1. 1 Probability Chapter 6

2. 2 6.2Assigning probabilities to Events Random experiment –a random experiment is a process or course of action, whose outcome is uncertain. Examples ExperimentOutcomes Flip a coinHeads and Tails Record a statistics test marksNumbers between 0 and 100 Measure the time to assemblenumbers from zero and above a computer

3. 3 6.2Assigning probabilities to Events Performing the same random experiment repeatedly, may result in different outcomes, therefore, the best we can do is consider the probability of occurrence of a certain outcome. To determine the probabilities we need to define and list the possible outcomes first.

4. 4 Determining the outcomes. –Build an exhaustive list of all possible outcomes. –Make sure the listed outcomes are mutually exclusive. A list of outcomes that meets the two conditions above is called a sample space. Sample Space

5. 5 Sample Space: S = {O 1, O 2,…,O k } Sample Space a sample space of a random experiment is a list of all possible outcomes of the experiment. The outcomes must be mutually exclusive and exhaustive. Simple Events The individual outcomes are called simple events. Simple events cannot be further decomposed into constituent outcomes. Event An event is any collection of one or more simple events Our objective is to determine P( A ), the probability that event A will occur. Our objective is to determine P( A ), the probability that event A will occur. O1O1 O2O2

6. 6 –Given a sample space S={ O 1,O 2,…,O k }, the following characteristics for the probability P( O i ) of the simple event O i must hold: –Probability of an event: The probability P( A ) of event A is the sum of the probabilities assigned to the simple events contained in A. Assigning Probabilities

7. 7 Approaches to Assigning Probabilities and Interpretation of Probability Approaches –The classical approach –The relative frequency approach –The subjective approach Interpretation –If a random experiment is repeated an infinite number of times, the relative frequency for any given outcome is the probability of this outcome.

8. 8 6.3 Joint, Marginal, and Conditional Probability We study methods to determine probabilities of events that result from combining other events in various ways. There are several types of combinations and relationships between events: –Intersection of events –Union of events –Dependent and independent events –Complement event

9. 9 Intersection The intersection of event A and B is the event that occurs when both A and B occur. The intersection of events A and B is denoted by (A and B). The joint probability of A and B is the probability of the intersection of A and B, which is denoted by P(A and B)

10. 10 Example 6.1 –A potential investor examined the relationship between the performance of mutual funds and the school the fund manager earned his/her MBA. –The following table describes the joint probabilities. Intersection Mutual fund outperform the market Mutual fund doesn’t outperform the market Top 20 MBA program.11.29 Not top 20 MBA program.06.54

11. 11 Example 6.1 – continued –The joint probability of [mutual fund outperform…] and […from a top 20 …] =.11 –The joint probability of [mutual fund outperform…] and […not from a top 20 …] =.06 Intersection Mutual fund outperforms the market (B 1 ) Mutual fund doesn’t outperform the market (B 2 ) Top 20 MBA program (A 1 ).11.29 Not top 20 MBA program (A 2 ).06.54 P(A 1 and B 1 )

12. 12 Example 6.1 – continued –The joint probability of [mutual fund outperform…] and […from a top 20 …] =.11 –The joint probability of [mutual fund outperform…] and […not from a top 20 …] =.06 Intersection Mutual fund outperforms the market (B 1 ) Mutual fund doesn’t outperform the market (B 2 ) Top 20 MBA program (A 1 ).11.29 Not top 20 MBA program (A 2 ).06.54 P(A 2 and B 1 ) P(A 1 and B 1 ) Intersection

13. 13 Marginal Probability These probabilities are computed by adding across rows and down columns Mutual fund outperforms the market (B 1 ) Mutual fund doesn’t outperform the market (B 2 ) Marginal Prob. P(A i ) Top 20 MBA program (A 1 ) Not top 20 MBA program (A 2 ) Marginal Probability P(B j ) P( A 1 and B 1 )+ P( A 1 and B 2 ) = P( A 1 ) P( A 2 and B 1 )+ P( A 2 and B 2 ) = P( A 2 )

14. 14 Marginal Probability These probabilities are computed by adding across rows and down columns Mutual fund outperforms the market (B 1 ) Mutual fund doesn’t outperform the market (B 2 ) Marginal Prob. P(A i ) Top 20 MBA program (A 1 ).11.29.40 Not top 20 MBA program (A 2 ).06.54.60 Marginal Probability P(B j ) + = + =

