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Chapter 6 Introduction to Probability 1 Introduction • In this chapter we discuss the likelihood of occurrence for events with uncertain outcomes. • We define a scale to measure and describe the chance that different outcomes of an uncertain event will take place. • This ‘measure of uncertainty’ called “Probability” serves as the basis for the analysis and results discussed at the rest of this course. 2 6.1 Assigning probabilities to Events • Random experiment – a random experiment is a process or course of action, whose outcome is uncertain. • Examples Experiment Outcomes • Flip a coin • The marks of a statistics test • The time to assemble a computer Heads and Tails Numbers between 0 and 100 Non-negative numbers 3 6.1 Assigning probabilities to Events • Because of the random nature of the experiment, repeated experiments may result in different outcomes. Consequently, one cannot refer to the experiment outcome in certain terms, only in terms of the probability of each outcome to occur. • To determine the probabilities we need to define and list the possible outcomes first. Such a list should be – exhaustive (list of all the possible outcomes). – the outcomes are mutually exclusive (outcome do not overlap). • A list of outcomes that meet the two conditions above, is called a sample space. 4 Sample Space: S = {O1, O2,…,Ok} O1 O2 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 Event An event is any collection of one or more simple events The individual outcomes are called simple events. Simple events cannot be further decomposed into constituent outcomes. 5 Sample Space: S = {O1, O2,…,Ok} Our objective is to determine P(A), the probability that event A will occur. 6 Sample Space: S = {O1, O2,…,Ok} Example 1: Build the sample space for the two random experiments described below – Case 1: A ball is randomly selected from an urn containing blue and red balls. An outcome is considered the ball color. The sample space: {B, R} – Case 2: Two balls are randomly selected from an urn that contains blue and red balls. An outcome is considered the colors of the two balls drawn. The sample space: {BB, BR, RB, RR} 7 Sample Space: S = {O1, O2,…,Ok} • Example 2: Build the sample space for the random experiment defined by the random selection of two numbers from the set 1, 2, 3, 4, 5 (a number cannot be selected twice). An outcome is defined by the sum of the two numbers. 1+3 1+5, 2+4 3+5 The sample space: {3, 4, 5, 6, 7, 8, 9} 1+2 1+4, 2+3; 2+5, 3+4 4+5 8 6.1 Assigning probabilities to Events • Given a sample space S={O1,O2,…,Ok}, the following characteristics for the probability P(Oi) of the experimental outcome Oi must hold: 1. 0 P(Oi ) 1 for each i 2. n P(Oi ) 1 i 1 • The 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. 9 Approaches to Assigning Probabilities and interpretation of Probability • Approaches – The classical approach – The relative frequency approach – The subjective approach 10 Events and their Probabilities • An event is a collection of sample points. – Example 3: A dice is rolled once. The following events are defined. Specify the sample points that belong to each event. Event A = the number facing up is 6 Event B = The number facing up is odd Event C = The number facing up is not greater than 4 Solution: A = {6}; B = {1, 3, 5}; C = {1, 2, 3) – Example 4: A dice is rolled once and the outcome is 3. Which event takes place? Solution: Event B and event C take place because the outcome ‘3’ belongs to both events. 11 Probability of an Event Example 5: A dice is rolled once. Each number is equally likely to face up. Find the probabilities of the following events: Event A = the number facing up is 6 Event B = The number facing up is odd Event C = The number facing up is not greater than 4 Solution: P(A) = P{6} = 1/6; P(B) = P(1 or 3 or 5) = 3/6 P(C) = P(1 or 2 or 3) = 3/6 12 Probability of an Event Example 6: A dice is rolled twice: Define the following events: D = The numbers facing up are the same and even E = The sum of the two numbers facing up is 5. Define the sample space and calculate the probabilities of events D and E. Solution The sample space: S = {(1,1); (1,2);….(6,5); (6,6)} P(D) = P{(2,2) or (4,4) or (6,6)} = 3/36 P(E) = P{(1,4) or (2,3) or (3,2) or (4,1)} = 4/36. Note: there are 36 points in the sample space. 