Phil 105 Exam 5

13) t or f: if A is certain and P(B) = 30% then P(A&B) =30%

13. 30%, since you'll multiply P(B) by 1.

15) t or f: if A is certain and B is certain then P(A&B)= P(A)

15. True, since each has probability 1.

ambiguous and vague questions

poorly worded questions simply in that its not clear what they are asking

example probability vs odds the outcome of getting a 6 on one roll of a die

probability 1/6 5:1 since 5 unfavorable losing and 1 getting a 6 unfavorable outcomes:favorable outcomes when odds are 1:1 we say odds are even (tossing a fair coin)

Complex questions

questions that ask at least two things at once but to which respondents are allowed to give only one answer ex) do you think that the mayor should be fired and put in jail for his conduct a)yes, b) no, c) no opinion there are two questions here, if he should be fired and if he should go to jail, but the boxes only allow one answer for both

Representative sample versus biased samples

representative: 1) the size of the sample 2) whether the sample was selected randomly Respresentative sample: How do we draw a representative sample? By drawing as large a sample as is feasible, and by randomizing our drawing of the sample. Chapter 78 is about the size of the sample. How large does our sample need to be? As the text notes, there is no single answer to this question. It really depends on several factors in addition the population size, mentioned on p.215. How precise a statistic we want (as measured by the length of the confidence interval or the margin of error). What level of confidence we want. How variable (i.e. how diverse) the population is (this would be measured by the standard deviation; we'll ignore this particular criterion here and focus on the first two criteria above).

False alternatives

some are poorly constructed in the sense that they present respondents with too few response options- a car of false alternatives ex) how do you feel about the mayors response to the recent crisis at city hall, check one a) the mayor overreacted b) the mayors respond was too little too late c) no opinion these do not exhaust the range of response one might have

if i make two random draws from a standard deck of cards with replacement of the first card then (3 of hearts on the first draw) and (king of spades on the second draw) are a) mutually exclusive b)jointly exhaustive c) independent d) none

(c) If the card is replaced after the first draw, what happens on the first draw does not affect what happens on the second draw.

1) explain why the questions are poorly worded a) do you favor abortion? b) are you religious or athiest? c) why do you think the rest of the world hates america? d) do you support a patients natural human right to refuse life extending medical treatment? e) is america headed in the right direction? f) are you a heavy drinker? g) do you consider yourself a liberated spirit? h) do you believe that global warming is man made? please answer yes no or no opinion i) do you object to geneticists playing god by tinkering with human dna? j) do you believe that the minimum voting age and minimum age at which one may purchase alcohol should be the same, so that every 18 yr old would be able to buy alcohol? answer yes no or no opinion 2) you want to know how happy people are with their most recent automobile purchase. what question might you ask first would tend to diminish their reported satisfaction with the car? 3) could the order in which questions are asked on a math exam influence student performance? 4) if you have 2 question to ask and believe placing either one before the other could influence the response to the second, how might you try to minimize or counteract the effect? 5) a pollster is going to ask people to estimate their own IQs, What circumstances could be manipulated to diminish the average self rating? 6) product marketing consultant surveying ppl about new snack, what can effect a more positive response?

1) a)This question is tendentious, since no one "favors abortion." It pushes people in the direction of giving a "no" answer. It should have been worded to ask whether respondents support giving women the legal option to abort (and probably should be even more specific than this in respect to the circumstances). b) This is a case of false alternatives, since "agnostic" (taking no position) is an obvious third choice that was omitted. c) This is a complex question, because it really is two questions that should have been asked separately: Do you think other people hate Americans? If so, why do they hate Americans? d) This is tendentious, because the question already assumes that there is a "natural human right" here. Who would not support a natural human right? e) This is an ambiguous question, because a country can be headed in the right/wrong direction in respect to many different things (economic conditions, foreign policy, etc.) f) This also is ambiguous (and probably tendentious as well, since no one wants to characterize him/herself as a heavy drinker). What does "heavy" mean? It could mean three drinks per day to one person, and five drinks a day to another. g) This is vague, since it's not clear what "liberated spirit" could possibly mean. (The question in (f) was ambiguous rather than vague because one can imagine some fairly definite meanings for "heavy drinker" defined in terms of number of drinks per day. But it's hard to think of any definite meanings for "liberated spirit.") h) This is a case of complex question and false alternatives. It's complex because two questions really are presented: Do you believe in global warming? If so, do you believe it's man-made? And it's a case of false alternatives because rather than give a simple "yes," for example, one might want to say that it is (for example) 60% man-made. i) The question is tendentious, since "playing God" has a clear negative connotation for most people. j) It's a case of complex question and false alternatives. We first need to know whether you think the drinking and voting age should be the same. Then, if so, we need to know what the age should be, and 18 is just one possibility. 2) Example: what has been the biggest problem you've had with the car so far? 3) Yes, since asking easy questions first can put students at ease and build confidence for the later questions. 4) You could tell them about the question-ordering effect, so that they are on their guard against it. Or you could pause a while before the next question, or ask some neutral question in between in order to clear their minds of any influence from the first question. 5) Give them some very difficult brain-teaser and let them fail to solve it before you ask them to estimate their own IQs. 6) Ask them to rate it while they're hungry.

