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Lesson 3:

Lesson 3:. A Survey of Probability Concepts. Outline. Learning exercise 1: University Demographics. Current enrollments by college and by sex appear in the following table. If we select a student at random, what is the probability that the student is :

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Lesson 3:

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  1. Lesson 3: A Survey of Probability Concepts

  2. Outline

  3. Learning exercise 1: University Demographics • Current enrollments by college and by sex appear in the following table. • If we select a student at random, what is the probability that the student is : • A female or male, i.e., P(Female or Male). • Not from Agricultural and Forestry, i.e., P(not-Ag-For) • A female given that the student is known to be from BusEcon, i.e., P(Female |BusEcon). • A female and from BusEcon, i.e., P(Female and BusEcon). • From BusEcon, i.e., P(BusEcon).

  4. Learning exercise 1: University Demographics P(Female or Male) =(4500 + 5500)/10000 = 1 P(not-Ag-For) =(10000 – 1400) /10000 = 0.86 P(Female | BusEcon) = 400 /900 = 0.44 P(Female and BusEcon) = 400 /10000 = 0.04 P(BusEcon) = 900 /10000 = 0.09 P(Female and BusEcon) = P(BusEcon) P(Female | BusEcon)

  5. Learning exercise 2: Predicting Sex of Babies • Many couples take advantage of ultrasound exams to determine the sex of their baby before it is born. Some couples prefer not to know beforehand. In any case, ultrasound examination is not always accurate. About 1 in 5 predictions are wrong. • In one medical group, the proportion of girls correctly identified is 9 out of 10, i.e., applying the test to 100 baby girls, 90 of the tests will indicate girls. and • the number of boys correctly identified is 3 out of 4. i.e., applying the test to 100 baby boys, 75 of the tests will indicate boys. • The proportion of girls born is 48 out of 100. • What is the probability that a baby predicted to be a girl actually turns out to be a girl? Formally, find P(girl | test says girl).

  6. Learning exercise 2: Predicting Sex of Babies • P(girl | test says girl) • In one medical group, the proportion of girls correctly identified is 9 out of 10 and • the number of boys correctly identified is 3 out of 4. • The proportion of girls born is 48 out of 100. • Think about the next 1000 births handled by this medical group. • 480 = 1000*0.48 are girls • 520 = 1000*0.52 are boys • Of the girls, 432 (=480*0.9) tests indicate that they are girls. • Of the boys, 130 (=520*0.25) tests indicate that they are girls. • In total, 562 (=432+130) tests indicate girls. Out of these 562 babies, 432 are girls. • Thus P(girl | test says girl ) = 432/562 = 0.769

  7. Learning exercise 2: Predicting Sex of Babies With the information given, we can fill in the following table in sequence from (1) to (9), with the initial assumption of 1000 babies in total. For the question at hand, i.e., P(girl | test says girl ), we only need to fill in the cells from (1) to (6). P(girl | test says girl ) = 432/562 = 0.769

  8. Learning exercise 2: Predicting Sex of Babies • 480 = 1000*0.48 are girls • 520 = 1000*0.52 are boys • Of the girls, 432 (=480*0.9) tests indicate that they are girls. • Of the boys, 130 (=520*0.25) tests indicate that they are girls. • In total, 562 tests indicate girls. • Out of these 562 babies, 432 are girls. • Thus P(girls | test syas girls ) = 432/562 = 0.769 1000*P(girls) 1000*P(boys) 1000*P(girls)*P(test says girls|girls) 1000*P(boys)*P(test says girls | boys) 1000*[P(girls)*P(test says girls|girls) + P(boys)*P(test says girls|boys)] 1000*P(girls)*P(test says girls|girls) 1000*[P(girls)*P(test says girls|girls) + P(boys)*P(test says girls|boys)]

  9. Example 1(to be used to illustrate the definitions) • A fair die is rolled once. • Peter is concerned with whether the resulted number is even, i.e., 2, 4, 6. • Paul is concerned with whether the resulted number is less than or equal to 3, i.e., 1, 2, 3. • Mary is concerned with whether the resulted number is 6. • Sonia is concerned with whether the resulted number is odd, i.e., 1, 3, 5. • A fair die is rolled twice. • John is concerned with whether the resulted number of first roll is even, i.e., 2, 4, 6. • Sarah is concerned with whether the resulted number of second roll is even, i.e., 2, 4, 6.

