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Chapter 6 Probability

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**Suppose two six-sided die is rolled and they both land on sixes. **

Or a coin is flipped and it lands on heads. Or record the color of the next 20 cars to pass an intersection. These would be examples of chance experiments. Chance experiment – any activity or situation in which there is uncertainty about which of two or more plausible outcomes will result.

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**Suppose a six-sided die is rolled**

Suppose a six-sided die is rolled. The possible outcomes are that the die could land with 1 dot up or 2, 3, 4, 5, or 6 dots up. S = {1, 2, 3, 4, 5, 6} This would be an example of a sample space. The sum of the probabilities of the outcomes in the sample space equals ONE. “S” stands for sample space. We use set notation to list the outcomes of the sample space. Sample space - the collection of all possible outcomes of a chance experiment

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**Suppose two coins are flipped. The sample space would be: **

S = {HH, HT, TH, TT} Where H = heads and T = tails We can also use a tree diagram to represent a sample space. H H We follow the branches out to show an outcome. T HT H T T

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**We typically use capital letters to denote an event.**

Suppose a six-sided die is rolled. The outcome that the die would land on an even number would be E = {2, 4, 6} This would be an example of an event. We typically use capital letters to denote an event. Event - any collection of outcomes (subset) from the sample space of a chance experiment

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**What would the event be that is the die NOT landing on an even number? **

Suppose a six-sided die is rolled. The event that the die would land on an even number would be E = {2, 4, 6} What would the event be that is the die NOT landing on an even number? EC = {1, 3, 5} This is an example of complementary events. The sum of the probabilities of complementary events equals ONE. The superscript “C” stands for complement E’ and E also denote the complement of E Complement - Consists of all outcomes that are not in the event

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**These complementary events can be shown on a Venn Diagram. **

E = {2, 4, 6} and EC = {1, 3, 5} Let the circle represent event E. Let the rectangle represent the sample space. Let the shaded area represent event not E.

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**Suppose a six-sided die is rolled**

Suppose a six-sided die is rolled. The event that the die would land on an even number would be E = {2, 4, 6} The event that the die would land on a prime number would be P = {2, 3, 5} What would be the event E or P happening? E or P = {2, 3, 4, 5, 6} This is an example of the union of two events.

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The union of A or B - consists of all outcomes that are in at least one of the two events, that is, in A or in B or in both. Consider a marriage or union of two people – when two people marry, what do they do with their possessions ? This symbol means “union” The bride takes all her stuff & the groom takes all his stuff & they put it together! And live happily ever after! This is similar to the union of A and B. All of A and all of B are put together!

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**Let’s revisit rolling a die and getting an even or a prime number . . .**

E or P = {2, 3, 4, 5, 6} Another way to represent this is with a Venn Diagram. E or P would be any number in either circle. Even number Prime number Why is the number 1 outside the circles? 3 4 2 6 5 1

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**What would be the event E and P happening? E and P = {2}**

Suppose a six-sided die is rolled. The event that the die would land on an even number would be E = {2, 4, 6} The event that the die would land on a prime number would be P = {2, 3, 5} What would be the event E and P happening? E and P = {2} This is an example of the intersection of two events.

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**This symbol means “intersection”**

The intersection of A and B - consists of all outcomes that are in both of the events This symbol means “intersection”

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**E and P would be ONLY the middle part that the circles have in common**

Let’s revisit rolling a die and getting an even or a prime number . . . E and P = {2} To represent this with a Venn Diagram: E and P would be ONLY the middle part that the circles have in common 3 4 2 6 5 1

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**Suppose a six-sided die is rolled. Consider the following 2 events:**

A = {2} B = {6} On a single die roll, is it possible for A and B to happen at the same time? These events are mutually exclusive. Mutually exclusive (or disjoint) events -two events have no outcomes in common; two events that NEVER happen simultaneously

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**The intersection of A and B is empty!**

A Venn Diagram for the roll of a six-sided die and the following two events: A = {2} B = {6} A and B are mutually exclusive (disjoint) since they have no outcomes in common The intersection of A and B is empty! 3 4 6 2 1 5

