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# Introduction to Probability Experiments Counting Rules Combinations Permutations Assigning Probabilities.

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Introduction to Probability Experiments Counting Rules Combinations Permutations Assigning Probabilities

Experiments ExperimentExperimental Outcomes Toss a coinHead, tail Select a part for inspection Defective, nondefective Conduct a sales callPurchase, no purchase Roll a die1, 2, 3, 4, 5, 6 Play a football gameWin, lose, tie These are processes that generate well- defined outcomes

Sample Space 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 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 Probability: 01.0 0.5 The occurrence of the event is just as likely as it is unlikely

Basic Requirements for Assigning Probabilities Let E i denote the ith experimental outcome (elementary event) and P(E i ) 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 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 Selecting a part for inspection: Rolling a die: P(1) = P(2) = … = P(6) = 1/6 P(Defective) = ? This method of assigning probabilities is indicated if each experimental outcome is equally likely

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: n 1 defective and n 2 nondefective P(Defective) = n 1 /N, P(Nondefective) = n 2 /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? 1.Counting rule for multi-step experiments 2.Counting rule for combinations 3.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 0 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). 4/404/40 Probability Number of Cleaners Rented Number of Days 0 1 2 3 4 4 6 18 10 2 40.10.15.45.25.05 1.00

Subjective Method When economic conditions and a company’s When economic conditions and a company’s circumstances change rapidly it might be circumstances change rapidly it might be inappropriate to assign probabilities based solely on inappropriate to assign probabilities based solely on historical data. historical data. We can use any data available as well as our We can use any data available as well as our experience and intuition, but ultimately a probability experience and intuition, but ultimately a probability value should express our degree of belief that the value should express our degree of belief that the experimental outcome will occur. experimental outcome will occur. The best probability estimates often are obtained by The best probability estimates often are obtained by combining the estimates from the classical or relative combining the estimates from the classical or relative frequency approach with the subjective estimate. 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 n 1 possible outcomes on the first step, n 2 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 Investment Gain or Loss in 3 Months (in \$000) in 3 Months (in \$000) Markley Oil Collins Mining 10 5 0  20 8 222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: n 1 = 4 Collins Mining: n 2 = 2 Total Number of Experimental Outcomes: n 1 n 2 = (4)(2) = 8

Tree Diagram Gain 5 Gain 8 Gain 10 Gain 8 Lose 20 Lose 2 Even Markley Oil (Stage 1) Collins Mining (Stage 2) ExperimentalOutcomes (10, 8) Gain \$18,000 (10, -2) Gain \$8,000 (5, 8) Gain \$13,000 (5, -2) Gain \$3,000 (0, 8) Gain \$8,000 (0, -2) Lose \$2,000 (-20, 8) Lose \$12,000 (-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. 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 Complement of an Event Intersection of Two Events Intersection of Two Events Mutually Exclusive Events Mutually Exclusive Events Union of Two Events

The complement of A is denoted by A c. The complement of A is denoted by A c. The complement of event A is defined to be the event The complement of event A is defined to be the event consisting of all sample points that are not in A. consisting of all sample points that are not in A. The complement of event A is defined to be the event The complement of event A is defined to be the event consisting of all sample points that are not in A. consisting of all sample points that are not in A. Complement of an Event Event A AcAcAcAc Sample Space S Sample Space S VennDiagram

The union of events A and B is denoted by A  B  The union of events A and B is denoted by A  B  The union of events A and B is the event containing The union of events A and B is the event containing all sample points that are in A or B or both. all sample points that are in A or B or both. The union of events A and B is the event containing The union of events A and B is the event containing all sample points that are in A or B or both. all sample points that are in A or B or both. Union of Two Events Sample Space S 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 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) + P (0, 8) + P (  20, 8) =.20 +.08 +.16 +.26 +.10 +.02 =.82

The intersection of events A and B is denoted by A  The intersection of events A and B is denoted by A  The intersection of events A and B is the set of all The intersection of events A and B is the set of all sample points that are in both A and B. sample points that are in both A and B. The intersection of events A and B is the set of all The intersection of events A and B is the set of all sample points that are in both A and B. sample points that are in both A and B. Sample Space S Sample Space S Event A Event B Intersection of Two Events 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 and Collins Mining Profitable M  C = {(10, 8), (5, 8)} P ( M  C) = P (10, 8) + P (5, 8) =.20 +.16 =.36

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

Event M = Markley Oil Profitable Event C = Collins Mining Profitable M  C = Markley Oil Profitable or Collins Mining 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 Addition Law (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 Two events are said to be mutually exclusive if the events have no sample points in common. events have no sample points in common. Two events are said to be mutually exclusive if the Two events are said to be mutually exclusive if the events have no sample points in common. events have no sample points in common. Two events are mutually exclusive if, when one event Two events are mutually exclusive if, when one event occurs, the other cannot occur. occurs, the other cannot occur. Two events are mutually exclusive if, when one event Two events are mutually exclusive if, when one event occurs, the other cannot occur. occurs, the other cannot occur. Sample Space S Sample Space S Event A Event B

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

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 ). Conditional Probability

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

Multiplication Law The multiplication law provides a way to compute the The multiplication law provides a way to compute the probability of the intersection of two events. probability of the intersection of two events. The multiplication law provides a way to compute the The multiplication law provides a way to compute the probability of the intersection of two events. 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 )

Event M = Markley Oil Profitable Event C = Collins Mining Profitable We know: P ( M ) =.70, P ( C | M ) =.5143 Multiplication Law M  C = Markley Oil Profitable and Collins Mining Profitable and Collins Mining Profitable 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 If the probability of event A is not changed by the existence of event B, we would say that events A existence of event B, we would say that events A and B are independent. and B are independent. If the probability of event A is not changed by the If the probability of event A is not changed by the existence of event B, we would say that events A existence of event B, we would say that events A and B are independent. 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

The multiplication law also can be used as a test to see The multiplication law also can be used as a test to see if two events are independent. if two events are independent. The multiplication law also can be used as a test to see The multiplication law also can be used as a test to see if two events are independent. if two events are independent. The law is written as: 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 We know: P ( M  C ) =.36, P ( M ) =.70, P ( C ) =.48 But: P ( M)P(C) = (.70)(.48) =.34, not.36 But: P ( M)P(C) = (.70)(.48) =.34, not.36 Are events M and C independent? Does  P ( M  C ) = P ( M)P(C) ? Hence: M and C are not independent. 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 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 p k and (1 – p) n – k become too small to calculate exactly by the binomial distribution. In such cases, the Poisson distribution becomes useful. Let be the expected number of successes in a process consisting of n trials, i.e., = np. The probability of observing k successes is The mean and variance of a Poisson distributed variable are given by  = and V =, respectively.

Normal Distribution

Gamma Distribution

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