  10. Definitions: Experiment and outcome • A randomexperimentis the observation of some activity or the act of taking some measurement. • The experiment is rolling the one die in the first example, and rolling one die twice in the second example. • An outcome is the particular result of an experiment. • The possible outcomes are the numbers 1, 2, 3, 4, 5, and 6 in the first example. • The possible outcomes are number pairs (1,1), (1,2), …, (6,6), in the second example. • Sample Space– the collection of all possible outcomes of a random experiment.

  11. Definition: Event • An event is the collection of one or more outcomes of an experiment. • For Peter: the occurrence of an even number, i.e., 2, 4, 6. • For Paul: the occurrence of a number less than or equal to 3, i.e., 1, 2, 3. • For Mary: the occurrence of a number 6. • For Sonia: the occurrence of an odd number, i.e., 1, 3, 5. • For John: the occurrence of (2,1), (2,2), (2,3),…, (2,6), (4,1),…,(4,6), (6,1),…,(6,6) [John does not care about the result of the second roll].

  12. Definition: Intersection of Events • Intersection of Events – If A and B are two events in a sample space S, then the intersection, A ∩ B, is the set of all outcomes in S that belong to both A and B S A AB B • The intersection of Peter’s event and Paul’s event contains 2. • The intersection of Peter’s event and Mary’s event contains 6. • The intersection of Paul’s event and Mary’s event contains nothing. • The intersection of Peter’s event and Sonia’s event contains nothing.

  13. Definition: Mutually Exclusive events • A and B are Mutually Exclusive Events if they have no basic outcomes in common • i.e., the set A ∩ B is empty S A B

  14. Example 2: Intersections and Mutually Exclusive events • Peter’s event and Paul’s event are not mutually exclusive – both contains 2. • The intersection of Peter’s event and Paul’s event contains 2. • Peter’s event and Mary’s event are not mutually exclusive – both contains 6. • The intersection of Peter’s event and Mary’s event contains 6. • Paul’s event and Mary’s event are mutually exclusive – no common numbers. • The intersection of Paul’s event and Mary’s event contains nothing. • Peter’s event and Sonia’s event are mutually exclusive – no common numbers. • The intersection of Peter’s event and Sonia’s event contains nothing.

  15. Definition: Union • Union of Events – If A and B are two events in a sample space S, then the union, A U B, is the set of all outcomes in S that belong to either A or B S The entire shaded area represents A U B A B

  16. Definition: Exhaustive events • Events are collectively exhaustive if at least one of the events must occur when an experiment is conducted. • Peter’s event (even numbers) and Sonia’s event (odd numbers) are collectively exhaustive. • Peter’s event (even numbers) and Mary’s event (number 6) are not collectively exhaustive. • Events E1, E2, … Ek are Collectively Exhaustive events if E1 U E2 U . . . U Ek = S • i.e., the events completely cover the sample space

  17. Complement • The Complement of an event A is the set of all basic outcomes in the sample space that do not belong to A. The complement is denoted S A

  18. Example 3 Let the Sample Space be the collection of all possible outcomes of rolling one die: S = [1, 2, 3, 4, 5, 6] Let A be the event “Number rolled is even” Let B be the event “Number rolled is at least 4” Then the two events contains [2, 4, 6] [4, 5, 6] A = B =

  19. Example 3 S = [1, 2, 3, 4, 5, 6] A = [2, 4, 6] B = [4, 5, 6] Complements: Intersections: Unions:

  20. Example 3 • Mutually exclusive: • Are A and B mutually exclusive? • No. The outcomes 4 and 6 are common to both • Collectively exhaustive: • Are A and B collectively exhaustive? • No. A U B does not contain 1 and 3 S = [1, 2, 3, 4, 5, 6] A = [2, 4, 6] B = [4, 5, 6]

  21. Probability • Probability – the chance that an uncertain event will occur (always between 0 and 1) 1 Certain .5 0 ≤ P(A) ≤ 1 For any event A 0 Impossible

  22. Assessing Probability • There are three approaches to assessing the probability of an uncertain event: 1. classical probability (Assumes all outcomes in the sample space are equally likely to occur.)

