Chapter 6 Probability Assigning probabilities to Events Random

Chapter 6 Probability

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 Heads and Tails • Record a statistics test marks Numbers between 0 and 100 • Measure the time to assemble numbers from zero to computer infinite

Probabilities… • • • List the outcomes of a random experiment… This list must be exhaustive, i. e. ALL possible outcomes included. Die roll {1, 2, 3, 4, 5} Die roll {1, 2, 3, 4, 5, 6} The list must be mutually exclusive, i. e. no two outcomes can occur at the same time: Die roll {odd number or even number} Die roll{ number less than 4 or even number}

Assigning 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.

Sample Space • 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: S = {O 1, O 2, …, Ok} O 1 O 2 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 Event events. An event is any collection Simple events of one or more simple eventscannot be further decomposed Our objective is to into constituent determine P(A), the outcomes. probability that event A will occur.

Assigning Probabilities – Given a sample space S={O 1, O 2, …, Ok}, the following characteristics for the probability P(Oi) of the simple event Oi 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.

Approaches to Assigning Probabilities… • There are three ways to assign a probability, P(Oi), to an outcome, Oi, namely: • Classical approach: make certain assumptions (such as equally likely, independence) about situation. • Relative frequency: assigning probabilities based on experimentation or historical data. • Subjective approach: Assigning probabilities based on the assignor’s judgment.

Example 1(Textbook 6. 3) • A quiz contains multiple-choice questions with five possible answers, only one of which is correct. A student plans to guess the answers because he knows absolutely nothing about the subject. a. Produce the sample space for each question. b. Assign probabilities to the simple events in the sample space you produced. c. Which approach did you use to answer part b d. Interpret the probabilities you assigned in part b

Solution • a {a is correct, b is correct, c is correct, d is correct, e is correct} • b P(a is correct) = P(b is correct) = P(c is correct) = P(d is correct) = P(e is correct) =. 2 • c Classical approach • d In the long run all answers are equally likely to be correct.

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

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)

Intersection • Additional Example – 1 – The number of spots turning up when a six-side die is tossed is observed. Consider the following events. – A: The number observed is at most 2. – B: The number observed is an even number. – Determine the probability of the intersection event A and B. 3 1 A 2 4 A and B B 6 P(A and B) = P(2) = 1/6 5

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(Ai) Top 20 MBA program (A 1) P(A 1 and B 1)+ P(A 1 and B 2) = P(A Not top 20 MBA program (A 2) P(A 2 and B 1)+ P(A 2 and B 2) = P(A Marginal Probability P(Bj)

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

Conditional Probability • 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

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.

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 2 • Suppose we have the following joint probabilities A 1 A 2 A 3 B 1 0. 2 0. 15 0. 1 B 2 0. 25 0. 05 • Compute the marginal probabilities

Solution

Example 3 • • The following tables lists the joint probabilities associated with smoking and lung disease among 60 to 65 year-old-men. He is a smoker He is a nonsmoker He has lung disease 0. 1 0. 03 He does not have lung disease 0. 21 0. 66 One 60 to 65 year old man is selected at random. What is the probability of the following events a. He is a smoker b. He does not have lung disease c. He has lung disease given that he is a smoker d. He has lung disease given that he does not smoke

Solution

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.

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

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(A and B) = P(A)P(B)

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

Example 4 (Textbook 6. 52) • a. b. c. d. Approximately 10% of people are lefthanded. If two people are selected at random, what is the probability of the following events? Both are right-handed Both are left-handed One is right-handed and the other is lefthanded At least one is right-handed

Solution

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

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

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.

Bayes’ Law • Example – Medical tests can produce false-positive or falsenegative 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 • Solution – Define the following events • • D = Has a disease DC = Does not have the disease PT = Positive test results NT = Negative test results – Build a probability tree

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

Bayes’ Law Prior Likelihood probabilities |D T P ( P(D ) 4 0. = P(D C )= . 96 ) P P( N T|D 4 9. = )= C) T|D P(P P( N . 06 2 =. 0 T|D C )= . 98 Posterior probabilities

Bayes’ Theorem • To find the posterior probability that event Ai will occur given that event B has occurred we apply Bayes’ theorem. • Bayes’ theorem is applicable when the events for which we want to compute posterior probabilities are mutually exclusive and their union is the entire sample space.

Bayesian Terminology… • The probabilities P(A) and P(AC) are called prior probabilities because they are determined prior to the decision about taking the preparatory course. • The conditional probability P(A | B) is called a posterior probability (or revised probability), because the prior probability is revised after the decision about taking the preparatory course.