Introduction to Probability Experiments Counting Rules Combinations Permutations Assigning Probabilities

These are processes that generate well-defined outcomes Experiments Experiment Experimental Outcomes Toss a coin Head, tail Select a part for inspection Defective, nondefective Conduct a sales call Purchase, no purchase Roll a die 1, 2, 3, 4, 5, 6 Play a football game Win, lose, tie

Sample Space For a coin toss: Selecting a part for inspection: The sample space for an experiment is the set of all experimental outcomes For a coin toss: Selecting a part for inspection: Rolling a die:

Event Rolling a die: Events: Any subset of the sample space is called an event. Rolling a die: Events: { 1 } // the outcome is 1 (elementary event) { 1, 3, 5 } // the outcome is an odd number { 4, 5, 6 } // the outcome is at least 4.

Probability is a numerical measure of the likelihood of an event occurring 1.0 0.5 Probability: The occurrence of the event is just as likely as it is unlikely

Basic Requirements for Assigning Probabilities Let Ei denote the ith experimental outcome (elementary event) and P(Ei) is its probability of occurring. Then: The sum of the probabilities for all experimental outcomes must be must equal 1. For n experimental outcomes:

Principle of Indifference This method of assigning probabilities is indicated if each experimental outcome is equally likely We assign equal probability to elementary events if we have no reason to expect one over the other. For a coin toss: P(Head) = P(Tail) = 1/2 Rolling a die: P(1) = P(2) = … = P(6) = 1/6 Selecting a part for inspection: P(Defective) = ?

Relative Frequency Method This method is indicated when the data are available to estimate the proportion of the time the experimental outcome will occur if the experiment is repeated a large number of times. What if experimental outcomes are NOT equally likely. Then the Principle of Indifference is out. We must assign probabilities on the basis of experimentation or historical data. Selecting a part for inspection: N parts: n1 defective and n2 nondefective P(Defective) = n1/N, P(Nondefective) = n2/N

Counting Experimental Outcomes To assign probabilities, we must first count experimental outcomes. We have 3 useful counting rules for multiple-step experiments. For example, what is the number of possible outcomes if we roll the die 4 times? Counting rule for multi-step experiments Counting rule for combinations Counting rule for permutations

Example: Lucas Tool Rental Relative Frequency Method Ace Rental would like to assign probabilities to the number of carpet cleaners it rents each day. Office records show the following frequencies of daily rentals for the last 40 days. Number of Cleaners Rented Number of Days 1 2 3 4 4 6 18 10 2

Relative Frequency Method Each probability assignment is given by dividing the frequency (number of days) by the total frequency (total number of days). Number of Cleaners Rented Number of Days Probability 1 2 3 4 4 6 18 10 2 40 .10 .15 .45 .25 .05 1.00 4/40

Subjective Method When economic conditions and a company’s circumstances change rapidly it might be inappropriate to assign probabilities based solely on historical data. We can use any data available as well as our experience and intuition, but ultimately a probability value should express our degree of belief that the experimental outcome will occur. The best probability estimates often are obtained by combining the estimates from the classical or relative frequency approach with the subjective estimate.

Counting Rule for Multi-Step Experiments If an experiment can be described as a sequence of k steps with n1 possible outcomes on the first step, n2 possible outcomes on the second step, then the total number of experimental outcomes is given by:

Example: Bradley Investments Bradley has invested in two stocks, Markley Oil and Collins Mining. Bradley has determined that the possible outcomes of these investments three months from now are as follows. Investment Gain or Loss in 3 Months (in $000) Markley Oil Collins Mining 10 5 -20 8 -2

A Counting Rule for Multiple-Step Experiments Bradley Investments can be viewed as a two-step experiment. It involves two stocks, each with a set of experimental outcomes. Markley Oil: n1 = 4 Collins Mining: n2 = 2 Total Number of Experimental Outcomes: n1n2 = (4)(2) = 8

Tree Diagram Markley Oil Collins Mining Experimental (Stage 1) Outcomes Gain 8 (10, 8) Gain $18,000 (10, -2) Gain $8,000 Lose 2 Gain 10 Gain 8 (5, 8) Gain $13,000 (5, -2) Gain $3,000 Lose 2 Gain 5 Gain 8 (0, 8) Gain $8,000 Even (0, -2) Lose $2,000 Lose 2 Lose 20 Gain 8 (-20, 8) Lose $12,000 Lose 2 (-20, -2) Lose $22,000

Counting Rule for Combinations This rule allows us to count the number of experimental outcomes when we select n objects from a (usually larger) set of N objects. Counting Rule for Combinations The number of N objects taken n at a time is where And by definition

Example: Quality Control An inspector randomly selects 2 of 5 parts for inspection. In a group of 5 parts, how many combinations of 2 parts can be selected? Let the parts de designated A, B, C, D, E. Thus we could select: AB AC AD AE BC BD BE CD CE and DE

Iowa Lottery Iowa randomly selects 6 integers from a group of 47 to determine the weekly winner. What are your odds of winning if you purchased one ticket?

Counting Rule for Permutations Sometimes the order of selection matters. This rule allows us to count the number of experimental outcomes when n objects are to be selected from a set of N objects and the order of selection matters.

