from Line 2. • Line 1 has a defect rate of 8%. Line 2 has a defect rate of 10%. • For randomly selected item, find probability the item is not defective. A: the selected item is not defective Problem 2.86 • A: the selected item is not defective. S A B1 B2 A ( A B1 ) ( A B2 ) • B1: item came from Line 1. B2: item came from Line 2. Problem 2.86 • So we may write A ( A B1 ) ( A B2 ) • Since this is the union of disjoint sets, the Additive Law yields P( A) P( A B1 ) P( A B2 ) • Or, in terms of conditional probabilities P( A) P( A | B1 ) P( B1 ) P( A | B2 ) P( B2 ) P( A) (0.08)(0.40) (0.10)(0.60) 0.092 The Decision Tree defective Line 1 not defective Line 2 defective not defective Problem 2.94 • Must find blood donor for an accident victim in the next 8 minutes or else… • Checking blood types of potential donors requires 2 minutes each and may only be tested one at a time. • 40% of the potential donors have the required blood type. • What is the probability a satisfactory blood donor is identified in time to save the victim? Finding a Donor • A: blood donor is found within 8 minutes • Some sample points: “B bad, G good” A = { (G), (B,G), (B,B,G), (B,B,B,G) } • Let Ai: ith donor has correct blood type A ( A1 ) ( A1 A2 ) ( A1 A2 A3 ) ( A1 A2 A3 A4 ) 4 mutually exclusive events Finding a Donor A ( A1 ) ( A1 A2 ) ( A1 A2 A3 ) ( A1 A2 A3 A4 ) • Trials are independent and each P(Ai) = 0.40, and so P( A) P( A1 ) P( A1 ) P( A2 ) P( A1 ) P( A2 ) P( A3 ) P( A1 ) P( A2 ) P( A3 ) P( A4 ) P( A) 0.4 (0.6)(0.4) (0.6)(0.6)(0.4) (0.6)(0.6)(0.6)(0.4) .8704 Finding a Donor A1 saved! A2 saved! A1 A3 saved! A2 A4 saved! A3 Or, more simply, P( A) 1 P(donor is not found) A4 too 1 P( A1 ) P( A2 ) P( A3 ) P( A4 ) late! 1 (0.6) 4 Problem 2.96 • Of 6 refrigerators, 2 don’t work. • The refrigerators are tested one at a time. • When tested, it’s clear whether it works! A. What is the probability the last defective refrigerator is found on the 4th test? B. What is the probability no more than 4 need to be tested to identify both defective refrigerators? Problem 2.96 C. Given that exactly one defective refrigerator was found during the first 2 tests, what is the probability the other one is found on the 3rd or 4th test? Partition of the Sample Space S B1 B2 … Bk A collection of sets {B1 , B2 , , Bk } such that 1. S B1 B2 Bk , and 2. Bi B j , for i j , is called a partition of S . Union of Disjoint Sets S A B1 B2 … Bk For the partition {B1 , B2 , , Bk } of the sample space S , we may write A ( A B1 ) ( A B2 ) ( A Bk ) and so P(A) P( A B1 ) P( A B2 ) P( A Bk ). Recall Problem 2.86 • A: the selected item is not defective. S A Not defective B1 B2 A ( A B1 ) ( A B2 ) • B1: item came from Line 1. B2: item came from Line 2. Law of Total Probability S A If P ( Bi ) 0, then … B1 B2 Bk P ( A Bi ) P ( A | Bi ) P ( Bi ) Hence, for the partition {B1 , B2 , , Bk } of the sample space S , we have P(A) P( A B1 ) P( A B2 ) P( A Bk ) or equivalently, P(A) P( A | B1 ) P( B1 ) P( A | B2 ) P( B2 ) P( A | Bk ) P( Bk ). Total Probability A P(A|B1)P(B1) B1 A A P(A|B2)P(B2) B2 B3 A A P(A|B3)P(B3) A P(A) P( A | B1 ) P( B1 ) P( A | B2 ) P( B2 ) P( A | B3 ) P( B3 ). Bayes’ Theorem follows… Since P(A) P( A | B1 ) P( B1 ) P( A | Bk ) P( Bk ), we also have P( A B j ) P( B j | A) P( A) P( A B j ) P( A | B1 ) P( B1 ) P( A | Bk ) P( Bk ) or simply, P( A B j ) P( B j | A) k P( A | B ) P( B ) i 1 i i Bayes’ A P(A|B1)P(B1) B1 A A P(A|B2)P(B2) B2 B3 A A P(A|B3)P(B3) A P( A B2 ) P( B2 | A) P( A | B1 ) P( B1 ) P( A | B2 ) P( B2 ) P( A | B3 ) P( B3 ) Making Resistors • Three machines M1, M2, and M3 produce “1000- ohm” resistors. • M1 produces 80% of resistors accurate to within 50 ohms, M2 produces 90% to within 50 ohms, and M3 produces 60% to within 50 ohms. • Each hour, M1 produces 3000 resistors, M2 produces 4000, and M3 produces 3000. • If all of the resistors are mixed together and shipped in a single container, what is the probability a selected resistor is accurate to within 50 ohms? Making Resistors • Define A: resistor is accurate to within 50 ohms. • M1 produces 80% of resistors accurate to within 50 ohms, M2 produces 90% to within 50 ohms, and M3 produces 60% to within 50 ohms. P( A | M 1 ) 0.80, P( A | M 2 ) 0.90, and P( A | M 3 ) 0.60 • Each hour, M1 produces 3000 resistors, M2 produces 4000, and M3 produces 3000. P( M 1 ) 0.3, P( M 2 ) 0.40, and P( M 3 ) 0.30. Using Total Probability Since P(A) P( A | M 1 ) P( M 1 ) P( A | M 2 ) P ( M 2 ) P ( A | M 3 ) P ( M 3 ) we have P(A) (0.8)(0.3) (0.9)(0.4) (0.6)(0.3) 0.78 That is, 78 % are expected to be accurate to within 50 ohms. The Tree A (0.8)(0.3) M1 P (A) (0.8)(0.3) A (0.9)(0.4) A (0.9)(0.4) M2 (0.6)(0.3) A 0.78 M3 A (0.6)(0.3) A …given it’s within 50 ohms… • Determine the probability that, given a selected resistor is accurate to within 50 ohms, it was produced by M1. P( M1 | A) = ? • Determine the probability that, given a selected resistor is accurate to within 50 ohms, it was produced by M3. P( M3 | A) = ? Given A… A (0.8)(0.3) P (M1 | A) M1 A P(M1 A) A (0.9)(0.4) M2 P ( A) (0.8)(0.3) M3 A (0.6)(0.3) 0.78 A A Arthritis • A test detects a particular type of arthritis for individuals over 50 years old. • 10% of this age group suffers from this arthritis. • For individuals in this age group known to have the arthritis, the test is correct 85% of the time. • For individuals in this age group known to NOT have the arthritis, the test indicates arthritis (incorrectly!) 4% of the time. • P( has arthritis | tests positive ) = ? Arthritis • 10% of this age group suffers from this arthritis. P(have arthritis) = 0.10 • For individuals in this age group known to have the arthritis, the test is correct 85% of the time. P( tests positive | have arthritis ) = 0.85 • For individuals in this age group known to NOT have the arthritis, the test indicates arthritis (incorrectly!) 4% of the time. P(tests positive | no arthritis ) = 0.04 • P( have arthritis | tests positive ) = ? P(has arthritis | tests positive ) = ? positive 0.85 Has arthritis negative 0.1 0.9 No arthritis 0.04 positive negative The 3 Urns • Three urns contain colored balls. Urn Red White Blue 1 3 4 1 2 1 2 3 3 4 3 2 • An urn is selected at random and one ball is randomly selected from the urn. • Given that the ball is red, what is the probability it came from urn #2 ? The Lost Labels • A large stockpile of cases of light bulbs, 100 bulbs to a box, have lost their labels. • The boxes of bulbs come in 3 levels of quality: high, medium, and low. • It’s known 50% of the boxes were high quality, 25% medium, and 25% low. • Two bulbs will be tested from a box to check if they’re defective. Lost Labels… • The likelihood of finding defective bulbs is dependent on the bulb quality: Number of defects Low Medium High 0 .49 .64 .81 1 .42 .32 .18 2 .09 .04 .01 • Given neither bulb is found to be defective, what is the probability the bulbs came from a box of high quality bulbs?

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posted: | 9/16/2011 |

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