1. 1 Chapter 4 – Probability An Introduction

2. 2 Chapter Outline – Part 1  Experiments, Counting Rules, and Assigning Probabilities  Events and Their Probability  Some Basic Relationships of Probability  Conditional Probability

3. 3 Understand the Uncertainties A decision-making process often involves analyses of uncertainties, such as:  What are the likelihood that the Federal Reserve will change its current monetary policy?  What are the chances that the oil price will go up by 5% in the summer?  What are the odds that a company’s investment in R&D will be profitable?

4. 4 Probability Likelihood, chances, and odds are all referring to probability. In statistics, is a numerical measure of the likelihood that an event will occur.  Probability is a numerical measure of the likelihood that an event will occur.  The value range of probability is always between 0 and 1;  A probability near zero indicates that an event is very much unlikely to occur.  A probability near one indicates that an event is almost certain to occur. certain to occur.

5. 5 Probability as A Numerical Measure of the Likelihood of Occurrence 0 1.5 The event is very unlikely to occur. The event is very unlikely to occur. The occurrence of the event is just as likely as just as likely as it is unlikely. The occurrence of the event is just as likely as just as likely as it is unlikely. The event is almost certain to occur. The event is almost certain to occur. Probability: Increasing Likelihood of Occurrence

6. 6 Statistical Experiments  Statistical experiments investigate the odds of a particular outcome of a study.  Experiments are conducted repeatedly and in the same manner, but the outcomes may vary.  Therefore, we call the outcomes of statistical experiments random.  Each outcome has its own probability to occur.

7. 7 An Experiment and Its Sample Space  An experiment is any process that generates well- defined outcomes.  The sample space for an experiment is the set of all experimental outcomes.  An experimental outcome is also called a sample point.

8. 8 An Experiment and Its Sample Space Experiment Toss a coin Inspect a part Conduct a sales call Roll a die Play a football game Experiment Outcomes Head, tail Defective, non-defective Purchase, no purchase 1, 2, 3, 4, 5, 6 Win, lose, tie

9. 9 An Experiment and Its Sample Space  Example: Travel Routes  Joe is to travel from city A, stop in city B, and then arrive at his destination city C. There are two ways to travel from city A to city B, and three ways from city B to city C. Question: how many different routes can Joe take to travel from city A to city C? City A City B City C 1 2 1 2 3

10. 10 An Experiment and Its Sample Space  Example: Travel Routes  We can use a Tree Diagram to list all the possible routes (outcomes) of Joe’s travel. A to B Step 1 B to C Step 2 Routes AB1 AB2 AB3 BC1 BC2 BC3 So, the total number of routes is 6.

11. 11 A Counting Rule for Multiple-Step Experiments If an experiment consists of a sequence of k steps in which there are n 1 possible results for the first step, n 2 possible results for the second step, and so on, then the total number of experimental outcomes is given by n 1  n 2...  n k. If an experiment consists of a sequence of k steps in which there are n 1 possible results for the first step, n 2 possible results for the second step, and so on, then the total number of experimental outcomes is given by n 1  n 2...  n k. Apply this rule to our example, the total # of routes Joe can take is calculated as follows: Apply this rule to our example, the total # of routes Joe can take is calculated as follows: n 1  n 2 = 2  3 = 6 n 1  n 2 = 2  3 = 6

12. 12 Q4.1 – Student ID A student ID code consists of 5 digits. The first digit (read from the left) is a letter, and the other digits can be any number from 0 to 9. How many possible student ID codes are there? What if it is required that no repetitions are allowed among the four numbers?

13. 13 Q4.1 – Student ID The first digit can be any letter from A to Z, and any of the remaining 4 digits can be any number from 0 to 9. So, applying the multiple-step rule, the total number of student ID codes is 26  10  10  10  10=260,000. If no repetitions are allowed for the 4 numbers, the total number of student ID codes is 26  10  9  8  7=131,040.

14. 14 Counting Rule for Combinations  This rule calculates the number of combinations of picking n from N objects at a time when the order of selection does not matter. where: N ! = N ( N  1)( N  2)... (2)(1) n ! = n ( n  1)( n  2)... (2)(1) n ! = n ( n  1)( n  2)... (2)(1) 0! = 1 0! = 1 ! is called factorial. ! is called factorial.