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**Practice with Venn Diagrams**

On the following four slides you will find Venn Diagrams representing the students at your school. Some students are enrolled in Statistics, some in Calculus, and some in Computer Science. For the next four slides, indicate what relationships the shaded regions represent. (use complement, intersection, and union)

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**Calculus or Computer Science**

Statistics Calculus Computer Science Calculus or Computer Science

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**(Statistics or Computer Science) and not Calculus**

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**Statistics and Computer Science and not Calculus**

Com Sci Statistics and Computer Science and not Calculus

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**Statistics and not (Computer Science or Calculus)**

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**Three different approaches to probability**

What is Probability? Three different approaches to probability

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**The Classical Approach**

When the outcomes in a sample space are equally likely, the probability of an event E, denoted by P(E), is the ratio of the number of outcomes favorable to E to the total number of outcomes in the sample space. Examples: flipping a coin, rolling a die, etc.

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On some football teams, the honor of calling the toss at the beginning of the football game is determined by random selection. Suppose this week a member of the 11-player offensive team will be selected to call the toss. There are five interior linemen on the offensive team. If event L is defined as the event that an interior linemen is selected to call the toss, what is probability of L? P(L) = 5/11

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**The classical approach doesn’t work for every situation.**

Consider an archer shooting arrows at a target. The probability of getting a bulls’ eye should be the ratio of the area of the inner circle to the area of the entire target. What if a very experienced archer were shooting the arrows? Would the probability of a bull’s eye still be the same? The classical approach doesn’t work for every situation.

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**The Relative Frequency Approach**

The probability of event E, denoted by P(E), is defined to be the value approached by the relative frequency of occurrence of E in a very long series of trials of a chance experiment. Thus, if the number of trials is quite large,

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**What do you notice in the graph at the right?**

Consider flipping a coin and recording the relative frequency of heads. When the number of coin flips is small, there is a lot of variability in the relative frequency of “heads” (as shown in this graph). What do you notice in the graph at the right? Discuss what is happening in the graph

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**What do you notice in this graph at the right?**

Consider flipping a coin and recording the relative frequency of heads. The graph at the right shows the relative frequency when the coin is flipped a large number of times. What do you notice in this graph at the right? Discuss what is happening in the graph

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Law of Large Numbers Notice how the relative frequency of heads approaches ½ the larger the number of trials! As the number of repetitions of a chance experiment increase, the chance that the relative frequency of occurrence for an event will differ from the true probability by more than any small number approaches 0. OR in other words, after a large number of trials, the relative frequency approaches the true probability.

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**The Subjective Approach**

Probability can be interpreted as a personal measure of the strength of belief that a particular outcome will occur. The problem with a subjective approach is that different people could assign different probabilities to the same outcome based on their subjective viewpoints. Example: An airline passenger may report that her probability of being placed on standby (denied a seat) due to overbooking is She arrived at this through personal experience and observation of events.

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Probability Rules!

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**Fundamental Properties of Probability**

Property 1. Legitimate Values For any event E, 0 < P(E) < 1 Property 2. Sample space If S is the sample space, P(S) = 1

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**If two events E and F are disjoint, P(E or F) = P(E) + P(F)**

Properties Continued . . . Property 3. Addition If two events E and F are disjoint, P(E or F) = P(E) + P(F) Property 4. Complement For any event E, P(E) + P(not E) = 1

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**Probabilities of Equally Likely Outcomes**

Consider an experiment that can result in any one of N possible outcomes. Denote the simple events by O1, O2, …, ON. If these simple events are equally likely to occur, then 1. 2. For any event E,

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**Suppose you roll a six-sided die once**

Suppose you roll a six-sided die once. Let E be the event that you roll an even number. P(E) = P(2 or 4 or 6) = 3/6 Number of outcomes in E Over N

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**Addition Rule for Disjoint Events**

If events E1, E2, . . ., Ek are disjoint (mutually exclusive) events, then P(E1 or E2 or or Ek) = P(E1) + P(E2) P(Ek) In words, the probability that any of these k disjoint events occurs is the sum of the probabilities of the individual events.

