Motivation. Stat Camp for the MBA Program. Probability. Experiments and Outcomes. Daniel Solow 5/10/2017

Transcription

1 Stat Camp for the MBA Program Daniel Solow Lecture 2 Probability Motivation You often need to make decisions under uncertainty, that is, facing an unknown future. Examples: How many computers should I produce this month? What premium should I charge a class of customers for a particular type of insurance policy? The answers to such questions requires knowledge of probability that is, the study of likelihood of certain events occurring Probability Probability is a number that measures the likelihood that an event will happen. Useful as an indicator of the uncertainty associated with an event. Scale from 0 to 1. Probability = 0: the event will certainly not happen. Probability = 1: the event will certainly happen. Probability = 0.5: the event is equally likely to happen or not happen. 68 Experiments and Outcomes Experiment: A situation in which an action could be repeated many times, each resulting in one of many possible outcomes or sample points. Exactly one of these outcomes will occur, but it is not known which. For example: Experiment Outcomes Toss a coin Head, Tail Roll a die 1, 2, 3, 4, 5, 6 Sales Call Sale, No sale Dow Jones tomorrow All positive numbers 69 1

2 Assigning Probabilities Assume an experiment has n possible outcomes E 1, E 2,, E n. The probability assigned to each outcome must be a number between 0 and 1. 0 P(E i ) 1 The sum of the probabilities of all outcomes must equal 1. P(E 1 ) + + P(E n ) = 1 Assigning Probabilities Depending on the situation, you can obtain the probabilities of the outcomes of an experiment from: Assumption that all outcomes are equally likely (classical method) Experience from past data (relative frequency method) Experience, intuition or personal judgment (subjective method) Classical Method In many situations it is reasonable to assume that all n outcomes of an experiment are equally likely to occur. Then each outcome has probability equal to 1/n (why)? Examples: Toss a coin: P(H) = P(T) = 1/2. Roll a die: P(1) = P(2) = = P(6) = 1/6. Relative Frequency Method In some experiments, there are past data available, from which you can estimate the proportion of time each outcome has occurred if the experiment is repeated a large number of times. This proportion is used as an estimate of the probability of the outcome

3 Example When asking about a person s attitude on a new law, the outcome could be: disagree (D), neutral (N), agree (A), or uninformed (U). One way to assign probabilities to these four outcomes is to use the results of a survey, such as the following: Attitude Number of Responses Relative Frequency Disagree (D) 52 52/127=0.41 Neutral (N) 12 12/127=0.09 Agree (A) 37 37/127=0.29 Uninformed (U) 26 26/127=0.21 Total P(D) = 0.41, P(N) = 0.09, P(A) = 0.29, P(U) = Subjective Method Appropriate when Equally-likely assumption is not appropriate. No data are available. One can use other available information, experience, intuition, judgment. In this case, the probability of each outcome expresses the person s subjective degree of belief about the likelihood of the outcome. 75 Example Events What is the probability that I will get that job offer? My personal belief of the chance is 50%. P(offer) = 0.5, P(no offer) = 0.5. The career office believes that the chance is 70%. P(offer) = 0.7, P(no offer) = 0.3 The outcomes are the simplest elements associated with an experiment. However, you are often interested in finding the probabilities of more complicated events related with this experiment. In probability language, events are collections of outcomes

4 Example 1: Rolling a Die Experiment: Roll a die Outcomes: 1, 2,, 6 Events: A = the outcome is less than 3 = {1, 2} B = the outcome is even = {2, 4, 6} C = the outcome is not less than 4 = {4, 5, 6} Example 2: Sporting Event Experiment: One of the following eight cities will be chosen to host a sports contest: New York (N), London (L), Paris (P), Tokyo (T), Beijing (B), Sydney (S), Madrid (M), Chicago (C) Events: A = contest is in Asia = {T, B} B = contest is in an English-speaking city = {N, L, S, C} C = contest is not in Europe = {N, T, B, S, C} Example 3: Playing Cards Example 3 (cont) A deck of cards consists of 52 cards, arranged in four suits: spades (S), clubs(c), diamonds (D), and hearts (H). Spades and clubs are black, diamonds and hearts are red. Each suit has 13 cards arranged in order: 1(ace), 2, 3,, 10, J(ack),Q(ueen), K(ing) face cards A card is denoted by the card number and the suit: 1C = ace of clubs, JH = jack of hearts, and so on. 80 Experiment: Draw a card at random 52 Outcomes: 1S,,KS,1C,,KC,1D,,KD,1H,,KH Events: A = draw a king = {KS, KC, KD, KH} B = draw a red two = {2H, 2D} C = draw a club face card = {JC, QC, KC} 81 4