HW

1) a. P(AvB)=1 It is certain that either A or B will happen. b. P(AIB&C) = 0.25 The probability that A will happen, if both B and C happen, is .25. c. P(~B) =0 The probability that B will not happen is 0 (which means that B is certain to happen). d. 0< P(C) <1 C is a contingent event. e. P(AIB) =P(A) The probability that A will happen if B does is equal to the unconditional probability of A, which means that B has no effect on A's probability.

1) if my sample indicates that 20% of registered voters are unmarried and my margin of error is +- 5 points, then what is the confidence interval? 2) in the preceding exercise, is it possible that the true population statistic falls outside my confidence interval? 3) t or f: to say that someone commits the fallacy of hasty generalization means that the persons estimate of the prevalence of some trait in the whole population, based on a small sample, is wrong 4) can a sample be representative of a population in respect to one characteristic while failing to be very representative in respect to a different characteristic? consider for ex whether the students in the class you are now taking are representative of all the students at your school in some respects but not others 5) what is meant by the term statistical dead heat in the context of an election? 6) t or f: if a poll with a +-5% margin of error indicates that candidate A is leading candidate B by 2 % points, then it is possible that candidate B is ahead. is it nevertheless highly probable in such cases that the candidate with the reported lead really is leading?

1) 15% - 25%. 2) Yes, it's possible. The confidence level will tell us how probable this is. If the confidence level is 95%, then there is a 5% chance that the true figure will fall outside my confidence interval. 3) False; it's not necessarily going to be wrong - it's just that there is a good chance it will be wrong. 4) Yes, a sample can be representative in some ways and unrepresentative in others. The students in your class probably are representative when it comes to height (there's no reason to think your classmates are much taller or shorter on average than a typical student). But if it's a freshman course, then they probably are unrepresentative when it comes to their total accumulated credit hours so far: they probably have fewer credit hours than an average student. 5) If a race is a statistical dead heat (a term that statisticians actually don't use very often; it's mostly used by the media), then this means that the lead one candidate has on the other is within the sample's margin of error. So, perhaps candidate A has 42% support and candidate B has 40% in the latest poll (in a race with 3 or more candidates), but the margin of error is +/- 5 percentage points. So, A's 2% lead is within the margin of error. 6) Yes, it is likely that A is ahead. Statisticians have ways of calculating this probability. The point is just this: to say a lead is within the margin of error does not mean that we really can't be confident about who is leading. It could be more than 90% probable that A really is ahead, even if the race is a statistical dead heat right now. (This is why statisticians don't much care for the term "statistical dead heat": it makes the public think that the race is completely up for grabs, when it might be very likely that one candidate really is ahead of the other.)

1) dear abby once asked readers to write in and tell her whether they have had an extramarital affair. a startling number of people said they have in fact most of those who wrote in had at least one. how might a selection effect be at work here? are faithful and unfaithful spouses as well as readers of dear abby and those who don't read her column equally likely to write in? 2) if i am conducting a phone survey to find out who ammericans will vote for in the next election and i only call people living in major cities, why is my sample likely to be unrepresentative? 3) if i want to poll people regarding their intended voting in the next presidential election and i set up pollen stations at various large cities and small towns all over the US (i selected them by throwing darts to us map) and in each city and town i set up my station outside pet grooming salons then should i worry that this choice of specific location will bias my sample? 4) the cia is criticized when it fails to prevent some terrorist plot giving the impression that the people running the org must be incompetent. how might this impression be caused by a selection effect? are we equally likely to hear about their successes vs their failures? 5) many polls now ask people to text their responses. how might this method introduce bias into the sample? 6) if most early hominid fossils are found in caves then it might be that our hominid ancestors spent most of time in caves, or it might be a selection effect at work here, explain 7) if the fox new channel and sub each conduct the same poll, by asking viewers to text their responses, why might one expect different results in the two polls? 8) could a selection effect be at work if you generalize about the average skill level of professional tennis players based on what you've seen from watching nationally televised tennis matches?

