3. 1. Try to estimate the likelihood of a result by actually observing the random phenomenon many times and calculating the relative frequency of the results. 2. Develop a probability model and use it to calculate a theoretical answer. 3. Start with a model that, in some fashion, reflects the truth about the random phenomenon, and then develop a plan for imitating – or simulating – a number of repetitions of the procedure.

4.  Method 1 is slow, can be costly, and often impractical or difficult.  Method 2 requires that we know something about the rules of probability and may not be possible.  Method 3 is usually quicker than repeating the real procedure, especially when we can use a calculator or computer. Simulation also allows us to get reasonably accurate results and answer questions that are more difficult when done with formal mathematical analysis.

5.  We are interested in estimating the likelihood that a couple will have a girl among their first four children. Develop a method for imitating births using › A coin › A six-sided die › A deck of cards › A Random Digit Table  What else could we use?

6.  Simulation is an effective tool for finding the likelihood of complex results once we have developed a trustworthy model.  We can use Random Digit Tables, graphing calculators, computers or other items (cards, coins, dice) to simulate many repetitions quickly.

7. 1. State the problem or describe the random phenomenon. 2. State the assumptions. 3. Assign digits to represent outcomes (i.e. odd number is a head or a boy and even number is a tail or a girl. This depends on the problem you are trying to simulate). 4. Simulate many repetitions (or trials). 5. State your conclusions.

8.  Events are independent if the results of one trial do not effect the results of the next.  For example, if you toss a coin and get a head, that has nothing to do with what you will get when you toss the coin again.  If you roll a die and get a 3, that has nothing to do with what you will get on the next roll.

9.  We want to choose a person at random from a group of which 70% are employed. How would we assign numbers to individuals. › 0, 1, 2, 3, 4, 5, 6  employed › 7, 8, 9  unemployed › We could assign any three digits to unemployed as long as they are distinct › We could also do the following:  00,01,…, 69  employed  70, 71,…99  unemployed › This is less efficient because it uses twice as many digits and 10 times as many numbers.

10.  We want to choose a person at random from a group of which 73% are employed. How would we assign numbers to individuals. › 00, 01, …, 72  employed › 73, 74, …, 99  unemployed › We had to assign 73 of the 100 numbers to be employed to get 73%. We could also do other assignments, but doing it by groups is easiest.

11.  We want to choose a person at random from a group of which 50% are employed, 20% are unemployed, and 30% are not in the labor force. How would we assign numbers to individuals. › 0, 1, 2, 3, 4  employed › 5, 6  unemployed › 7, 8, 9  not in labor force › The assignment of digits is not important. What is important is how many digits are assigned to each outcome.

12.  Orders of frozen yogurt flavors (based on sales) have the following relative frequencies: 43% chocolate, 38% vanilla and 19% strawberry. How could we simulate customers entering the store and buying yogurt? › 00 to 42 would represent the outcome chocolate. › 43 to 80 would represent the outcome vanilla. › 81 to 99 would represent the outcome strawberry.

13.  A couple will have children until they have 4 children or a girl, whichever comes first.  What is the likelihood that they would have a girl?  We will assume that each child is equally likely to be a boy or a girl and that the sex of each successive child is independent.  How should we assign the digits? › 0, 1, 2, 3, 4  girl or even  girl › 5, 6, 7, 8, 9  boy or odd  boy

14.  Let’s do the simulation letting girls be 0-4! Start at line 130 in Table B and lets do 14 simulations. 690516481 7871 740 BBGBG BG BG BBBG BG G 951784 53 4 064 8987 BBG BBG BG G GBG BBBB  So 13 of the 14 trials resulted in the birth of a girl. Hence this strategy will produce a girl 13/14 or about 93% of the time. The actual probability of having a girl would be 93.75%.

15. Using the randInt Command!

16.  Let’s try problems 6.11, 6.12, 6.13, 6.16, 6.20

18.  Chance behavior is unpredictable in the short run but has a regular and predictable pattern in the long run.  Probability is the long-run proportion of repetitions on which an event occurs.

19.  Count Buffon in the 1700's › Tossed a coin 4040 times and got 2048 heads for a proportion of 0.5069.  John Kerrich imprisoned in WWII › Tossed a coin 10,000 times and got 5067 heads for a proportion of 0.5067  Karl Pearson around 1900 › Tossed a coin 24,000 times and got 12,012 heads for a proportion of 0.5005

21.  We can never observe a probability exactly.  Mathematical probability is an idealization based on imagining what would happen in a long series of trials.  Remember the following: › We must have a long series of independent trials, i.e. the outcome of one trial must not influence the outcome of another trial. › The idea of probability is empirical, i.e. we must observe many trials to estimate a probability. › Computer simulation is very useful because it allows us to see the results of many trials since it often takes several hundred trials to determine the probability of an outcome.

23.  What is the sample space when we roll two dice?  Sometimes the sample space varies depending on what exactly you're asking for!  What is the sample space when we record four coin tosses?  What is the sample space when we toss a coin and roll a die?