  23. Counting the Possible Outcomes • Use the Combinations formula to determine the number of combinations of n things taken k at a time • where • n! = n(n-1)(n-2)…(1) • 0! = 1 by definition

  24. Example 4: Combination • Suppose there are 2 stocks in our portfolio. We would like to select 1 stocks and sell them. What are all the possible combinations. • [1], [2] • C(2,1) = 2!/(1!(2-1)!)=2

  25. Example 4: Combination • Suppose there are 3 stocks in our portfolio. We would like to select 1 stocks and sell them. What are all the possible combinations. • [1], [2], [3] • C(3,1) = 3!/(1!(3-1)!)=3*2*1/[1*(2*1)]=3 • Suppose there are 3 stocks in our portfolio. We would like to select 2 stocks and sell them. What are all the possible combinations. • [1,2], [1,3], [2,3] • C(3,2) = 3!/(2!(3-2)!)=3*2*1/[(2*1)*1]=3

  26. Example 5: Combination • Suppose there are 4 stocks in our portfolio. We would like to select 1 stocks and sell them. What are all the possible combinations. • [1], [2], [3], [4] • C(4,1) = 4!/(1!(4-1)!)=4*3*2*1/[1*(3*2*1)]=4 • Suppose there are 4 stocks in our portfolio. We would like to select 2 stocks and sell them. What are all the possible combinations. • [1,2], [1,3], [1,4], [2,3], [2,4], [3,4] • C(4,2) = 4!/(2!(4-2)!)=4*3*2*1/[(2*1)*(2*1)]=6

  27. Assessing Probability Three approaches (continued) 2. relative frequency probability • the limit of the proportion of times that an event A occurs in a large number of trials, n 3. subjective probability an individual opinion or belief about the probability of occurrence

  28. Probability Postulates 1. If A is any event in the sample space S, then 2. Let A be an event in S, and let Oi denote the basic outcomes (mutually exclusive). Then (the notation means that the summation is over all the basic outcomes in A) 3. P(S) = 1

  29. Probability Rules • The Complement rule: • The Addition rule: • The probability of the union of two events is

  30. A Probability Table Probabilities and joint probabilities for two events A and B are summarized in this table:

  31. Example 5: Addition Rule Consider a standard deck of 52 cards, with four suits: ♥ ♣ ♦ ♠ Let event A = card is an Ace Let event B = card is from a red suit

  32. Example 5: Addition Rule P(Red UAce) = P(Red) + P(Ace) - P(Red∩Ace) = 26/52 + 4/52 - 2/52 = 28/52 Don’t count the two red aces twice! Color Type Total Red Black 2 2 4 Ace 24 24 48 Non-Ace 26 26 52 Total

  33. Conditional Probability • A conditional probability is the probability of one event, given that another event has occurred: The conditional probability of A given that B has occurred The conditional probability of B given that A has occurred

  34. Example 6: Conditional Probability • Of the cars on a used car lot, 70% have air conditioning (AC) and 40% have a CD player (CD). 20% of the cars have both. • What is the probability that a car has a CD player, given that it has AC ? i.e., we want to find P(CD | AC)

  35. Example 6: Conditional Probability • Of the cars on a used car lot, 70% have air conditioning (AC) and 40% have a CD player (CD). 20% of the cars have both. CD No CD Total .2 .5 .7 AC .2 .1 No AC .3 .4 .6 1.0 Total

  36. Example 6: Conditional Probability • Given AC, we only consider the top row (70% of the cars). Of these, 20% have a CD player. 20% of 70% is 28.57%. CD No CD Total .2 .5 .7 AC .2 .1 No AC .3 .4 .6 1.0 Total

  37. Multiplication Rule • Multiplication rule for two events A and B: • also • Examples: • P(test says girl and girl) = P(girls) * P(test says girls | girls) • P(test says boy and boy) = P(boys) * P(test says boys | boys)

  38. Example 7: Multiplication Rule P(Red ∩Ace) = P(Red| Ace)P(Ace) Color Type Total Red Black 2 2 4 Ace 24 24 48 Non-Ace 26 26 52 Total

  39. Example 8: IndependenceShould I go to a party without my girlfriend? • The probability of the my going to party is 0.7 (i.e., I go to 70 out of 100 parties on average). • If I tend to go to whichever party my girlfriend (Venus) goes, my party behavior depends on Venus’s. That is, my probability of going to a party conditional on Venus’s presence is larger than 0.7 (extreme case being 1.0). • If I tends to avoid going to whichever party Venus goes, my party behavior also depends on Venus. That is, my probability of going to a party conditional on Venus’s presence is less than 0.7 (extreme case being 0.0). • If in making the party decision, I never consider whether Venus is going to a party, my party behavior does not depends on Venus’s. That means, the probability of going to a party conditional on Venus’s presence is 0.7.