5 Probabilities of Events The probability of an event can be computed by adding up the probabilities of all outcomes included in that event. Example 1: Rolling a Die Pr(1) = Pr(2) = = Pr(6) = 1/6. A = the outcome is less than 3 = {1, 2} P(A) = P(1) + P(2) = 1/6 + 1/6 = 1/3. B = the outcome is even = {2, 4, 6} P(B) = P(2) + P(4) + P(6) = ½. C = the outcome is not less than 4 = {4, 5, 6} P(C) = P(4) + P(5) + P(6) = ½ Example 2: Sporting Event Assume: P(N) = 0.2, P(L) = 0.1, P(P) = 0.1, P(T) = 0.1, P(B) = 0.05, P(S) = 0.15, P(M) = 0.05, P(C) = Then A = contest is in Asia = {T, B} P(A) = P(T)+P(B) = = B = contest in an English-speaking city = {N, L, S, C} P(B) = = C = contest not in Europe = {N, T, B, S, C} P(C) = = Example: Playing Cards P(any outcome) = 1/52. A = draw a king = {KS, KC, KD, KH} P(A) = 4/52 = 1/13 = B = draw a red two = {2H, 2D} P(B) = 2/52 = 1/26 = C = draw a club face = {JC, QC, KC} P(C) = 3/52 =

6 Complement of an Event The complement of an event A is the event that A does not happen and thus contains all outcomes that are not contained in A. The complement of A is written as A c. If A happens, A c does not happen and vice versa. Complement Law: P(A) = 1 P(A c ). Note: If the event A you are interested in has many outcomes but A c does not, then, to compute P(A), it is easier to find P(A c ) and then Examples Example 1: Suppose A is the event that the weekly sales exceed $2,000 and P(A) = Then A c is the event that the weekly sales do not exceed $2,000 and P(A c ) = Example 2: When you pick a card at random, what is the probability that you do not pick an ace? Answer: Let A = not pick and ace so A c = pick an ace = {AC, AD, AH, AS} and so P(A) = 1 P(A c ) = 1 (4/52) = 48/52. P(A) = 1 P(A c ) Intersection of Two Events If A and B are two events, you are often interested in the probability that both A and B occur simultaneously. The event A and B is called the intersection of A and B, written A B, and consists of outcomes that are in both A and B simultaneously. A A B When you see the word AND think of. B Intersection of Two Events Example: What is the probability of drawing a red king? A = draw a king = {KS, KC, KH, KD} B = draw a red card = {1H,2H,3H,4H,5H,6H,7H,8H,9H,10H,JH,QH,KH, 1D,2D,3D,4D,5D,6D,7D,8D,9D,10D,JD,QD, KD}. A B = draw a king and red card = {KD, KH}. P(drawing a red king) = P(drawing a red card and a king) = P(A B) = {KD, KH} = 1 /

7 Mutually Exclusive Events If A B contains no outcomes, then the events A and B are called mutually exclusive. If A and B are mutually exclusive it is not possible that both A and B will happen. If A and B are mutually exclusive, then Example A = draw a king = {KS, KC, KH, KD} B = draw a queen = {QS, QC, QH, QD} A B contains no outcomes. If you draw a single card, it is not possible that the card will be both a king and a queen. P(A B) = Union of Two Events If A and B are two events, you are often interested in probability that either A or B (or both) occur simultaneously. In math terms, this is called the union of A and B and is denoted by A B. Example: Roll a die. A = the outcome is less than 3 = {1, 2}. B = the outcome is even = {2, 4, 6}. A B = the outcome is less than 3 or even (or both) = {1, 2, 4, 6}. 92 The Addition Law The event A B consists of all the outcomes that belong to both A and B. When you see the word OR think of. The probability of A B can be computed by adding the probabilities of the individual outcomes in A B, OR with the following addition law (whichever is easier): P(A B) = P(A) + P(A B) A B: A A B B 93 7

