Lecture 4

1. Lecture 4 Conditional Probability, Total Probability Rule Instructor: Kaveh Zamani Course material mainly developed by previous instructors: Profs. Mokhtarian and Kendall, Ms. Reardon Lecture 4: Conditional Probability

2. Reminder • Previous Lecture: • - Axioms of probability • P(S) =1 • 0P(A) 1 • A1 and A2 with A1 ∩ A2 = ∅ • P(A1 ∪ A2 ) = P(A1) + P(A2) • - Probability of multiplication • - Mutually exclusive events • Next session: • Reading: section 2.7 text book • HW #2: posted on SmartSite problems: Lecture 4: Conditional Probability

3. Definition • Sometimes probabilities need to be reevaluated as additional information becomes available • The probability of an event B under the knowledge that the outcome will be in event A is denoted as: P(B|A) • This is called the conditional probability of B given A Lecture 4: Conditional Probability

4. Example 1,2 • A: event of rainy day in May • B: event of day colder than 40 °F in May • P(A|B) > P(A) • Chance of rain in a cold weather is higher than the average chance of rain! • A: event of certain heart disease in people older than 60 • B: event of abdominal obesity in people older than 60 • P(A)= 10% Probability of the heart disease • P(B)= 30% Probability of abdominal obesity • P(B’)=1-P(B) Probability of being slim • P(A|B)= 20% probability of the heart disease given that the person is fat P(A|B) > P(A), P(A|B’)< P(A) Lecture 4: Conditional Probability

5. Example 3: Welding (Page 41) • Automatic welding devices have error rates of 1/1000 • Errors are rare, but when they occur, because of wearing of the device, they tend to occur in groups that affect many consecutive welds • If a single weld is performed, we might assume the probability of an error as 1/1000 • However, if the previous welding was wrong, because of the wearing, we might believe that the probability that the next welding is wrong is greater than 1/1000, [P(Wi+1|Wi)>1/1000] Lecture 4: Conditional Probability

6. Example 4: Manufacturing (Page 42) D : a part of a steel column is defective F : a part of a steel column has a surface defect, P(F) = 0.10, P(D|F) = 0.25 and P(D|F’) = 0.05 Lecture 4: Conditional Probability

7. Conditional Probability • The conditional probability of an event B given an event A, denoted as P(B|A), is • P(B|A) = P(B ∩ A) / P(A) for P(A)>0 • Therefore, P(B|A) can be interpreted as the relative frequency of event B among the trials that produce an outcome in event A • It is like scaling down to a smaller sample space Lecture 4: Conditional Probability

8. Example 4 • D = a part is defective • F = a part has a surface flaw [ P(F) = 0.10 ] • P(D|F) = 0.25 and P(D|F’) = 0.05 • P(D|F) = P(D ∩ F) / P(F) = 0.025 / 0.1 = 0.25 Lecture 4: Conditional Probability

9. Example 5: (2-78, Page 45) • 100 samples of a cast aluminum part are summarized as: P(A) = 82/100 = 0.82 P(B) = 90/100 = 0.90 P(B|A)= 80/82 = 0.9756 P(A|C)= 2/10 = 0.2 P(A|B)= 80/90 = 0.889 Lecture 4: Conditional Probability

10. Exercise 2-85 (Page 46) • A batch of 350 steel bars contains 8 that are defective, 2 are selected, at random, without replacement from the batch • What is the probability that … : • 1) both are defective? • P(D1 ∩ D2) = P(D1|D2) P(D2) = P(D1) P(D2) = 8/350 ⋅ 7/349 = 0.000458 • 2) the second one selected is defective given that the first one was defective? • P(D2|D1) = P(D2 ∩ D1) / P(D1) = 0.000458/(8/350) = 0.020057 = (7/349) • 3) both are acceptable? • P(D1’ ∩ D2’) = P(D1’|D2’) P(D2’) = P(D1’) P(D2’)= 342/350 ⋅ 341/349 = 0.954744 Lecture 4: Conditional Probability