15. 15 Counting Rule for Combinations  Example: Pick 3 out of 8 students to form a team for a group project. Question: how many different teams can be formed?

16. 16 Counting Rule for Permutations  This rule calculates the number of permutations of picking n from N objects at a time when the order of selection does matter. where: N ! = N ( N  1)( N  2)... (2)(1) n ! = n ( n  1)( n  2)... (2)(1) n ! = n ( n  1)( n  2)... (2)(1) 0! = 1 0! = 1 ! is called factorial. ! is called factorial.

17. 17 Counting Rule for Permutations  Example: Pick 3 out of 8 students to form a committee which is comprised of three different positions – President, Secretary, and Treasurer. Question: how many different committees can be formed?

18. 18 Assigning Probabilities  Basic requirements for assigning probabilities 1. The probability assigned to each experimental outcome must be between 0 and 1, inclusively. 1. The probability assigned to each experimental outcome must be between 0 and 1, inclusively. 0 < P ( E i ) < 1 for all i where: E i is the i th experimental outcome and P ( E i ) is its probability

19. 19 Assigning Probabilities  Basic requirements for assigning probabilities 2. The sum of the probabilities for all experimental 2. The sum of the probabilities for all experimental outcomes must equal 1. outcomes must equal 1. where: n is the number of experimental outcomes P ( E 1 ) + P ( E 2 ) +... + P ( E n ) = 1

20. 20 Assigning Probabilities Classical Method Relative Frequency Method Subjective Method Assigning probabilities based on the assumption Assigning probabilities based on the assumption of equally likely outcomes of equally likely outcomes Assigning probabilities based on experimentation Assigning probabilities based on experimentation or historical data or historical data Assigning probabilities based on judgment Assigning probabilities based on judgment

21. 21 Classical Method  Experiment: Rolling a die Sample Space: S = {1, 2, 3, 4, 5, 6} Sample Space: S = {1, 2, 3, 4, 5, 6} Probabilities: Each sample point has a Probabilities: Each sample point has a 1/6 chance of occurring 1/6 chance of occurring If an experiment has n possible outcomes, the If an experiment has n possible outcomes, the classical method would assign a probability of 1/n to each outcome.

22. 22 Relative Frequency Method  Example: Monthly Sales Volume of 50 Starbucks Stores in NYC  The probability that a Starbuck store’s sales volume falls into a certain range is given by dividing the frequency by the sample points. As learned from Chapter 2, the assigned probability is simply the relative frequency. 6/506/50

23. 23 Subjective Method  In the business world, as economic conditions and a firm’s circumstances change, the patterns revealed from historical data might not apply onto forecasting the future outcomes.  Experience, intuition, and judgment should be utilized to evaluate the likelihood that a particular outcome will occur.  For instance, the performance of stock market is difficult, if not impossible, to predict based solely on historical data. The subjective evaluations from analysts are often combined with the classical or relative frequency methods to form better probability estimates.

24. 24 Events and Their Probabilities  An event is a collection of sample points.  The probability of any event is equal to the sum of the probabilities of the sample points in the event.  If we can identify all the sample points of an experiment and assign a probability to each, we can compute the probability of an event.

25. 25 Events and Their Probabilities Example: Monthly Sales Volume of 50 Starbucks Stores Event M = Monthly sales no more than $100,000 M = {(61,70),(71,80),(81,90),(91,100)} M = {(61,70),(71,80),(81,90),(91,100)} P(M) = P (61,70)+ P (71,80)+ P (81,90)+ P (91,100) P(M) = P (61,70)+ P (71,80)+ P (81,90)+ P (91,100) = 0.04+0.12+0.22+0.34 = 0.04+0.12+0.22+0.34 = 0.72 = 0.72 Event M = Monthly sales no more than $100,000 M = {(61,70),(71,80),(81,90),(91,100)} M = {(61,70),(71,80),(81,90),(91,100)} P(M) = P (61,70)+ P (71,80)+ P (81,90)+ P (91,100) P(M) = P (61,70)+ P (71,80)+ P (81,90)+ P (91,100) = 0.04+0.12+0.22+0.34 = 0.04+0.12+0.22+0.34 = 0.72 = 0.72