13 6.2 Joint, Marginal, and Conditional Probability • Determining the probability of an event involves the counting process of its simple events. Often this may become quite cumbersome. • Understanding relationships among events may reduce the computational effort involved in determining the probability of combined events. • Combined events and there probabilities are discussed next 14 Intersection and Joint Probability • The intersection of event A and B is the event that occurs when both A and B occur. A C B 15 Intersection and Joint Probability • The intersection of events A and B is denoted by “A and B” (A B ). • The probability of the intersection of A and B is called also the joint probability of A and B = P(A and B). 16 Intersection and Joint Probability • Example 9 – 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. 17 Probabilities of Joint Events • Example 9 – continued – The joint probability of P(A1 and B1) [Mutual fund outperforms…] and [Top 20 MBA …] = .11 – The joint probability of [Mutual fund outperform…] and […not top 20 MBA …] = .06 Mutual fund outperforms Mutual fund doesn’t the market outperform the market (B2) (B1) Top 20 MBA program (A1) .11 .29 Not top 20 MBA program (A2) .06 .54 18 Probabilities of Joint Events • Example 9 – continued – The joint probability of P(A1 and B1) [Mutual fund outperform…] and […Top 20 MBA …] = .11 – The joint probability of P(A2 and B1) [Mutual fund outperform…] and […not from a top 20 …] = .06 Mutual fund outperforms Mutual fund doesn’t the market outperform the market (B2) (B1) Top 20 MBA program (A1) .11 .29 Not top 20 MBA program (A2) .06 .54 19 Marginal Probabilities To better understand the concept of marginal probabilities, observe first the following demonstration Let us separate event C into Event intersect two sub event: “ACand C”, with and “Bboth and event C” A and B A C B 20 Marginal Probabilities The intersection events “A and C”, and “B and C” are indicated as two triangles, and for clarity are separated for a moment (click). A A and C B and C B 21 Marginal Probabilities As the two intersection events are brought back to there original location, it becomes clear (we hope) that the probability of event C can be calculated as the sum of the two joint probabilities. A C B P(C) = P(A and C) + P(B and C)) 22 Marginal Probabilities • Applying this concept to the table of joint probabilities, we notice that marginal probabilities are determining by adding joint probabilities across rows and columns. Click to continue. • These probabilities are computed by adding across the rows and down the columns and appear in the margins of the table. Watch. 23 Marginal Probabilities Mutual fund Mutual fund Marginal outperforms the doesn’t outperform Prob. market (B1) the market (B2) P(Ai) Top 20 MBA program (A1) P(A1 and B1) + P(A1 and B2) = P(A1) Not top 20 MBA program (A2) P(A2 and B1) + P(A2 and B2) = P(A2) Marginal Probability P(Bj) 24 Marginal Probabilities Mutual fund Mutual fund Marginal outperforms the doesn’t outperform Prob. market (B1) the market (B2) P(Ai) Top 20 MBA program (A1) .11 + .29 = .40 Not top 20 MBA program (A2) .06 + .54 = .60 Marginal Probability P(Bj) 25 Marginal Probabilities Mutual fund Mutual fund Marginal outperforms the doesn’t outperform Prob. market (B1) the market (B2) P(Ai) Top 20 MBA program (A1) P(A1 and B1) + Not top 20 MBA program (A2) P(A2 and B1 = Marginal Probability P(Bj) P(B1) P(A1 and B2) + P(A2 and B2 = P(B2) .40 .60 26 Marginal Probabilities Mutual fund Mutual fund Marginal outperforms the doesn’t outperform Prob. market (B1) the market (B2) P(Ai) Top 20 MBA program (A1) .11 .29 .40 Not top 20 MBA program (A2) .06 .17 .54 .83 .60 Marginal Probability P(Bj) + + 27 Conditional Probability • Frequently, information about the occurrence of event A changes the probability of event B. Here is a short demonstration of one possible such situation 28 Conditional Probability The sample space is S. S The probability event B takes place is roughly: This is event B A B {The area of B} {The area of S}. The probability event B takes place given ‘A’ took place is roughly: {The area of B} {The area of A}. Now, assume event ‘A’ takes place first followed by event B. Note that event B is contained in event A. Explanation: Since it is known ‘A’ took place, we can reduce the sample space from ‘S’ to ‘A’. Obviously P(B) is not equal to P(B given A) 29 Conditional Probability • The probability of an event given the information about the occurrence of another event is called conditional probability. • Specifically, the conditional probability of event B given that event A has occurred is calculated as follows: P(A and B) P(B|A) = P(A) 30 