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**A large auto center sells cars made by many different manufacturers**

A large auto center sells cars made by many different manufacturers. Three of these are Honda, Nissan, and Toyota. Consider a chance experiment that consist of observing the make of the next car sold. Suppose that P(H) = 0.25, P(N) = 0.18, P(T) = 0.14. Are these disjoint events? yes P(H or N or T) = = .57 P(not (H or N or T)) = = .43

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**Let D = the event that a person has the disease P(D) = .001**

Sometimes the knowledge that one event has occurred changes our assessment of the likelihood that another event occurs. Consider the following example: Suppose that 0.1% of all the individuals in a population have a certain disease. The presence of the disease is not discernable from appearances, but there is a screening test for the disease. Let D = the event that a person has the disease P(D) = .001

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**This is an example of conditional probability.**

Disease example continued . . . Suppose that 0.1% of all the individuals in a population have a certain disease. 80% of those with positive test results actually have the disease. 20% of those with positive test results actually do NOT have the disease (false positive) Let P = the event that a person tests positive for the disease P(D|P) = 0.80 This is an example of conditional probability. Knowing that event P, the person tested positive, has occurred, changes the probability of event D, the person having the disease, from to 0.80. Read: Probability that a person has the disease “GIVEN” the person tests positive

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**Conditional Probability**

A probability that takes into account a given condition has occurred

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The article “Chances Are You Know Someone with a Tattoo, and He’s Not a Sailor” (Associated Press, June 11, 2006) included results from a survey of adults aged 18 to 50. The accompanying data are consistent with summary values given in the article. At Least One Tattoo No Tattoo Totals Age 18-29 18 32 50 Age 30-50 6 44 24 76 100 Assuming these data are representative of adult Americans and that an adult is selected at random, use the given information to estimate the following probabilities.

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**P(tattoo) = 24/100 = 0.24 Tattoo Example Continued . . .**

At Least One Tattoo No Tattoo Totals Age 18-29 18 32 50 Age 30-50 6 44 24 76 100 What is the probability that a randomly selected adult has a tattoo? P(tattoo) = 24/100 = 0.24

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**P(tattoo|age 18-29) = 18/50 = 0.36 Tattoo Example Continued . . .**

At Least One Tattoo No Tattoo Totals Age 18-29 18 32 50 Age 30-50 6 44 24 76 100 This is a condition! How many adults in the sample are ages 18-29? How many adults in the sample are ages AND have a tattoo? What is the probability that a randomly selected adult has a tattoo if they are between 18 and 29 years old? P(tattoo|age 18-29) = 18/50 = 0.36

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**P(age 18-29|tattoo) = 18/24 = 0.75 Tattoo Example Continued . . .**

At Least One Tattoo No Tattoo Totals Age 18-29 18 32 50 Age 30-50 6 44 24 76 100 How many adults in the sample have a tattoo? How many adults in the sample are ages AND have a tattoo? What is the probability that a randomly selected adult is between 18 and 29 years old if they have a tattoo? This is a condition! P(age 18-29|tattoo) = 18/24 = 0.75

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**These are independent events.**

Sometimes the knowledge that one event has occurred does NOT change our assessment of the likelihood that another event occurs. Consider the genetic trait, hitch hiker’s thumb, which is the ability to bend the last joint of the thumb back at an angle of 60° or more. Whether or not an offspring has hitch hiker’s thumb is determined by two random events: which gene is contributed by the father and which gene is contributed by the mother. Which gene is contributed by the father does NOT affect which gene is contributed by the mother These are independent events.