15. 15 Marginal Probability These probabilities are computed by adding across rows and down columns Mutual fund outperforms the market (B 1 ) Mutual fund doesn’t outperform the market (B 2 ) Marginal Prob. P(A i ) Top 20 MBA program (A 1 ).40 Not top 20 MBA program (A 2 ).60 Marginal Probability P(B j ) P(A 1 and B 1 ) + P(A 2 and B 1 = P( B 1 ) P(A 1 and B 2 ) + P(A 2 and B 2 = P( B 2 )

16. 16 Marginal Probability These probabilities are computed by adding across rows and down columns Mutual fund outperforms the market (B 1 ) Mutual fund doesn’t outperform the market (B 2 ) Marginal Prob. P(A i ) Top 20 MBA program (A 1 ).11.29.40 Not top 20 MBA program (A 2 ).06.54.60 Marginal Probability P(B j ).17.83 + +

17. 17 Example 6.2 (Example 6.1 – continued) –Find the conditional probability that a randomly selected fund is managed by a “Top 20 MBA Program graduate”, given that it did not outperform the market. Solution P(A 1 |B 2 ) = P(A 1 and B 2 ) =.29 =.3949 P(B 2 ).83 Conditional Probability

18. 18 Example 6.2 –Find the conditional probability that a randomly selected fund is managed by a “Top 20 MBA Program graduate”, given that it did not outperform the market. Solution P(A 1 |B 2 ) = P(A 1 and B 2 ) P(B 2 ) =.29/.83 =.3949 Conditional Probability Mutual fund outperforms the market (B 1 ) Mutual fund doesn’t outperform the market (B 2 ) Marginal Prob. P(A i ) Top 20 MBA program (A 1 ).11.29.40 Not top 20 MBA program (A 2 ).06.54.60 Marginal Probability P(B j ).17.83.29.83 New information reduces the relevant sample space to the 83% of event B 2.

19. 19 Before the new information becomes available we have P(A 1 ) = 0.40 After the new information becomes available P(A 1 ) changes to P(A 1 given B 2 ) =.3494 Since the the occurrence of B 2 has changed the probability of A 1, the two event are related and are called “ dependent events”. Conditional Probability

20. 20 Independence Independent events –Two events A and B are said to be independent if P(A|B) = P(A) or P(B|A) = P(B) –That is, the probability of one event is not affected by the occurrence of the other event.

21. 21 Dependent and independent events Example 6.3 (Example 6.1 – continued) –We have already seen the dependency between A 1 and B 2. –Let us check A 2 and B 2. P(B 2 ) =.83 P(B 2 |A 2 )=P(B 2 and A 2 )/P(A 2 ) =.54/.60 =.90 –Conclusion: A 2 and B 2 are dependent. Dependent and Independent Events

22. 22 Union The union event of A and B is the event that occurs when either A or B or both occur. It is denoted “A or B”. Example 6.4 (Example 6.1 – continued) Calculating P(A or B)) –Determine the probability that a randomly selected fund outperforms the market or the manager graduated from a top 20 MBA Program.

23. 23 Solution Union Mutual fund outperforms the market (B 1 ) Mutual fund doesn’t outperform the market (B 2 ) Top 20 MBA program (A 1 ).11.29 Not top 20 MBA program (A 2 ).06.54 A 1 or B 1 occurs whenever either: A 1 and B 1 occurs, A 1 and B 2 occurs, A 2 and B 1 occurs. P(A 1 or B 1 ) = P(A 1 and B 1 ) + P(A 1 and B 2 ) + P(A 2 and B 1 ) =.11 +.29 +.06 =.46 Comment: P(A 1 or B 1 ) = 1 – P(A 2 and B 2 ) = 1 –.46 =.54 Union of events

24. 24 6.4 Probability Rules and Trees We present more methods to determine the probability of the intersection and the union of two events. Three rules assist us in determining the probability of complex events from the probability of simpler events.

25. 25 Complement Rule The complement of event A (denoted by A C ) is the event that occurs when event A does not occur. The probability of the complement event is calculated by P(A C ) = 1 - P(A) A and A C consist of all the simple events in the sample space. Therefore, P(A) + P(A C ) = 1 Complement event

26. 26 For any two events A and B When A and B are independent P(A and B)= P(A)P(B|A) = P(B)P(A|B) P(A and B)= P(A)P(B|A) = P(B)P(A|B) P(A and B)= P(A)P(B) Multiplication Rule

27. 27 Example 6.5 What is the probability that two female students will be selected at random to participate in a certain research project, from a class of seven males and three female students? Solution –Define the events: A – the first student selected is a female B – the second student selected is a female –P(A and B) = P(A)P(B|A) = (3/10)(2/9) = 6/90 =.067 Multiplication Rule