26. 26 Events and Their Probabilities Example: Roll a die Event K = The number of dots on the side facing facing up is more than 3. facing up is more than 3. K = {4,5,6} K = {4,5,6} P(K) = P (4)+ P (5)+ P (6) P(K) = P (4)+ P (5)+ P (6) = 1/6+1/6+1/6 = 1/6+1/6+1/6 = 3/6 = 3/6 = 0.5 = 0.5 Event K = The number of dots on the side facing facing up is more than 3. facing up is more than 3. K = {4,5,6} K = {4,5,6} P(K) = P (4)+ P (5)+ P (6) P(K) = P (4)+ P (5)+ P (6) = 1/6+1/6+1/6 = 1/6+1/6+1/6 = 3/6 = 3/6 = 0.5 = 0.5

27. 27 Q4.2 – Black and White Balls In a bag, there are one white ball and two black balls. Randomly pick two balls at a time out of the bag, what is the probability that the two balls are one white and one black?

28. 28 Q4.2 – Black and White Balls First, let’s list all the sample points of two out of three balls (W, B 1, B 2 ): –W & B 1 –W & B 2 –B 1 & B 2 So, the probability of one white and one black is 2/3.

29. 29 Basic Relationships of Probability There are some basic probability relationships that can be used to compute the probability of an event without knowledge of all the sample point probabilities. There are some basic probability relationships that can be used to compute the probability of an event without knowledge of all the sample point probabilities.  Complement of An Event  Union of Two Events  Intersection of Two Events (joint events)  Mutually Exclusive Events

30. 30 Venn Diagram  A Venn diagram is to graphically illustrate the probability relationships.  The rectangular represents the whole sample space (  ) and a circle inside the rectangular represents an event. Event A  Sample Space

31. 31 Complement of An Event  The complement of event A is defined to be the event consisting of all sample points that are not in A.  The complement of A is denoted by A c.  P(A c ) = P(  P(A c ) = P(  ) – P(A) = 1 – P(A) A AcAcAcAc

32. 32 Complement of An Event  The seemingly very simple Complement rule proves to be very useful.  When it’s more difficult to figure out P(A) than P(A C ), we can use the Complement rule to calculate P(A). P(A) = 1 – P(A C ) A AcAcAcAc

33. 33 Complement of An Event  Example:  Example: Monthly Sales Volume of 50 Starbucks Stores  When it’s more difficult to figure out P(A) than P(A C ), we can use the Complement rule to calculate P(A). P(A) = 1 – P(A C ) Event M = Monthly sales more than $70,000 M c = Monthly sales less than or equal to $70,000 M c = Monthly sales less than or equal to $70,000 = {(61,70)} = {(61,70)} P(M) = 1 – P ( M c ) P(M) = 1 – P ( M c ) = 1 – 0.04 = 1 – 0.04 = 0.96 = 0.96 Event M = Monthly sales more than $70,000 M c = Monthly sales less than or equal to $70,000 M c = Monthly sales less than or equal to $70,000 = {(61,70)} = {(61,70)} P(M) = 1 – P ( M c ) P(M) = 1 – P ( M c ) = 1 – 0.04 = 1 – 0.04 = 0.96 = 0.96

34. 34 Union of Two Events  The union of events A and B is the event containing all sample points that are in A or B or both.  The union of events A and B is denoted by A  B ( ‘  ’ is read as ‘or’)  AA A B 

35. 35 Intersection of Two Events  The intersection of events A and B is the set of all sample points that are in both A and B.  The intersection of events A and B is denoted by A  ( ‘  ’ is read as ‘and’)  ( ‘  ’ is read as ‘and’)  AA A B  Intersection of A and B

36. 36 Law of Addition  The addition law provides a way to compute the probability of event A, or B, or both A and B occurring. P ( A  B ) = P ( A ) + P ( B )  P ( A  B  Note: Because both P ( A ) and P ( B ) include the common part [ P ( A  B , subtracting P ( A  B  from the sum of P ( A ) and P ( B ) is to avoid double-counting.