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**Let’s consider a bank that offers different types of loans:**

The bank offers both adjustable-rate and fixed-rate loans on single-family dwellings, condominiums and multifamily dwellings. The following table, called a joint-probability table, displays probabilities based upon the bank’s long-run loaning practices. Single Family Condo Multifamily Total Adjustable .40 .21 .09 .70 Fixed .10 .11 .30 .50 .20 P(Adjustable loan) = .70

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**Bank Loan’s Continued . . . P(Adjustable loan) = .70**

Single Family Condo Multifamily Total Adjustable .40 .21 .09 .70 Fixed .10 .11 .30 .50 .20 P(Adjustable loan) = .70 P(Adjustable loan|Condo) = .21/.30 = .70 Knowing that the loan is for a condominium does not change the probability that it is an adjustable-rate loan. Therefore, the event that a randomly selected loan is adjustable and the event that a randomly selected loan is for a condo are independent.

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Independent Events Two events are independent if knowing that one will occur (or has occurred) does not change the probability that the other occurs. Two events, E and F, are said to be independent if P(E|F) = P(E). If P(E|F) = P(E), it is also true that P(F|E) = P(F). If two events are not independent, they are said to be dependent events.

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**Multiplication Rule for Two Independent Events**

Two events E and F are independent, if and only if,

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**Hitch Hiker’s Thumb Revisited**

Suppose that there is a 0.10 probability that a parent will pass along the hitch hiker’s thumb gene to their offspring. What is the probability that a child will have a hitch hiker’s thumb? Since these are independent events, we just multiply the probabilities together. This would happen if the mother contributes a hitch hiker’s gene (H+) AND if the father contributes a hitch hiker’s gene (H+). P(H+ from mom AND H+ from dad) = 0.1 × 0.1 = 0.01

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**Multiplication Rule for k Independent Events**

Events E1, E2, . . ., Ek are independent if knowledge that some number of the events have occurred does not change the probabilities that any particular one or more of the other events occurred. This relationship remains valid if one or more of the events are replaced by their complement (not E).

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Suppose that a desktop computer system consist of a monitor, a mouse, a keyboard, the computer processor itself, and storage devices such as a disk drive. Most computer system problems due to manufacturer defects occur soon in the system’s lifetime. Purchasers of new computer systems are advised to turn their computers on as soon as they are purchased and then to let the computer run for a few hours to see if any problems occur. Let E1 = event that a newly purchased monitor is not defective E2 = event that a newly purchased mouse is not defective E3 = event that a newly purchased disk drive is not defective E4 = event that a newly purchased processor is not defective Suppose the four events are independent with P(E1) = P(E2) = P(E3) = P(E4) = .99

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Let E1 = event that a newly purchased monitor is not defective E2 = event that a newly purchased mouse is not defective E3 = event that a newly purchased disk drive is not defective E4 = event that a newly purchased processor is not defective Suppose the four events are independent with P(E1) = P(E2) = P(E3) = P(E4) = .99 What is the probability that none of these components are defective? In the long run, 89% of such systems will run properly when tested shortly after purchase. (.98)(.98)(.94)(.99) = .89

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Let E1 = event that a newly purchased monitor is not defective E2 = event that a newly purchased mouse is not defective E3 = event that a newly purchased disk drive is not defective E4 = event that a newly purchased processor is not defective Suppose the four events are independent with P(E1) = P(E2) = P(E3) = P(E4) = .99 What is the probability that all these components will run properly except the monitor? (.02)(.98)(.94)(.99) = .018

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**Probability of a spade given I drew a spade on the first card.**

Suppose I will pick two cards from a standard deck. This can be done two ways: Pick a card at random, replace the card, then pick a second card Pick a card at random, do NOT replace, then pick a second card. If I pick two cards from a standard deck without replacement, what is the probability that I select two spades? Sampling without replacement – the events are typically dependent events. Sampling with replacement – the events are typically independent events. Probability of a spade given I drew a spade on the first card. Are the events E1 = first card is a spade and E2 = second card is a spade independent? NO P(E1 and E2) = P(E1) × P(E2|E1) =

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Suppose the manufacturer of a certain brand of light bulbs made 10,000 of these bulbs and 500 are defective You randomly pick a package of two such bulbs off the shelf of a store. What is the probability that both bulbs are defective? Are the events E1 = the first bulb is defective and E2 = the second bulb is defective independent? What would be the probability of selecting a defective light bulb? To answer this question, let’s explore the probabilities of these two events? 500/10,000 = .05