1) It could be that those who have had an affair are more likely to write in than those who have not, simply because these people have a story to tell and perhaps want to unburden themselves. So, it would not be surprising if Abby's sample is biased toward cheaters. ("Dear Abby" is the title of an advice column in daily papers.) 2) You will only reach urban dwellers with this sample, and so rural dwellers won't have the same chance to be included in the sample. Since urbanites and those in rural areas often vote differently, this will be a problem for your sample. 3) Pet-grooming is a luxury, and so wealthy people are more likely to be included in the sample than poor people. If rich and poor people vote differently, this will be a problem. (Of course, you're more likely to include pet owners than non-pet-owners as well. Is this a problem? It depends on whether these groups of people vote differently.) 4) Your "sample" likely includes only CIA operations reported in the media, and the media might be more likely to report failures than successes, simply because failures are more sensational. And the CIA might deliberately conceal some of its successes in order to protect the assets it used to achieve them. So, it could be that we are much more likely to hear about failures than successes, even if successes actually outnumber failures. 5) Since mainly young people know how to text, your sample will be biased toward young people. 6) Perhaps they spent equal time outdoors and in caves, but the conditions for preserving their fossils are better in caves, since these fossils are shielded from weathering, predation, etc. So, the fossils left inside might be more likely to endure and end up in our sample than the fossils that were left outside, biasing our sample toward fossils found in caves. 7) Since the liberal/conservative biases of the audience for each channel are different, each will get a sample biased toward one end of the political spectrum. 8) The networks that televise tennis matches generally will televise the matches of the more elite players (or they will only televise the later rounds of the tournament, when inferior players have been eliminated). So, your sample (gathered from watching matches on television) will be biased toward the most skilled players, and this could lead you to overestimate the skill level of an average pro tennis player.

1) for the values of 5,20,45,50,50 find the mean, median, mode and range 2) what type of average is misleading when a data set has a few unusually high or low values? 3) what type of average is misleading when the most common value is the lowest and or highest value? 4) what does it mean to say that a set of values like scores on an exam are normally distributed? 5) for qualitative data like hair color, can there be a thing like averages in any of the 3 senses we introduced? 6) sometimes instead of providing a mean average statisticians will provide a trimmed mean that drops some number of highest ad lowest values before the mean is computed. why do this? 7) why does knowing the mean average score on an exam was 75% not tell you everything that a teacher would want to know about how well the students are doing as a group? if the mean score is 75% can we infer from this alone that a graph of the scores would take the familiar shape of a bell curve? 8) for the data below what sort of average would give misleading impression about annual salary of typical employee? 765,000 51,442 48,000 52,446 50,000 49,555 50,342, 51,000 a)median average b)mean average c) both d) neither 9) can you validly infer that at least half of the men in a certain group wear a size 40 suit if 40 is the average suit size in all 3 senses of the term average we discussed? 10) you are told that the average age of people who watch a certain tv show is 40. answer the questions: a) is it possible that no 40 year old watches the show if this is the average viewer age in the sense of mean? b) is it possible that no 40 year old watches the show if this is the average viewer age in the sense of median? c) mode? 11) in any sense of the term average, is it possible for most people to be above average in respect to some attribute or other? 12) it is common in many contexts to divide the total number of times something happens in a year by 365 to get the number of times it happens on an average day. give 2 examples that this would be misleading 13) if we get told that 200 murders occur every year in a certain state A and that only 100 occurs per year in state B, then would it seem as if state A is much more violent than state B. but these figures could be misleading. what we need to look at is the per captia murder rate for each state. Explain what that means