11. Exercise 2-85 (Cont.) • Reminder: De Morgan’s rule P(A’)=1-P(A) • The probability that both are acceptable can also be found as follows: • P(D2 ∩ D1)=P[(D2∪ D1)’]=1-P(D2∪ D1) • =1-[P(D1)+P(D2)-P(P(D2 ∩ D1)] = • 1-[8/350+8/350-0.000458]=0.954744 • We are looking at the event (D1 or D2), so P(D2) depends on the outcome of the first selection, which can be either defective or acceptable • Therefore, we can use the TOTAL PROBABILITY RULE, to obtain: • P(D2)=P(D2|D1)P(D1)+P(D2|D’1)P(D’1) • =7/349.8/350+8/349.342/350=8/350 Lecture 4: Conditional Probability

12. Multiplication Rule • When the probability of the intersection is needed: • P(A ∩ B) = P(B|A) P(A) = P(A|B) P(B) • Example: a concrete batch passes compressive tests with • P(A) = 0.90; a second concrete batch is known to pass the • tests if the first already does, with P(B|A) = 0.95 • What is the probability P(A ∩ B) that both pass the tests? • Ans: P(A ∩ B) = P(B|A) P(A) = 0.95⋅0.90 = 0.855 • Note: it is also true that P(A ∩ B) = P(A|B) P(B), but the • Information provided in the question does not match this second formulation Lecture 4: Conditional Probability

13. Total Probability Rule • Sometimes, the probability of an event can be recovered by summing up a series of conditional probabilities. • Everyday life example: • If a student is undergrad there is 60% chance he/she passes ECI-114, and if he/she is a grad student there is 70% chance he/she passes Eci-114. • P(A): Student is undergrad (55%) • P(A’): Student is grad (45%) • P(B): he/she passes ECI-114 • P(B)= P(B ∩ A) + P(B ∩ A’)= P(B|A) P(A) + P(B|A’) P(A’) • = 60/100.55/100+70/100.45/100 • = 33/100+31.5/100=64.5/100 Lecture 4: Conditional Probability

14. Total Probability Rule For two events we have: P(B) = P(B ∩ A) + P(B ∩ A’) = P(B|A) P(A) + P(B|A’) P(A’) Lecture 4: Conditional Probability

15. Total Probability Rule for Multiple Events For several mutually exclusive and exhaustive events we have: P(B) = P(B ∩ E1) + P(B ∩ E2) + ...+ P(B ∩ Ek) = P(B|E1)P(E1) + P(B|E2) P(E2) + … + P(B|Ek) P(Ek) • Exhaustive Events: • E1 ∪ E2 ∪ … ∪ Ek = S • Mutually Exclusive Event: • can NOT happen at the same time • P(Ei ∩ Ej )=0 (i ≠ j) Lecture 4: Conditional Probability

16. Example 1 (Page 48) • A member fails when subjected to various stress levels, with the following probability Sum is 1 • Let F: Failure and H: member has high stress level • What is the probability of failure of the member? • Ans. P(F) = P(F|H) P(H) + P(F|H’) P(H’) • = 0.10 ⋅ 0.20 + 0.005 ⋅ 0.80 = 0.024 Lecture 4: Conditional Probability

17. Example 2 (Page 48) • A water treatment unit fails when subjected to various contamination levels, with the following probability • Let H, M, L = member has high/medium/low contamination level Ans. P(F) = P(F|H) P(H) + P(F|M) P(M) + P(F|L) P(L) = 0.10 ⋅ 0.20 + 0.01 ⋅ 0.30 + 0.001 ⋅ 0.50 = 0.0235 Lecture 4: Conditional Probability

18. Example 2 (Cont.) • Tree diagram for the same example can be used too: Lecture 4: Conditional Probability

19. Exercise 2-96 (Page 50) • Building failures are due to either natural actions N (87%) or M man-made causes (13%) • Natural actions include: earthquakes E (56%), wind W • (27%), and snow S (17%) • Man-made causes include: construction errors C (73%) or • design errors D (27%) • What is the probability of failure due to construction errors? • Ans. P(F) = P(C|M) P(M) = 0.73 ⋅ 0.13 = 0.0949 • 2) What is the probability of failure due to wind or snow? • Ans. P(F) = [P(W|N) + P(S|N)] P(N) • = [0.27 + 0.17] ⋅ 0.87 = 0.3828 Lecture 4: Conditional Probability

20. Monday before class • Problems HW set 2 • 2-51 • 2-62 • 2-68 • 2-69 • 2-75 • 2-88 • 2-95 • Reading • Bayes’ Theorem (Pages 55-59) Lecture 4: Conditional Probability