25.  What is the sample space when we record four coin tosses?  What is the sample space when we toss a coin and roll a die?  Being able to properly enumerate the outcomes in a sample space is critical to determining probabilities. Sometimes it's helpful to use a tree diagram.

27.  For example: › How many ways can I toss a coin and roll a die? › How many ways can I toss a coin and draw a card from a standard 52-card deck? › How many ways can I roll a die and draw a card from a standard 52-card deck?

28.  How do we describe probabilities mathematically? Rather than try to give “correct” probabilities by recording large numbers of trials, we start by laying down facts that must be true for any assignment of probabilities.

30. Disjoint (mutually exclusive) events A and B The Complement of Event A

31. Intersection of A and B ( A⋂B)

32.  Union – set of all outcomes in A or B › Written A ⋃B and read “A union B.”  Intersect or intersection – set of all outcomes that are in A and B. › Written A⋂B and read “A intersect B” or the intersection of A and B.”  Let’s do some examples. › Let A = {0, 1, 2, 5, 7, 9}, B = {1, 3, 4, 5, 6, 8, 9} and C = {1, 3, 5, 7, 9}  A ⋃B =  A⋂B =  A ⋂C =  B ⋃C =  B ⋂C =  A ⋃ C=

33.  Union – set of all outcomes in A or B › Written A ⋃B and read “A union B.”  Intersect or intersection – set of all outcomes that are in A and B. › Written A⋂B and read “A intersect B” or the intersection of A and B.”  Let’s do some examples. › Let A = {0, 1, 2, 5, 7, 9}, B = {1, 3, 4, 5, 6, 8, 9} and C = {1, 3, 5, 7, 9}  A ⋃B = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}  A⋂B = {1, 5, 9}  A ⋂C = {1, 5, 7, 9}  B ⋃C = {1, 3, 4, 5, 6, 7, 8, 9}  B ⋂C = {1, 3, 5, 9}  A ⋃ C= {0, 1, 2, 3, 5, 7, 9}

34.  What is the probability for rolling a 5 with a pair of dice?

36.  Find the following probabilities. › The first digit is one › The first digit is odd › The first digit is more than six › Compare this last probability with the probability that a random first digit is greater than six

38.  Successive coin tosses  A single coin toss  Taking an IQ test twice  Dealing playing cards  Spinning a number on a spinner

39.  A rapid HIV test has probability about 0.004 of producing a false-positive.  If 200 hundred people are tested, what is the probability that at least 1 false-positive will occur? › Based on what we are given, we know there is a 0.996 probability that we will not get a false positive for a single person using the complement rule. › There are too many ways that at least 1 person could receive a false-positive. So we need to use the complement rule. Hence: P(at least 1 false-positive) = 1 – P(no false-positives out of 200) =1-0.996 200 =1-0.4486=0.5514

40.  6.40, 6.41, 6.50, 6.51, 6.53, 6.55, 6.57, 6.59

44.  Let A be the event that Joseph gets a hit, let B be the event that Joseph gets a walk, let C be the event that Joseph flies out, let D be the event that Joseph grounds out, and let E be the event that Joseph gets on base because of an error. If P(A) = 0.27, P(B) = 0.10, P(C) = 0.31, P(D) = 0.28, and P(E) = 0.04, then find the following.  P(A or B or E) =  P(C or D) =  P(C or D or E) =

45.  Let A be the event that Joseph gets a hit, let B be the event that Joseph gets a walk, let C be the event that Joseph flies out, let D be the event that Joseph grounds out, and let E be the event that Joseph gets on base because of an error. If P(A) = 0.27, P(B) = 0.10, P(C) = 0.31, P(D) = 0.28, and P(E) = 0.04, then find the following.  P(A or B or E) = P(A) +P(B) + P(E) =0.27+0.10+0.04 = 0.41  P(C or D) = P(C) + P(D) = 0.31+0.28 = 0.59  P(C or D or E) = P(C) + P(D) + P(E) = 0.31+0.28+0.04 = 0.63  This works because the events are disjoint!

46.  Suppose Event A has a 0.2 chance of occurring and Event B has a 0.4 chance of occurring. If we add them, we would get that there is a 0.6 chance of A or B occurring, but that is wrong. Why?  Because both events occur at the same time sometimes. That means that the area where they overlap is counted twice and must be taken out!  So if we know that the Event that A and B both occur has a chance of 0.12, then we know that the Chance of A or B occurring would be 0.2 + 0.4 – 0.12 = 0.48. See the diagram!

47.  Suppose Event A has a 0.2 chance of occurring and Event B has a 0.4 chance of occurring. If we add them, we would get that there is a 0.6 chance of A or B occurring, but that is wrong. Why?  Because both events occur at the same time sometimes. That means that the area where they overlap is counted twice and must be taken out!  So if we know that the Event that A and B both occur has a chance of 0.12, then we know that the Chance of A or B occurring would be 0.2 + 0.4 – 0.12 = 0.48. See the diagram!

48. If A and B are disjoint, P(A and B) (also written as P(A⋂B)) has no outcomes in it. Since it has no outcomes, it has a probability of 0. So this rule works for all events and is a more generalized form of rule 3.