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Light Bulbs Continued . . . What would be the probability of selecting a defective light bulb? Having selected one defective bulb, what is the probability of selecting another without replacement? If a random sample of size n is taken from a population of size N, then the outcomes of selecting successive items from the population without replacement can be treated as independent when the sample size n is at most 5% of the population size N. 500/10,000 = .05 499/9999 = .0499 These values are so close to each other that when rounded to three decimal places they are both For all practical purposes, we can treat them as being independent.

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**Light Bulbs Continued . . . (0.05)(0.05) = .0025**

What is the probability that both bulbs are defective? Are the selections independent? We can assume independence. (0.05)(0.05) = .0025

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**General Rule for Addition**

Since the intersection is added in twice, we subtract out the intersection. For any two events E and F, E F

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**Musical styles other than rock and pop are becoming more popular**

Musical styles other than rock and pop are becoming more popular. A survey of college students finds that the probability they like country music is .40. The probability that they liked jazz is .30 and that they liked both is What is the probability that they like country or jazz? = .6

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**Yes Ask yourself, “Are the events mutually exclusive?” No**

Here is a process to use when calculating the union of two or more events. Ask yourself, “Are the events mutually exclusive?” In some problems, the intersection of the two events is given (see previous example). Yes No In some problems, the intersection of the two events is not given, but we know that the events are independent. If independent

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**How can you find the probability of A and B?**

Suppose two six-sided dice are rolled (one white and one red). What is the probability that the white die lands on 6 or the red die lands on 1? How can you find the probability of A and B? Let A = white die landing on 6 B = red die landing on 1 Are A and B disjoint? NO, independent events cannot be disjoint

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**General Rule for Multiplication**

For any two events E and F,

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**Here is a process to use when calculating the intersection of two or more events.**

Ask yourself, “ Are these events independent?” Yes No

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**There are seven girls and eight boys in a math class**

There are seven girls and eight boys in a math class. The teacher selects two students at random to answer questions on the board. What is the probability that both students are girls? Are these events independent? NO

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**= (.05)(.95) + (.95)(.05) = .095 Light Bulbs Revisited . . .**

A certain brand of light bulbs are defective five percent of the time. You randomly pick a package of two such bulbs off the shelf of a store. What is the probability that exactly one bulb is defective? Let D1 = first light bulb is defective D2 = second light bulb is defective = (.05)(.95) + (.95)(.05) = .095

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**This can happen in one of two ways: **

An electronics store sells DVD players made by one of two brands. Customers can also purchase extended warranties for the DVD player. The following probabilities are given: Let B1 = event that brand 1 is purchased B2 = event that brand 2 is purchased E = event that extended warranty is purchased P(B1) = .7 P(B2) = .3 P(E|B1) = .2 P(E|B2) = .4 If a DVD customer is selected at random, what is the probability that they purchased the extended warranty? This can happen in one of two ways: They purchased the extended warranty and Brand 1 DVD player OR 2) They purchased the extended warranty and Brand 2 DVD player

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**These are disjoint events**

DVD Player Continued . . . Let B1 = event that brand 1 is purchased B2 = event that brand 2 is purchased E = event that extended warranty is purchased P(B1) = .7 P(B2) = .3 P(E|B1) = .2 P(E|B2) = .4 If a DVD customer is selected at random, what is the probability that they purchased the extended warranty? These are disjoint events Use the General Multiplication Rule:

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**P(E) = (.2)(.7) + (.4)(.3) = .26 DVD Player Continued . . .**

Let B1 = event that brand 1 is purchased B2 = event that brand 2 is purchased E = event that extended warranty is purchased P(B1) = .7 P(B2) = .3 P(E|B1) = .2 P(E|B2) = .4 If a DVD customer is selected at random, what is the probability that they purchased the extended warranty? P(E) = (.2)(.7) + (.4)(.3) = .26 This is an example of the Law of Total Probabilities.