Example: Quality Control Again An inspector randomly selects 2 of 5 parts for inspection. In a group of 5 parts, how many permutations of 2 parts can be selected? Again let the parts be designated A, B, C, D, E. Thus we could select: AB BA AC CA AD DA AE EA BC CB BD DB BE EB CD DC CE EC DE and ED

Some 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. Complement of an Event Union of Two Events Intersection of Two Events Mutually Exclusive Events

Complement of an Event Venn Diagram 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 Ac. Sample Space S Event A Ac Venn Diagram

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 Sample Space S Event A Event B

Union of Two Events Event M = Markley Oil Profitable Event C = Collins Mining Profitable M C = Markley Oil Profitable or Collins Mining Profitable M C = {(10, 8), (10, -2), (5, 8), (5, -2), (0, 8), (-20, 8)} P(M C) = P(10, 8) + P(10, -2) + P(5, 8) + P(5, -2) + P(0, 8) + P(-20, 8) = .20 + .08 + .16 + .26 + .10 + .02 = .82

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  Sample Space S Event A Event B Intersection of A and B

Intersection of Two Events Event M = Markley Oil Profitable Event C = Collins Mining Profitable M C = Markley Oil Profitable and Collins Mining Profitable M C = {(10, 8), (5, 8)} P(M C) = P(10, 8) + P(5, 8) = .20 + .16 = .36

Addition Law The addition law provides a way to compute the probability of event A, or B, or both A and B occurring. The law is written as: P(A B) = P(A) + P(B) - P(A  B

Addition Law Event M = Markley Oil Profitable Event C = Collins Mining Profitable M C = Markley Oil Profitable or Collins Mining Profitable We know: P(M) = .70, P(C) = .48, P(M C) = .36 Thus: P(M  C) = P(M) + P(C) - P(M  C) = .70 + .48 - .36 = .82 (This result is the same as that obtained earlier using the definition of the probability of an event.)

Mutually Exclusive Events Two events are said to be mutually exclusive if the events have no sample points in common. Two events are mutually exclusive if, when one event occurs, the other cannot occur. Sample Space S Event A Event B

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’s no need to include “- P(A  B”

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

Conditional Probability Event M = Markley Oil Profitable Event C = Collins Mining Profitable = Collins Mining Profitable given Markley Oil Profitable We know: P(M C) = .36, P(M) = .70 Thus:

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

Multiplication Law Event M = Markley Oil Profitable Event C = Collins Mining Profitable M C = Markley Oil Profitable and Collins Mining Profitable We know: P(M) = .70, P(C|M) = .5143 Thus: P(M  C) = P(M)P(M|C) = (.70)(.5143) = .36 (This result is the same as that obtained earlier using the definition of the probability of an event.)

Independent Events If the probability of event A is not changed by the existence of event B, we would say that events A and B are independent. Two events A and B are independent if: P(A|B) = P(A) P(B|A) = P(B) or

Multiplication Law for Independent Events The multiplication law also can be used as a test to see if two events are independent. The law is written as: P(A B) = P(A)P(B)

Multiplication Law for Independent Events Event M = Markley Oil Profitable Event C = Collins Mining Profitable Are events M and C independent? DoesP(M  C) = P(M)P(C) ? We know: P(M  C) = .36, P(M) = .70, P(C) = .48 But: P(M)P(C) = (.70)(.48) = .34, not .36 Hence: M and C are not independent.

Terminology Events may or may not be mutually exclusive. If E and F are mutually exclusive events, then P(E U F) = P(E) + P(F) If E and F are not mutually exclusive, then P(E U F) = P(E) + P(F) – P(E n F). All elementary events are mutually exclusive.

The birth of a son or a daughter are mutually exclusive events. The event that the outcome of rolling a die is even and the event that the outcome of rolling a die is at least four are not mutually exclusive.

Simple probabilities If A and B are mutually exclusive events, then the probability of either A or B to occur is the union Example: The probability of a hat being red is ¼, the probability of the hat being green is ¼, and the probability of the hat being black is ½. Then, the probability of a hat being red OR black is ¾.

Simple probabilities If A and B are independent events, then the probability that both A and B occur is the intersection

Simple probabilities Example: The probability that a US president is bearded is ~14%, the probability that a US president died in office is ~19%. If the two events are independent, the probability that a president both had a beard and died in office is ~3%. In reality, 2 bearded presidents died in office. (A close enough result.) Harrison, Taylor, Lincoln*, Garfield*, McKinley*, Harding, Roosevelt, Kennedy* (*assassinated)

Conditional probabilities What is the probability of event A to occur given that event B did occur. The conditional probability of A given B is Example: The probability that a US president dies in office if he is bearded 0.03/0.14 = 22%. Thus, out of 6 bearded presidents, 22% are expected to die in office. In reality, 2 died. (Again, a close enough result.)

Probability Distribution The probability distribution refers to the frequency with which all possible outcomes occur. There are numerous types of probability distribution.

The Uniform Distribution A variable is said to be uniformly distributed if the probability of all possible outcomes are equal to one another. Thus, the probability P(i), where i is one of n possible outcomes, is

The Binomial Distribution A process that has only two possible outcomes is called a binomial process. In statistics, the two outcomes are frequently denoted as success and failure. The probabilities of a success or a failure are denoted by p and q, respectively. Note that p + q = 1. The binomial distribution gives the probability of exactly k successes in n trials

The Binomial Distribution The mean and variance of a binomially distributed variable are given by

The Poisson distribution Siméon Denis Poisson 1781-1840 Poisson d’April

The Poisson distribution When the probability of “success” is very small, e.g., the probability of a mutation, then pk and (1 – p)n – k become too small to calculate exactly by the binomial distribution. In such cases, the Poisson distribution becomes useful. Let l be the expected number of successes in a process consisting of n trials, i.e., l = np. The probability of observing k successes is The mean and variance of a Poisson distributed variable are given by m = l and V = l, respectively.

Normal Distribution

Gamma Distribution