8 Union of Mutually Exclusive Events If A and B are mutually exclusive, you know that P(A B) = 0. Therefore, for mutually exclusive events, P(A B) = P(A) + P(B) P(A B) = P(A) + P(B) 94 Example Roll a die A = the outcome is less than 3 = {1, 2} B = the outcome is even = {2, 4, 6} A B={2} A B = {1, 2, 4, 6} Probabilities P(A) = 1/3 P(B) = 1/2 P(A B) = 1/6 P(A B) = 4/6 = 2/3. From the addition law: P(A B) = P(A) + P(B) P(A B) = 1/3 + 1/2 1/6 = 2/3 which agrees with the direct calculation. 95 Example Answer You draw a single card from a deck. You win if you draw a king or a red card. What is the probability you win? A = draw a king P(A) = 4/52. B = draw a red card P(B) = 26/52. A B = red king P(A B) = 2 /52 A B = king or red card P(A B) = P(A) + P(B) P (A B) = 4/ /52 2/52 = 28/52 =

9 Conditional Probability Sometimes the probability of an event changes when you get information about another related event. Example: You roll a die and don t see the outcome. What is the probability of the outcome being a 2? 1/6 If I tell you that the outcome was odd, what is the probability of a 2 now? 0 If I tell you that the outcome was even, what is the probability of a 2 now? 1/3 This is the conditional probability of an event given that another event happened. 98 Example: Police Force A police force consists of 960 men an 240 women officers. Last year 288 men and 36 women were promoted. Women officers complained of discrimination. The administration said that this was due to the low number of women officers in the force. Do you think the discrim. complaint is justified? Approach: Compare P(promotion given a man) and P(promotion given a woman). Probability Model Experiment: Select an officer at random Outcomes: Any one of 1200 officers, all equally likely. 99 Events and Their Probabilities Events: M = man officer, P(M) = 960/1200 = 0.8 W = woman officer, P(W) = 240/1200 = 0.2 A = promoted officer, P(A) = 324/1200 = 0.27 A c = non-promoted officer, P(A c ) = 0.73 M A = male prom. off., P(M A) = 288/1200 = 0.24 Men Women Totals Promoted Not Totals Conditional Probability Men Women Totals Promoted Not Totals Select an officer at random. P(selected officer is promoted) = P(A) = 324/1200 = 0.27 Suppose the selected off. is a man. P(A M) = 288/960 = 0.3 Suppose the selected off. is a woman. P(A W) = 36/240 =

10 Conclusion We have found that P(A M) = 0.3: 30% of men officers were promoted. P(A W) = 0.15: 15% of women officers were promoted. Is the complaint justified in your opinion? YES! Computing Conditional Prob. The general formula (Baye s Theorem) for the conditional probability of an event A given that event B has occurred is: Police Example: 102 This is consistent with our original calculation. 103 Independent Events Two events A and B are independent if the probability of A does not change with information about B, that is, if P(A B) = P(A) and vice versa P(B A) = P(B). If this relation is not true, then the two events are dependent. 104 Example: Police Force You have found that P(A M) = 0.3 P(A) = 0.27 Therefore, P(A M) P(A), and the events A and M are dependent. This means that the probability of an officer being promoted is influenced by whether this officer is a man. This justifies the discrimination claim