1) Mean: 34 Median: 45 Mode: 50 Range: 45 2) Mean, since an unusually high or low score in a list (an "outlier") can skew the mean up or down and give a misleading impression as a result (consider again the case of the CEO who makes a million dollars a year in the text example). 3) mode 4) It means they take on the shape of a bell curve, as in the image above (there actually is a family of bell-shaped curves; they are all left-right symmetric in shape). 5) Yes: modal average (most common hair color, for example) 6) For the reason mentioned in (2): an outlier can skew the mean up or down. 7) The scores might not take on a bell-shaped curve. The mean would be 75 if everyone scored 75, but it would also be 75 if half the class scored 50, and the other half scored 100. Those would present two very different pictures to a teacher. 8) (a), since the 765K salary will skew the mean way up. 9) No. Suppose there are six mean who wear sizes 44, 42, 40, 40, 38, and 36. Then 40 is the mean, median, and modal average size, but only two of the men wear that size. 10) a (yes); b (yes); c (no, because 40 would have to be in the list if it's the modal average) 11) Yes. Consider the values 34, 34, 40, 42, and 44. (Imagine they are suit sizes.) The mode is 34, but three of the five sizes are bigger, and so most are above the modal average. The mean size is around 39, and three of the five sizes are bigger. So, when it comes to mode and mean, it is possible for most people in a group to be above the average for that group. (It is not possible for this to happen when it comes to median, because by definition there have to be an equal number of values above and below the median.) 12) For anything seasonal, such as fireworks injuries, this would be misleading. 13) The per capita rate is computed simply by dividing the total number of murders by the population size. The resulting figure can be compared between two states with different population sizes.

1) a study shows that cocaine use among teens is down 15% from last year, even if this is accurate stat, doesn't necessarily mean that anti drug programs aimed at teens are working. list 3 other things it might mean 2) a survey indicates that 10% fewer teens gave birth to children out of wedlock in the us last year than in previous years. but from this alone we cannot draw conclusion that fewer teens are having sex or even that more teens practicing safe sex. why? 5) a poll shows that half of americans believe pro athletes are overpaid. why not conclude that most americans don't respect what pro athletes do for a living? 8) Mean household income has risen by 5% over last year in a certain country. give two reasons why this info alone tells us little about how well off people in this nation are compared to last year 10) suppose a study shows that the IQs of children who's parents exposed them to mozarts music while children were still in the womb were on mean average 6 points higher than those whose parents did not expose them to this music. why does it not necessarily follow that prenatal exposure to mozarts music increases a childs IQ.

1) Perhaps demand for drugs has not diminished, but teens are switching to some other drug instead of cocaine. Or perhaps the supply of cocaine has dipped. 2) Perhaps the number of abortions has increased, or perhaps more teens are getting married, so that their children are not born out of wedlock. 5) Perhaps most Americans believe murderers deserve execution, but they worry about the risk of executing an innocent person, and they think the risk is too high. 8) Perhaps the cost of living has risen as well, so that no one really is better off. (Notice also that the statistic is a mean value; perhaps the richest 5% of the population got so much richer that they skewed the mean upward, while everyone else remained at the same level or even got worse.) 10) Perhaps only highly intelligent parents try the Mozart experiment, and they simply are passing on their superior intelligence genetically. Or perhaps the same parents who try the Mozart experiment also provide other sources of intellectual stimulation, and some other source accounts for the higher IQs.

how big should sample be? depends on 3 things in addition to population size

1) desired margin of error 2) desired confidence level 3) variance of the population in respect to the characteristic in which you're interesed

familiarize yourself with the basic rules of probability on p.185

1) the probability of any event is a number between 0 and 1, inclusive. We can express probabilities as decimals, fractions, or percentages. ex) a probability of 0.25 can also be 25% or 1/4 2)if A is an event that is certain to happen, then P(A)=1. If A is an impossible event, then P(A)=0 3) if A is neither necessary nor impossible, i.e. it is contingent, then 0<P(A)<1 4)According to the negation rule, P(A)=1-P(~A). That is, the probability that A happens is equal to one minus the probability that A does not happen. The event ~A is called the complement of A

two important factor when it comes to whether the sample is representative

1) the size of the sample 2) whether the sample was selected randomly introducing randomness helps

Schematically, the Gambler's Fallacy looks like this:

1. The unconditional probability of event E is n%. 2. But in the trials so far, E has happened at a rate higher (or lower) than n% of the time. 3. So, in the next trial, E's probability must be lower (or higher) than n%. When the trials are independent (as when tossing coins or rolling dice, for example), this inference is fallacious. The probabilities do not change in independent trials, no matter what has happened in previous trials.

use the conjunction rule to compute the prob of getting tails 3 times on 3 tosses

3. Every time you toss a coin it's like you're doing it for the first time: the trials are independent. So, we compute P(T1 & T2 & T3) with ½ x ½ x ½ = 1/8.

if i wager 2 dollars on snake eyes in the roll of two fair dice, and the betting odds are fair, what should be my winnings if i win?