28. 28 Example 6.6 What is the probability that a female student will be selected at random from a class of seven males and three female students, in each of the next two class meetings? Solution –Define the events: A – the first student selected is a female B – the second student selected is a female –P(A and B) = P(A)P(B) = (3/10)(3/10) = 9/100 =.09 Multiplication Rule

29. 29 For any two events A and B P(A or B) = P(A) + P(B) - P(A and B) A B P(A) =6/13 P(B) =5/13 P(A and B) =3/13 A or B + _ P(A or B) = 8/13 Addition Rule

30. 30 B When A and B are mutually exclusive, P(A or B) = P(A) + P(B) Addition Rule A B P(A and B) = 0

31. 31 Example 6.7 –The circulation departments of two newspapers in a large city report that 22% of the city’s households subscribe to the Sun, 35% subscribe to the Post, and 6% subscribe to both. –What proportion of the city’s household subscribe to either newspaper? Addition Rule

32. 32 Solution –Define the following events: A = the household subscribe to the Sun B = the household subscribe to the Post –Calculate the probability P(A or B) = P(A) + P(B) – P(A and B) =.22+.35 -.06 =.51 Addition Rule The Addition and Multiplication Rules

33. 33 This is a useful device to calculate probabilities when using the probability rules. Probability Trees

34. 34 Example 6.5 revisited ( dependent events ). –Find the probability of selecting two female students (without replacement), if there are 3 female students in a class of 10. Probability Trees First selection P(F) = 3/10 P( M) = 7/10 Second selection P(F|M) = 3/9 P(F|F) = 2/9 P( M|M) = 6/9 P( M|F) = 7/9 P(FF)=(3/10)(2/9) P(FM)=(3/10)(7/9) P(MF)=(7/10)(3/9) P(MM)=(7/10)(6/9) Joint probabilities Dependent events

35. 35 Example 6.6 – revisited (independent events) –Find the probability of selecting two female students (with replacement), if there are 3 female students in a class of 10. F MFMF M FMFM First selection P(F) = 3/10 P( M) = 7/10 Second selection Second selection P(F|M) = 3/10 P(F|F) = 3/10 P( M|M) =7/10 P( M|F) = 7/10 P(FF)=(3/10)(3/10) P(FM)=(3/10)(7/10) P(MF)=(7/10)(3/10) P(MM)=(7/10)(7/10) = P(F) = = P(M) = Probability Trees Independent events

36. 36 Example 6.8 (conditional probabilities) –The pass rate of first-time takers for the bar exam at a certain jurisdiction is 72%. –Of those who fail, 88% pass their second attempt. –Find the probability that a randomly selected law school graduate becomes a lawyer (candidates cannot take the exam more than twice). Probability Trees

37. 37 Solution Probability Trees P(Pass) =.72 P(Fail and Pass)=. 28(.88)=.2464 P(Fail and Fail) = (.28)(.(12) =.0336 First exam P(Pass) =.72 P( Fail) =.28 Second exam P(Pass|Fail) =.88 P( Fail|Fail) =.12 P(Pass) = P(Pass on first exam) + P(Fail on first and Pass on second) =.9664

38. 38 Bayes’ Law Conditional probability is used to find the probability of an event given that one of its possible causes has occurred. We use Bayes’ law to find the probability of the possible cause given that an event has occurred.

39. 39 Example 6.9 –Medical tests can produce false-positive or false-negative results. –A particular test is found to perform as follows: Correctly diagnose “Positive” 94% of the time. Correctly diagnose “Negative” 98% of the time. –It is known that 4% of men in the general population suffer from the illness. –What is the probability that a man is suffering from the illness, if the test result were positive? Bayes’ Law

40. 40 Solution –Define the following events D = Has a disease D C = Does not have the disease PT = Positive test results NT = Negative test results –Build a probability tree Bayes’ Law

41. 41 Solution – Continued –The probabilities provided are: P(D) =.04P(D C ) =.96 P(PT|D) =.94P(NT|D)=.06 P(PT|D C ) =.02P(NT|D C ) =.98 –The probability to be determined is Bayes’ Law

42. 42 D D P(PT|D C ) =.02 P( NT|D C ) =.98 P(PT|D) =.94 P( NT|D) =.06 PT|D PT PT|D PT D D D D| PT P(D C ) =.96 P(D) =.04 PT|D P(D and PT ) =.0376 P(D C and PT ) =.0192 P( PT ) =.0568 + Bayes’ Law

43. 43 P(PT|D C ) =.02 P( NT|D C ) =.98 P(PT|D) =.94 P( NT|D) =.06 P(D C ) =.96 P(D) =.04 Bayes’ Law Prior probabilities Likelihood probabilities Posterior probabilities