11 Multiplication Law Assume A and B are independent. Then P(A B) = P(A). However, you also know that P(A B) = P(A B) / P(B). Then you find that P(A B) / P(B) = P(A B) = P(A), or P(A B) = P(A) P(B). Independence It is often reasonable to assume that two events are independent, because of their nature. For example, if you roll two dice. Does knowing what happened on the first roll tell you anything about what will happen on the second roll? No! Because there is no relationship between the first and the second roll, you may assume that A and B are independent. Then the multiplication law implies 106 P(A B) = P(A) P(B) = (1/6) (1/6) = 1/ Summary Example Bob and Jon live together and each has a car that works, respectively, 60% and 90% of the time. A potential employer has said she will hire them if they have one working car at least 95% of the time. Will they get the job? State what you are looking for as a prob. question. Is P(that at least one car is working) 0.95? Use probability theory to answer the question. Events: P( A = Bob s car works ) = 0.6 P( B = Jon s car works) = 0.9 P(at least one car is working) = P(A OR B) = P(A B) = P(A) + P(B) P(A B) (Addition Law) = (0.9) (A & B are ind.) = They get the job. 108 Random Variables A random variable (rv) is a quantity of interest: Whose value is uncertain (by uncertain is meant that there are many (at least two) possible values and you do not know which value will occur). You cannot control the value that occurs. Example 1: Y = the outcome of flipping a coin. Random variables are used to help make decisions in a problem involving uncertainty. Example 2: Roll a die once. If the outcome is 1 or 2 you lose $5, if 3 you lose $1, if 4 you win $2, if 5 or 6 you win $4. Do you want to play this game? To decide, you need to identify appropriate rvs

12 Identifying Random Variables and Their Possible Values The first two steps involved in working with a random variable are: Step 1: Identify the random variable. Use a symbol and write the meaning of the variable, including units. Example : Let X = $ earned in the dice game. Step 2: List all possible values the r.v. can have. Example: X = 5, 1, 2, or Types of Random Variables Discrete Random Variable: A r.v. whose possible values you can count (either finite or an infinite set of countable numbers, for example, 0,1,2, ). Example Roll-and-Earn: X = 5, 1, 2, 4 Can you count these possible values? Yes! Continuous Random Variable: A r.v. whose value is any number (including decimals and fractions) in an interval or a collection of intervals (infinite uncountable number of values). Poss. Val. Example: X = liters of water I drink today. [0, 5] Can you count these possible values? No! 111 Examples of Random Variables Number of heads in 50 tosses of a coin. All possible values: 0, 1,, 50 (discrete) Number of customers who enter a store in a day. All possible values: 0, 1, 2, (discrete) Number of cm of rain next month. All possible values: [0, 30]. (continuous) Time in minutes between two customers arriving at a bank. All possible values: [0, ). (continuous) 112 Examples of Random Variables Number of defective products in a shipment of 100. All possible values: 0, 1, 2,, 100. (discrete) Quantity x of liquid inside a 12 oz can All possible values: [0, 12]. (continuous) Percentage x of a project completed by the deadline. All possible values: [0, 100]. (continuous) $ sales in a retail store tomorrow. All possible values: $ $ (discrete)

13 Which of the Following are RVs? The temperature at noon yesterday. (No) Time is The temperature at noon tomorrow. (Yes) critical! The age of a person chosen at random in this class. (It depends on timing.) If you have already selected the person: If you have not yet selected the person: (No: there is only one value for that person s age.) (Yes) All possible values: The (finite) list of ages of everyone in this class. (discrete) 114 Which of the Following are RVs? The average of a population. (No, µ only has one val.) The average of a sample of size 2. (Depends on time.) If you have already (No, there is only one value selected the sample: for that average.) If you have not yet selected the sample: (Yes) All possible values of The (finite) list of averages of every group of size 2 in the population. Groups of size 2: G1 G2 G3 for the group: A1 A2 A3 (discrete, but ) Warning: When a discrete rv has too many possible values, it might not be practical to work with that rv. 115 RVs, Populations, and Sampling For any population, you can create the following two discrete random variables: Y = the value of an item that will be chosen randomly from the population. = the average of a sample of size n before taking the sample. Note: The average of the pop. is not a RV. For any RV, you can create a sample of size n by observing and recording the value of the RV n separate times. 116 When Something is Unknown Question: What can you do when a quantity of interest such as the average of a population is unknown? Alternative 1: Determine the value, however If doing requires too much time, effort, money, then Alternative 2: Estimate the value, for example, by Building a model! For example: Model 1: Take a sample of size n (the model) and use the average from the sample as your best estimate of. Model 2: Think of as a discrete random variable (the model) with possible values: 20, 21,, 30 Model 3: Think of as a continuous random variable (the model) with possible values: [20, 30]