Conditional Probability • Example 13 – 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. Mutual fund outperforms the market (B1) • Solution Mutual fund Marginal doesn’t Prob. outperform the P(Ai) market (B2) Top 20 MBA program (A1) .11 .29 .40 Not top 20 MBA program (A2) .06 .54 .60 Marginal Probability P(Bj) .17 .83 31 Conditional Probability • Example 13 - 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(A1 and B2) = .29 P(B2) = .83 This information reduces the relevant sample space to the 83% of event B2. Mutual fund outperforms the market (B1) Mutual fund Marginal doesn’t Prob. outperform the P(Ai) market (B2) Top 20 MBA program (A1) .11 .29 .29 .40 Not top 20 MBA program (A2) .06 .54 .60 Marginal Probability P(Bj) .17 .83 .83 32 Conditional Probability • Example 13 (Solution – continued) P(A1|B2) = P(A1 and B2) = .29 = .3949 P(B2) .83 33 Dependent Events • Before the new information becomes available the probability for the occurrence of A1 is P(A1) = 0.40 • After the new information becomes available P(A1) changes to P(A1 given B2) = .3949 • Since the occurrence of B2 has changed the probability of A1, the two events are related and are called “dependent events”. 34 Independent Events • Two events A and B are said to be independent if or P(A|B) = P(A) P(B|A) = P(B) • That is: The probability of one event is not affected by the occurrence of the other event. 35 Dependent and independent events • Example 13 – continued – We have already seen the dependency between A1 and B2. – Let us check A2 and B2. • P(B2) = .83 • P(B2|A2)=P(B2 and A2)/P(A2) = .54/.60 = .90 – Conclusion: A2 and B2 are dependent. 36 Mutually Exclusive Events • Two events are said to be mutually exclusive if the occurrence of one precludes the occurrence of the other one. • If A and B are mutually exclusive, by definition, the probability of their intersection is equal to zero. • Example: When rolling a dice once the event “The number facing up is 6” and the event “The number facing up is odd” are mutually exclusive. 37 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” (A U B). A C B 38 Union • Example 9 – 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. 39 Union • Solution Mutual fund outperforms the market (B1) Mutual fund A1 or B1 occurs doesn’t outperform the whenever either: A1 and B1 occurs, market (B2) Top 20 MBA program (A1) .11 .29 A1 and B2 occurs, Not top 20 MBA program (A2) .06 .54 A2 and B1 occurs. P(A1 or B1) = P(A1 and B1) + P(A1 and B2) + P(A2 and B1) = .11 +.29 + .06 = .46 40 6.3 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. – The complement rule – The multiplication rule – The addition rule 41 Complement of an Event The complement of event A (denoted by AC) is the event that occurs when event A does not occur. A AC 42 Computing Probability using the Complement • The probability of event A can be calculated using the probability of its complement event by the complement rule P(A) = 1 - P(AC) Why is this true? Because A and AC consist of all the simple events in the sample space. Therefore, P(A) + P(AC) = 1 43 Multiplication Rule • For any two events A and B P(A and B) = P(A)P(B|A) = P(B)P(A|B) • When A and B are independent P(B|A) = P(B), so P(A and B) = P(A)P(B) 44 Multiplication Rule • Example 14 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 45 Multiplication Rule • Example 15 What is the probability that a female student will be selected at random in each of two classes of seven males and three female students? • Solution – Define the events: A – the student selected in one class is a female B – the student selected in the other class is a female – P(A and B) = P(A)P(B) = (3/10)(3/10) = 9/100 = .09 46 Addition Rule For any two events A and B P(A or B) = P(A) + P(B) - P(A and B) Skip P(A) =6/13 + A P(B) =5/13 _ P(A and B) =3/13 P(A or B) = 8/13 B 47 Addition Rule for Mutually Exclusive Events When A and B are mutually exclusive, P(A or B) = P(A) + P(B) – P(A and B) A B P(A and B) = 0 48 Addition Rule • Example 11 – 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? 