37. 37 Law of Addition  Example: From a standard deck of 52 playing cards, what is the probability that a card drawn from the deck is a King or a Heart? Event A = The card is a King; Event B = The card is a Heart Event (A  he card is a King or a Heart Event (A  he card is both a King and a Heart, i.e. the King of Heart P(A  P(A) + P(B) – P(A   –    Note: The King of Heart is included in both event A and event B.

38. 38 Mutually Exclusive Events  Two events are said to be mutually exclusive if the events have no sample points in common, i.e. when one event occurs, the other cannot occur. A B

39. 39 Mutually Exclusive Events  If events A and B are mutually exclusive, P ( A  B  = 0.  The addition law for mutually exclusive events is: P ( A  B ) = P ( A ) + P ( B ) There is no need to include “  P ( A  B  ”

40. 40 Conditional Probability The probability of an event given that another event The probability of an event given that another event has occurred is called a conditional probability. has occurred is called a conditional probability. The probability of an event given that another event The probability of an event given that another event has occurred is called a conditional probability. has occurred is called a conditional probability. A conditional probability is computed as follows : A conditional probability is computed as follows : The conditional probability of A given B is denoted The conditional probability of A given B is denoted by P ( A | B ). by P ( A | B ). The conditional probability of A given B is denoted The conditional probability of A given B is denoted by P ( A | B ). by P ( A | B ).

41. 41 Conditional Probability  Example – Investment in Two Stock Funds  The following table presents the joint and marginal probabilities of investing in two stock funds – Technology and Utility. Joint Probabilities (appear in the body of the table) Marginal Probabilities (appear in the margins of the table)

42. 42 Conditional Probability  Example – Investment in Two Stock Funds Event U = Gain in the Utility Fund Event T = Gain in the Technology Fund P(U|T) = Gain in the Utility Fund given the gain in the Technology Fund. Since P(U  and P(T) = 0.52,  P (U   P (  0.24 0.52 P (U|T) = = = 0.46

43. 43 Law of Multiplication The multiplication law provides a way to compute the probability of the intersection of two events. The multiplication law provides a way to compute the probability of the intersection of two events. The law is written as: The law is written as: P ( A  B ) = P ( B ) P ( A | B ) P ( A  B ) = P ( B ) P ( A | B ) The law is written as: The law is written as: P ( A  B ) = P ( B ) P ( A | B ) P ( A  B ) = P ( B ) P ( A | B )

44. 44 Law of Multiplication Event U = Gain in the Utility Fund Event T = Gain in the Technology Fund P(U|T) = Gain in the Utility Fund given the gain in the Technology Fund. Since P(U  and P(T) = 0.52,  P (U|   P (  P (U  T) = = 0.52  0.46 = 0.24 Example: Investment in Two Stock Funds

45. 45 Independent Events If the probability that event A happens is not affected by whether or not event B happens, we say that events A and B are independent. Two events A and B are independent if: Two events A and B are independent if: P ( A | B ) = P ( A ) P ( B | A ) = P ( B ) or

46. 46 Multiplication Law for Independent Events The following equation can be used to test if two events are independent. P ( A  B ) = P ( A ) P ( B )

47. 47 Multiplication Law for Independent Events  Example: Investment in Two Stock Funds Event U = Gain in the Utility Fund Event T = Gain in the Technology Fund Are events U and T independent? Let’s check if P(U  P(U) P(T) ? Because P(U  P(U) =0.6, P(T) = 0.52, P(U)P(T) = 0.6  0.52 = 0.312  0.24 Therefore: U and T are not independent.

48. 48 Mutually Exclusive Events vs. Independent Events  Two events with nonzero probabilities cannot be both mutually exclusive and independent.  For two mutually exclusive events with nonzero probabilities, when one occurs, the other cannot happen. Thus, the two events are dependent on each other.  Two events that are not mutually exclusive are also independent only if at least one of the two events has a zero probability to occur.

49. 49 Q4.3 If A and B are independent events with P(A) = 0.4 and P(B) = 0.6, then P(A  B) is: P(A  B) = P(A) + P(B) - P(A   P(A) + P(B) - P(A)P(  for A & B are independent  