3. Since the odds are 35:1, you should get $35 dollars for every $1 you wager if you win. So, a $2 wager should earn $70 in winnings (plus your original $2 stake back).

if a and b are mutually exclusive and the probability of a is 40% then what follows? a) the probability of b is 60% b) the probability of b is 40% c) the probability of b is 0% d) probability of b cannot be determined

4. The answer is (d), because we were not told whether the two events are also jointly exhaustive. If they are not exhaustive, then there is at least a third possible outcome. Maybe A is 40% probable, B is 30% probable, and some third event C is 30% probable also.

4) how might the inverse probability fallacy be committed in the context of scanning luggage for weapons and bombs when the scanner alert is sounded for a particular piece of luggage?

4. This problem is much like the others. Someone might mistakenly reason that since a scanner is 90% likely to sound an alarm if there really is a weapon in a piece of luggage, then, if the alarm goes off, it must be 90% likely that there is a weapon. But that conclusion does not follow. We'd have to know how often the alarm goes off even when there is no weapon (the "false positive" rate for the scanner), and how likely it is (the base rate) that any given piece of luggage would have a weapon in it.

7) t or f: for any events A and B, P(AvB)= P(BvA)

7. True (the order of the disjuncts makes no difference to the probability).

8) t or f: if i toss a coin 3 times then the outcome TTT has the same prob as the outcome HTH

8. True; each has probability 1/8.

9) if A is a necessary event and B is 15% probable then what is P(AvB)?

9. 100%, because if A is certain to happen, then it is certain that at least one will happen.

9) t or f: if P validly entails Q then P(P&Q) = P(P)

9. False; of the eight possible outcomes, only two (TTT and HHH) are monotonous, while the other six are mixed, such as HHT. So, the probability of a monotonous outcome is 2/8 = 1/4, while the probability of mixed = 6/8, or 3/4. (Note that "mixed" and "monotonous" are TYPES of outcome. Each SPECIFIC outcome, such as TTT and HHT, have an equal 1/8 probability.)

Distribution of the Data

Also as the text points out, just knowing the average(s) for a set of data is not by itself very informative. It's desirable to know also how the data are distributed: are the values tightly clustered together, or are they more spread out (i.e., more dispersed)? If one were to plot the data on a graph or histogram, what shape would the graph take?

Inverse Probability

As the text points out, for any conditional probability P(A|B) there is an inverse probability, P(B|A). In some cases, P(A|B) = P(B|A). But these probabilities do not have to be equal, and very often are not. So, the inference below is invalid: 1. P(A|B) = n% 2. So, P(B|A) = n% This fallacy has several names, including "Inverse Probability Fallacy" and "Prosecutor's Fallacy," among others. We'll use the first name here.

Chapter 81: Interpreting Statistics

Chapter 81 is a very simple chapter with no vocabulary you need to remember. It merely points out that sometimes, even when the statistic in question is an accurate one, people draw conclusions from it that the statistic does not really support (which is not, of course, to say that the conclusion is wrong - only that it is not entailed by the statistic). The exercises merely invite you to formulate some alternative explanations for the statistics in question. For illustration, try exercises 1, 2, 5, 8, and 10. Sample answers are on the following slides.

Drawing the Sample

Depending on how you select your sample, you could end up with a biased sample (for example, 80% of the people in your sample might be over the age of 65 when only 22% of people in the target population are over 65; hence, your sample is heavily biased toward senior citizens, perhaps because you drew the sample somewhere old people are likely to gather). Selecting a sample by some random method (so that each member of the target population has the same chance of being included in the sample) is one way to help ensure that you avoid getting a biased sample, although even in this case there is no guarantee it won't be biased.

Precision

If I tell you that somewhere between 0% and 100% of students are athletes, this statistic has no precision at all, and so it's not informative. If I tell you that between 10% and 20% of students are athletes, then this statistic has much more precision, and it's far more informative. If I narrow it down to 12%-18%, this statistic is still more precise. How much precision do you want? Or, to put it differently, how much imprecision can you accept? That's really up to you, and your tolerance for imprecision may vary by context. The point is that, all else being equal, you're going to need a bigger sample if you want a more precise statistic. At the limit, you can achieve full precision if you sample everyone in the population. But typically that's not feasible if the population is large.