49 Addition Rule • Solution – Define the following events: • A = the household subscribes to the Sun • B = the household subscribes to the Post – Calculate the probability P(A or B) = P(A) + P(B) – P(A and B) = .22+.35 -.06 = .51 50 Addition Rule • Example 12 (repeat of example 9) Calculating P(A or B) using the addition rule Determine the probability that a randomly selected fund outperforms the market or the manager graduated from a top 20 MBA Program. • Solution P(A1 or B1) = P(A1)+P(B1)-P(A1 and B1)= .4 + .17 - .11 = .46 Mutual fund outperfor ms the market (B1) Mutual fund Margi doesn’t nal outperform Prob. the market P(Ai) (B2) Top 20 MBA program (A1) .11 .29 .40 Not top 20 MBA program (A2) .06 .54 .60 Marginal Probability P(Bj) .17 .83 51 Addition, Multiplication, and the complementary rules • • Example A local photocopy shop has three black-and white (BW) copy machines and two color copiers (CC). It is known that a BW is down 10% of the time for repairs and a CC is down 20%. Assume machines break down independently. – What proportion of the time a customer cannot run a color photocopy job? Answer: The probability both CCs are down is (.2)(.2) = .04 52 Addition, Multiplication, and the complementary rules • Example - Continued – If a customer wants both a color copy and a black and white copy now, what is the probability he can complete the two jobs? Answer: Each job can be completed if at least one machine of each type is operational. • • • P(At least one CC is working) = 1 – P(Both CCs are down) = 1 – (.2)2 = .96. P(At least one BW is working) = 1 – (.1)3 = .999 P(At least one CC and at least one BW is working) = P(At least one CC is working)*P(At least one BW is working) = (.96)(.999). 53 Addition, Multiplication, and the complementary rules • Example - Continued – If a customer wants both a color copy and a black and white copy now, but this time a CC can perform also a BW job, what is the probability he can complete the two jobs? Answer: P(The two jobs can be completed) = P(The two CCs are working or one CC and at least one BW are working) = P(The two CCs are working)+ P(One CC is working)*P(At least one BW is working) = 2(.8)(.2) + [1 – (.1)3] 54 Probability Trees • This is an effective graphical tool that help apply probability rules. • The events are represented by the tree branches. • The given probabilities are indicated along these branches too. 55 Probability Trees • Example 14 revisited (dependent events). – Find the probability of selecting two female students (without replacement), if there are 3 female students in a class of 10. Joint probabilities FF P(FF)=(3/10)(2/9) Second selection First selection FM P(FM)=(3/10)(7/9) MF P(MF)=(7/10)(3/9) Second selection MM P(MM)=(7/10)(6/9) 56 Probability Trees • Example 15 – revisited (independent events) – Find the probability of selecting one female student in each of two classes if there are 3 female students in a class of 10. FF P(FF)=(3/10)(3/10) Second selection FM P(FM)=(3/10)(7/10) MF P(MF)=(7/10)(3/10) First selection Second selection MM P(MM)=(7/10)(7/10) 57 Probability Trees • Example 16 (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). 58 Probability Trees • Solution P(Pass1) = .72 .72 + .2464 First exam Second exam P(Fail1 and Pass2)=. .28(.88)=.2464 P(Fail1 and Fail2) = (.28)(.12) = .0336 P(Pass) = P(Pass on first exam) + P(Fail on first and Pass on second) = 59 .9664 Baye’s Law • We use Baye’s law to find the conditional probability of a possible cause for an event that is known to have occurred. 60 Baye’s Law If event B took place, either event A or AC could have occurred before. Baye’s law calculates the conditional probability P(A|B) and P(AC|B). 61 Baye’s Law P( A and B) P( A | B) P(B) 62 Baye’s Law • Example 17 – 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 men is suffering from the illness, if the test result were positive? 63 Baye’s Law • Solution – Define the following events • • • • D = Has the disease DC = Does not have the disease PT = Positive test results NT = Negative test results Called before: A Called before: AC Called before: B – Build a probability tree 64 Baye’s Law • Solution – Continued – The probabilities provided are: • P(D) = .04 • P(NT|DC) P(DC) = .96 = .98 • P(PT|D) = .94 P(PT|DC) = .02 P(NT|D)= .06 4% of the population has the disease. The test result is “Negative” 98% of the time given that a patient is not ill. The test result is “Positive” 94% of the time given that a patient is ill. • The probability to be determined is P(D | PT ) The patient is ill given that the test comes “Positive”. 65 Baye’s Law P(D and PT) =.0376 + P(D | PT ) P(DC and PT) =.0192 P(D and PT) P(PT) .0376 .6620 .0568 P(PT) =.0568 P(PT) = P(D and PT) + P(DC and PT) 66 Baye’s Law Prior probabilities Likelihood probabilities Posterior probabilities P(D | PT ) .0376 .6620 .0568 The prior probability of event ‘D’ was 4%, but because event ‘PT’ occurred the probability the cause had been ‘D’ increased to 66.2%. 67