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**Law of Total Probabilities**

If B1 and B2 are disjoint events with probabilities P(B1) + P(B2) = 1, for any event E More generally B1, B2, …, Bk are disjoint events with probabilities P(B1) + P(B2) + … + P(Bk) = 1, for any event E

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Bayes Rule (Theorem) A formula discovered by the Reverend Thomas Bayes, an English Presbyterian minister, to solve what he called “converse” problems. Let’s examine the following problem before looking at the formula . . .

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Lyme’s disease is the leading tick-borne disease in the United States and England. Diagnosis of the disease is difficult and is aided by a test that detects particular antibodies in the blood. The article, “Laboratory Consideration in the Diagnosis and Management of Lyme Borreliosis”, American Journal of Clinical Pathology, 1993, used the following notations: + represents a positive result on a blood test - represents a negative result on a blood test L represents the patient actually has Lymes LC represents the patient doesn’t have Lymes The article gave the following probabilities: P(L) = P(LC) = P(+|L) = P(-|L) = .063 P(+|LC) = P(-|LC) = .97

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**Lyme’s Disease Continued . . .**

The article gave the following probabilities: P(L) = P(LC) = P(+|L) = P(-|L) = .063 P(+|LC) = P(-|LC) = .97 Bayes’s converse problem poses this question: “Given that a patient test positive, what is the probability that he or she really has the disease?” written: P(L|+) This question is of primary concern in medical diagnosis problems!

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**Lyme’s Disease Continued . . .**

The article gave the following probabilities: P(L) = P(LC) = P(+|L) = P(-|L) = .063 P(+|LC) = .03 P(-|LC) = .97 Bayes reasoned as follows: Using the Law of Total Probabilities, the denominator becomes P(+|L)P(L) + P(+|LC)P(LC). Substitute values: Since we can use P(+|L) × P(L) for the numerator.

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Bayes Rule (Theorem) If B1 and B2 are disjoint events with probabilities P(B1) + P(B2) = 1, for any event E More generally B1, B2, …, Bk are disjoint events with probabilities P(B1) + P(B2) + … + P(Bk) = 1, for any event E

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**Estimating Probabilities using Simulation**

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Simulation Design a method that uses a random mechanism (such as a random number generator or table, tossing a coin or die, etc.) to represent an observation. Be sure that the important characteristics of the actual process are preserved. Generate an observation using the method in step 1 and determine if the outcome of interest has occurred. Repeat step 2 a large number of times Calculate the estimated probability by dividing the number of observations of the outcome of interest by the total number of observations generated. Simulation provides a means of estimating probabilities when we are unable to determine them analytically or it is impractical to estimate them empirically by observation.

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Suppose that couples who wanted children were to continue having children until a boy was born. Would this change the proportion of boys in the population? We will use simulation to estimate the proportion of boys in the population if couples were to continue having children until a boy was born. We will use a single-random digit to represent a child, where odd digits represent a male birth and even digits represent a female birth. Select random digits from a random digit table until a male is selected and record the number of boys and girls. Repeat step 2 a large number of times.

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**Boy Simulation Continued . . . **

Below are four rows from the random digit table at the back of our textbook. Continue this process a large number of times (at least 100 trials). Calculate the proportion of boys out of the number of children born. Row 6 9 3 8 7 5 2 4 1 Notice that with only 10 trials, the proportion of boys is 10/22, which is close to 0.5! Trial 1: girl, boy Trial 2: boy Trial 3: girl, boy Trial 4: girl, boy Trial 5: boy Trial 6: boy Trial 7: boy Trial 8: girl, girl, boy Trial 9: girl, boy Trial 10: girl, girl, girl, girl, girl, girl, boy

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Copyright ©2005 Brooks/Cole, a division of Thomson Learning, Inc. Understanding Probability and Long-Term Expectations Chapter 16.

Copyright ©2005 Brooks/Cole, a division of Thomson Learning, Inc. Understanding Probability and Long-Term Expectations Chapter 16.

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