Chapter 78 Hasty Generalization

Important terms: Statistical inference Representative versus biased samples Fallacy of hasty generalization Margin of error Confidence interval Confidence level Misleading vividness Statistical dead heat (from one of the exercises)

CHAPTER 80 FRAMING EFFECTS

Important terms: Tendentious questions False alternatives Complex questions Ambiguous questions Vague questions Question order Framing effects The term "Bradley Effect" on p.221 will not be on the exam. You should be able to determine what sort of framing effect could be caused by a particular survey question.

Distribution

In a course on statistics, you would learn about the various measures of dispersion, such as variance and standard deviation, and about the "normal" (bell-shaped) distribution and the Central Limit Theorem mentioned in the shaded box. Here, we will simplify matters by requiring only that you remember (in addition to mean, median, and mode) the concept of dispersion (how clustered or spread out the data [for example, scores on an exam] are, and one simple measure of dispersion: the range (the highest value in the list minus the lowest value).

Chapter 79 Selection Effects

Key terms: selection effects; simple random sampling; file-drawer effect You can ignore the term stratified sampling the shaded box on p.217 You should be able to determine when a selection effect may be at work

Type of averages

Mean or arithmetic mean: the sum of the values in a list divided by the number of values. For the values 4, 8, 11,22,22, the mean is 67/5= 13.4 Median: the value in the middle when the values are listed in order of size, or the mean of the two values in the middle when there is an even number of values. so for the values 10,22,48,57,112, the median value is 48, since it has the same number of values above and below it. for values 12, 17,23,31, the median is 20 ( the mean of 17 and 23) mode: the value that occurs most often. if no value appears more than once, then there is no mode. a list of values could have two or more modes

Misleading vividness

Often, we draw conclusions about a whole population based on a few extreme cases. For example, if we spend two days somewhere and the weather is extremely hot on one day and mild on the other, we're more likely to generalize by saying that the weather in this place is usually very hot. This phenomenon is called misleading vividness: it's the tendency to generalize from the most extreme cases. Why do we do it? Probably because extreme cases make deep impressions and stick in one's memory longer. So, when we think about a population, our attention fixates on the extreme cases we remember best. Be on your guard against this phenomenon. It can even happen when we draw conclusions about people: are you generalizing about this person's temperament based on just one extreme fit, when the person was actually quite mild-tempered most of the times you observed? Maybe it's because the fit is what stuck in your memory.

Unconditional probability

P (A) is read "the probability of A"

CHAPTER 70: GENERAL DISJUNCTION RULE addition

P(A ∨ B) = P(A) + P(B) - P(A & B) used to compute the probability that either or both of two events will occur the probability of either a or b occurring is equal to the probability of a added to the probability of b, minus the probability that both a and b occur

joint probability

P(A&B) the probability that both A and B happen

ex)

Students selling candy bars to raise money for new band uniforms had the following results: Alice: 10 bars Dwayne: 90 bars Bob: 30 bars Edward: 60 bars Chris: 20 bars Fiona: 30 bars Question: Give the mean, median, and modal average number of candy bars sold, and the range. (Answers on the next slide.) Mean: (10 + 30 + 20 + 90 + 60 + 30) / 6 = 240/6 = 40 Median: the values must first be arranged in order: 10, 20, 30, 30, 60, 90. There are an even number of values, so the median is the mean of the two in the middle: 30 + 30 / 2 = 30. Mode: 30 is the mode, because it's the only value that appears twice. Range: 90 - 10 = 80

Confidence Interval and Margin of Error

Suppose I find that 15% of the students in my sample are athletes (see the diagram above). I then generalize to the whole student body (the population) by saying that 15% of all students are athletes. But since I didn't sample everyone, there is bound to be some uncertainty in my estimate. So, I say that my estimate is 15% with a margin of error of (say) plus or minus 3 percentage points, or +/- 3. Basically, this means that the true percentage of students who are athletes should fall somewhere in the range (called the confidence interval) of 12%-18%. For the sample size I used (whatever it was), that's as precise as I'm able to be. The +/- 3 percentage points is called the margin of error. So, the length of the margin of error and the confidence interval measures the precision of my statistic. If the confidence interval were 13%-17% (i.e. a margin of error of +/- 2), my statistic would be even more precise.