  40. Example 8: IndependenceShould I go to a party without my girlfriend? • Define events: • A: a young man goes to a party • B: his girlfriend goes to the same party. • Assume P(A) =0.7 • His party behavior does not depend on his girlfriend’s only if P(A|B) =P(A) = 0.7. And, event A is said to be independent of event B. • P(the young man and his girlfriends shows up in a party) = P(A & B) = P(B)*P(A|B). • If he always goes to whichever party his girlfriend goes, P(A|B) = 1. Hence, P(A & B) = P(B)*P(A|B) = P(B). • If he always avoid to whichever party his girlfriend goes, P(A|B) = 0. Hence, P(A & B) = P(B)*P(A|B) = 0. • If in making the party decision, he never considers whether his girlfriend is going to a party, P(A|B) = 0.7. Hence, P(A & B) = P(B)*P(A|B) = P(B)*P(A) = P(B)*0.7.

  41. Statistical Independence • Two events are statistically independent if and only if: • Events A and B are independent when the probability of one event is not affected by the other event • If A and B are independent, then if P(B)>0 if P(A)>0

  42. Example 9: Statistical Independence • Of the cars on a used car lot, 70% have air conditioning (AC) and 40% have a CD player (CD). 20% of the cars have both. • Are the events AC and CD statistically independent? CD No CD Total .2 .5 .7 AC .2 .1 No AC .3 .4 .6 1.0 Total

  43. Example 9: Statistical Independence CD No CD Total .2 .5 .7 AC .2 .1 No AC .3 .4 .6 1.0 Total P(AC ∩ CD) = 0.2 P(AC) = 0.7 P(CD) = 0.4 P(AC)P(CD) = (0.7)(0.4) = 0.28 P(AC ∩ CD) = 0.2 ≠ P(AC)P(CD) = 0.28 So the two events are not statistically independent

  44. Bivariate Probabilities Outcomes for bivariate events:

  45. Joint and Marginal Probabilities • The probability of a joint event, A ∩ B: • Computing a marginal probability: • Where B1, B2, …, Bk are k mutually exclusive and collectively exhaustive events

  46. Example 10: Marginal Probability P(Ace) Color Type Total Red Black 2 2 4 Ace 24 24 48 Non-Ace 26 26 52 Total

  47. Tree Diagrams • A tree diagram is useful for portraying conditional and joint probabilities. It is particularly useful for analyzing business decisions involving several stages.

  48. EXAMPLE 11: Tree Diagram In a bag containing 7 red chips and 5 blue chips you select 2 chips one after the other without replacement. Construct a tree diagram showing this information. 6/11 R2 R1 7/12 B2 5/11 R2 7/11 B1 5/12 4/11 B2

  49. EXAMPLE 11: Tree Diagram The tree diagram is very illustrative about the relation between joint probability and conditional probability Let A(B) be the event of a red chip in the first(second) draw. P(B|A) = 6/11 6/11 6/11 R2 P(A) = 7/12 R1 7/12 7/12 P(A and B) = P(A)*P(B|A) = 6/11 * 7/12 B2 5/11 R2 7/11 B1 5/12 4/11 B2

  50. Example 12: Using a Tree Diagram Of the cars on a used car lot, 70% have air conditioning (AC) and 40% have a CD player (CD). 20% of the cars have both. Has CD (.2/.7) P(AC ∩ CD) = .2 P(AC)= .7 Has AC All Cars Does not have CD (.5/.7) P(AC ∩ CD) = .5 Does not have AC P(AC ∩ CD) = .2 Has CD (.2/.3) P(AC)= .3 Does not have CD (.1/.3) P(AC ∩ CD) = .1

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