49.  Let’s look at an example.  Deborah estimates that she has a 0.7 chance of being made a partner while Matthew has a 0.5 chance of being made a partner. She also estimates that there is a 0.3 chance that they would both be made partners. Answer the following:  P(at least one is made a partner) =  P(neither is made a partner) =

50.  Let’s look at an example.  Deborah estimates that she has a 0.7 chance of being made a partner while Matthew has a 0.5 chance of being made a partner. She also estimates that there is a 0.3 chance that they would both be made partners. Answer the following:  P(at least one is made a partner) = P(D or M) = = P(D) +P(M) –P(D ⋂M) = 0.7 + 0.5 – 0.3 = 0.9  P(neither is made a partner) = 1 – P(D or M) = = 1 – 0.9 = 0.1

52.  Fill in what we know!  Complete the table! Matthew PromotedNot PromotedTotal Deborah Promoted0.30.7 Not Promoted Total0.51 Matthew PromotedNot PromotedTotal Deborah Promoted0.30.40.7 Not Promoted0.20.10.3 Total0.5 1

53. Coffee ColaTea 5% 20%10% 5% 15% 20% NONE 20%

54.  What’s the probability of drawing an Ace out of a deck of cards?  You have already drawn four cards and you have a King, 2 Queens and a Jack in your hand. Now what is the probability of drawing an Ace out of the deck on the next draw?  This is called a conditional probability because it is finding the probability of one thing occurring (drawing an Ace) given that we know that another event (we already have a King, 2 Queens and a Jack) has already occurred.

55.  The notation for a conditional probability is P(A|B) and we read it “ the probability of A given that B has already occurred.”  Let’s look at an example on the next slide.

56.  The table below breaks down student grades awarded at a university by grade and the school that the student took the class in. Let A = the grade comes from an Engineering and Physical Sciences class, B = the grade is below a B, C = the grade is an A, and D = Liberal Arts class.  Find P(B) =  Find P(B|A) =  Find P(C) =  Find P(C|D) =

57.  The table below breaks down student grades awarded at a university by grade and the school that the student took the class in. Let A = the grade comes from an Engineering and Physical Sciences class, B = the grade is below a B, C = the grade is an A, and D = Liberal Arts class.  Find P(B) = 3,656/10,000 = 0.3656  Find P(B|A) = 800/1,600 = 0.5  Find P(C) = 3,392/10,000 = 0.3392  Find P(C|D) = 2,142/6,300 = 0.34

58.  Since P(A ⋂ B) = P(B ⋂ A), P(A ⋂ B) = P(B) P(A|B).  This rule is true because in order for both A and B to occur, one of them must occur, then we have to multiply by the probability that the other occurs given that the first already occurred.

59.  Downloading Music › 29% of internet users download music files and 67% of downloaders say they don’t care if the music is copyrighted. So the percent of internet users who download music (event A) and don’t care about copyright (event B) is 67% of the 29% who download or (0.67)(0.29) = 0.1943 = 19.43%. The multiplication rule expresses this as P(A and B) = P(A) P(B|A).  Slim wants two diamonds › Slim needs to draw two diamonds in a row. Slim can see 11 cards and 4 of these are diamonds. What’s the probability that Slim can draw two diamonds in a row? › P( drawing two diamonds ) = P( drawing one diamond ) P( drawing another diamond given he already drew one ) › So P( drawing two diamonds ) = (9/41)(8/40) = 72/1640 = 0.044

60.  We can rearrange the general multiplication rule [ P(A and B) = P(A) P(B|A) ]to get a formula for conditional probability.  We can also get

61.  Let A = grade is an A and B = a liberal arts course.  What’s the conditional probability that a grade is an A, given that it comes from a liberal arts course?  P(A|B)= P(A and B) / P(B) = (2,142/10,000) / (6,300/10,000) = 0.34

62.  Let A = grade is an A and B = not a liberal arts course.  What’s the conditional probability that a grade is an A, given that it does not come from a liberal arts course?  P(A|B)= P(A and B) / P(B) = (1,250/10,000) / (3,700/10,000) = 0.3378

64.  P(A and B and C) = P(A) P(B|A) P(C|A and B)  P(A and B and C and D) = P(A) P(B|A) P(C|A and B) P(D|A and B and C)  And so on…

65.  Online chat rooms are dominated by the young. If we look at adult Internet users only, 47% of the 18 to 29 age group chat, as do 21% of those aged 30 to 49 and just 7% of those 50 and over. Total 29% of adult Internet users are aged 18 to 29, another 47% are 30 to 49, and the remaining 24% are 50 or over.

67.  The probability of reaching the end of any complete branch is the product of the probabilities written on its segments.

68.  What percent of adult chat room participants are aged 18 to 29?

69.  What is the probability that a randomly chosen user of the Internet participates in chat rooms? 0.2518 is the probability because we would add these three up!

71.  6.80, 6.83, 6.86, 6.87, 6.88, 6.90, 6.91