Take care in drawing your sample

The point of the chapter is that before you draw any conclusions about a population (of people or things), you should consider the possibility that you have been led to this conclusion by examining a sample that is biased in some way because not everyone or everything in the target population had the same chance of being included in your sample - i.e., you selected the sample in a way that made some people less likely to be included than others (similar to reaching into a bag to draw a marble, but drawing only from the top and not mixing them up at all, so that marbles on the bottom have no chance of being drawn). As the text notes, it takes quite a bit of skill in some contexts to select a sample in a way that truly gives everyone the same chance of being picked for the sample.

Mean, Median, and Modal Averages

There are multiple types of average. You only need to learn the three mentioned in the body of this chapter: mean, median, and mode. Definitions for each are provided in the text. The differences are important because, as the examples in the text illustrate, one type of average or another can, depending on the data, give a misleading impression about the data, and often people don't tell you what kind of average they are quoting to you.

Independent Events

To say that A and B are independent outcomes or events just means that neither of them affects the probability that the other will happen. If you toss a coin twice, the outcome of the second toss is independent of the first. Every time you toss the coin, it's as if you're tossing it for the first time, and so the probabilities don't change. If you draw cards with replacement, then each draw is independent of the others; the probabilities do not change from one draw to the next. Mathematically, if A and B are independent, then both of the following are true: P(A|B) = P(A) P(B|A) = P(B) Thus, B does not change A's probability, and A does not change B's probability.

INDEPENDENT EVENTS

When the events you're dealing with are independent, just multiply the two probabilities together. So if, for example, you're rolling a fair die twice (or two dice at the same time), and you want to know the probability of getting 5 on the first roll, and 1 on the second roll, then you just multiply 1/6 x 1/6 = 1/36. We do this because the probability of any number 1-6 coming up on any given roll of the die is 1/6. If you're drawing two cards with replacement, and you want to know the probability of getting a spade on both draws, then you multiply ¼ x ¼ = 1/16. We do this because, when the first card is replaced before drawing again, the draws are independent, and so the probability of spade (1/4, since there are four suits and an equal number of each suit) is the same both times.

Translate

a) b is seventy five percent probable P(B)= 0.75 b) b is impossible P(B)=0 or P(~B)=1 c) it is certain that either a or its complement will occur P(A v ~A) = 1 d) if b occurs then a cannot occur P(A|B) = 0 (or, P(~A|B) = 1) e) a and b cannot occur at the same time P(A & B) = 0 (note: this just says they can't happen together; it does not mean that A is impossible and that B is impossible) f) neither a nor b can occur P(A v B) = 0 (or, P(~A & ~B) = 1) g) it is as probable as not that A and B will occur P(A & B) = ½ h) A's occurring increases the probability that B will occur P(B|A) > P(B)

6) assume an urn has 10 colored and numbered marbles: -5 red marbles numbered 1-5 -2 green marbles numbered 6-7 -3 blue marbles numbered 8-10 a) if one marble is randomly drawn, what is the prob that it will have the number 7 or be green b) if one marble is randomly drawn, what is the prob that it will either be green or have an even number on it? c) if two marbles are drawn with replacement, then what is the probability of getting a blue marble on both draws? d) if two marbles drawn with replacement, what is the prob that at least one marbles will be blue? e) if two marbles drawn without replacement, what is prob that both marbles will be red? f) if one marble drawn, what is prob that it will be either red or have an odd number? g) if one marble drawn, what is prob that it will be red or have 8 on it? h) what are the odds against drawing a green marble? What are the odds against drawing the marble with 1 on it?