14 Identifying Random Variables INSURANCE PREMIUMS What should the insurance premium be for a particular class of customers? Question: Is the annual premium a r.v.? Answer: No, because you can control its value. Let C = the $/year to be claimed by this type of customer with possible values: 0 100,000 (discrete) 118 Examples of Random Variables WARRANTIES GoodTire has just introduced a new tire in the car market. How many miles of warranty should the company offer? Qn: Is the warranty mileage a r.v.? Ans: No, because you can control its value. M = the number of miles such a tire is expected to last, with possible values: [0, ). (continuous) 119 Examples of Random Variables PERSONNEL PLANNING How many bank tellers should be working during the busiest time of the day? A = the total number of customers that arrive during that period, with possible values: 0, 1, 2, (discrete) W = the number of minutes it takes a teller to serve a customer, with possible values: [0, 30]. (continuous) Probability Distribution To work with a random variable, you must know that variable s probability distribution, which describes the probabilities of all the possible values of the random variable occurring. Note: Probability distributions are different for discrete RVs and continuous RVs

15 Discrete Distributions For a discrete random variable X, the probability distribution is described by a probability density function that consists of: The list of the possible values and, for each one, the probability of that value occurring. Notationally, if t is a possible value for the rv X then the density function is written as follows: f(t) = P(X = t) = the probability that the random variable takes the value t. 122 Example 1 Toss a coin once. X = number of heads. Possible values of X: t 0 1 P(X = t) = f(t) A valid probability density function for a discrete random variable must satisfy the following two properties: 0 f(t) 1, for each value of t. f(t) = Example 2 Toss a coin twice, and let X = number of heads. Values of X 0 1 Probability Density Function f(0) = P(X = 0) = P(no heads) = P(T and T) = P(T) * P(T) = 0.5 (0.5) = 0.25 f(1) = P(X = 1) = P(1 head) = P(HT or TH) = P(HT) + P(TH) = = f(2) = 1 f(0) f(1) = = 0.25 Example 2 (cont d) The probability density function can be shown in a table or graph: f(t) t

16 Example 3: Roll and Earn Find the probability density function of X, the amount earned in the following game: Roll a die once. If the outcome is 1 or 2 you lose $5, if 3 you lose $1, if 4 you win $2, if 5 or 6 you win $4. Possible values for X: Probabilities: 2/6 1/6 1 or 2 3 1/6 4 2/6 5 or Expected Value In addition to the density function, you also want to find the expected value (mean) of the discrete random variable, which is a measure of the central location of the value of the random variable. Computed as the sum of the products of the possible values x and the corresponding probabilities: X = E[X] = t f(t) Note: The mean of a r.v. is different from the mean of a population and the mean of a sample. 127 Why is the Expected Value Useful? Let the rv X = be your random variable. Example Toss a coin twice and let X = number of heads. The probability density function is: Law of Large Numbers: If you observe the value of the random variable X a large number of times, the average of the observed values will be very close to the expected value of the random variable X. The expected value is: E(X) = (0) (0.25) + (1)(0.5) +(2)(0.25) = 1 If the two tosses are repeated many times and the number of heads recorded each time, the average number of heads per two-tosses will be close to

17 Computing Exp.Value in Excel Roll and Earn Example: Car Dealership Now that the new car models are available, a dealership has lowered the prices on last year s models in order to clear its inventory. With prices slashed, a salesperson estimates the following probability distribution of X, the number of cars that person will sell next week. This means that, if you play the game many times and record Find the expected value of X. What does it mean? the results, on average, you will lose $0.16 each time Answer E(X) = (0)(0.05) + (1)(0.15) + (2)(0.35) + (3)(0.25) + (4)(0.2) = 2.40 If prices stay the same over several weeks, the average number of cars sold per week will be close to Variance and Standard Deviation of Random Variables You also want to find the variance of the r.v., which is a measure of how close the possible values are to the expected value,. µ Var. is small µ Var. is larger The variance of a discrete random variable X is: 2 = VAR(X) = (t ) 2 f(t) The standard deviation is the square root of the variance. Note: Variance and standard deviation for a r.v. are different from those for a population and sample

18 Example Toss a coin twice and let X = number of heads. Recall that the expected value is = 1. Example For the roll-and-earn game, find the variance and standard deviation. Recall that E(X) = = Variance: Var(X) = 2 = 0.50 Standard Deviation: = 0.50 = Variance: Var(X) = 2 = Standard deviation: = =