a) P(7 ∨ G) is what we want, and these are not mutually exclusive, since one of the green marbles in the urn has the #7 on it. Since there are ten marbles in the urn, and exactly one of them is green #7, it follows that P(7 & G) = 1/10. Since there are 2 green marbles and one marble with #7 on it, we have P(7 v G) = P(7) + P(G) - P(7 & G) = (1/10 + 2/10) - 1/10 = 2/10, or 1/5. 6b. We want P(G ∨ E). Since marble #6 is green, these outcomes are not mutually exclusive, and P(G & E) = 1/10. So, P(G ∨ E) = P(G) + P(E) - P(G & E). Since five of the ten marbles have even numbers on them, P(E) = 5/10, and we get: P(G ∨ E) = P(G) + P(E) - P(G & E) = (2/10 + 5/10) - (1/10) = 6/10, or 3/5. 6c. Getting blue on both draws means P(B1 & B2), which means this is a joint probability, and we're told that we'll replace the first marble, which means the probability will be the same on both draws (i.e. the draws are independent). Since there are three blue marbles, we get: P(B1 & B2) = P(B1) × P(B2|B1), with replacement = 3/10 × 3/10 = 9/100 6d. We've seen this type of "at least one" problem before. By the Negation Rule, P(blue at least once) = 1 - P(no blues), which is 1 - P(~B1 & ~B2), i.e. it's one minus the probability of getting a not-blue marble on draw 1, and again on draw 2, with replacement. Since there is replacement, the probability of getting not-blue is the same both times we draw, and it's 7/10, since seven of the ten marbles are not blue. So, 1 - P(~B1 & ~B2) = 1 - (7/10 × 7/10) = 51/100 [With GDR, we compute P(B1 ∨ B2) = P(B1) + P(B2) - P(B1 & B2), which is (3/10 + 3/10) - (3/10 × 3/10) = 51/100. We get the same answer, as we should.] 6e. This is a simple joint probability. The probability that both marbles will be red is P(R1 & R2), and we're told there is no replacement of the first marble. So, P(R1 & R2) = P(R1) × P(R2|R1), without replacement = 5/10 × 4/9 = 20/90 = 2/9. 6f. This is another disjunctive probability, P(R ∨ O). Since the marble could be red and also have an odd number on it, the outcomes are not mutually exclusive, and P(R & O) = 3/10, since three of the ten marbles in the urn are both red and have odd numbers on them. With GDR, we compute P(R ∨ O) = P(R) + P(O) - P(R & O) = (5/10 + 5/10) - 3/10 = 7/10 [Note: don't make the mistake of thinking that P(R & O) = 3/5. That's the conditional probability of a marble having an odd number on it, given that it's red - i.e. P(O|R). The joint probability that it's red and has an odd number on it is 3/10.) g) P(red v 8) = P(red) + P(8) - P(red&8) (5/10 + 1/10) -0 =6/10 6g. This is P(R ∨ 8). Notice that these outcomes are mutually exclusive. No marble in the urn is red and has #8 on it (the #8 marble is blue). So, P(R & 8) = 0. So, P(R v 8) = P(R) + P(8) - 0. We simply add the 5/10 probability of red to the 1/10 probability that the marble has #8 on it, and get 6/10, or 3/5. 6h. The odds against green are 8:2 = 4:1, since there are eight ways to fail to get green, and two ways to get a green marble. The odds against getting #1 are 9:1, since there is only one marble with #1 on it.

Principle of indifference applied would you use it?

a) five films are nominated for best pic, and so each has a prob of 1/5 of winning the oscar We would not use it here, because not all films are equally likely to win (they vote, rather than picking the winner randomly) b) there are 40,000 fans at a game, each fan has 1/40,000 prob chance of catching a foul ball We would not use it here, because depending on where they sit, some fans are more likely than others to catch a foul ball. c) there are 52 cards in a deck, each card has a 1/52 chance of being selected on a single random draw We do use it here, because if the draw is random then by definition each card has the same probability of being selected d) there rare 102 counties in this state and so if a famous person is from this state each country has a 1/102 chance of being this persons home county We would not use it here, because the counties have different population sizes e) each point on earth surface is land or water so a meteor striking earth has a 1/2 prob of landing in water We would not use it here, because 71% of the surface is water. f) a fair die has 6 faces, sand so each face has prob 1/6 of landing up on any given roll We would use it here; if the die is fair, then each face has the same 1/6 probability of landing up.

Statistical inference

an inference we draw about an entire population based on what we find in a sample drawn from that population it matters that sample is representative ex) 47% of people in my sample intend to vote for trump, so 47% of voting pop intends to vote for trump Statistical inference involves drawing conclusions about a whole population of things based on a sample from that population. In order to have confidence in our conclusion, we need a representative sample - i.e. a sample that represents a typical cross-section of the population. Otherwise, the sample is biased to some degree.

Simple random sample

each member of the study population has an equal probability of being selected

Question order

even when questions are not tendentious or complex, it is possible to influence things just by question order, like prefacing our main question with one or more questions that put the respondents in a certain frame of mind before we get into the main question ex) set 1: what is the administrations biggest failure? How do you think the administration is doing on scale of 1-10 set 2: what is the administrations biggest success? How do you think the administration is doing on scale of 1-10

Confidence level

if your sample technique has confidence level 95% then this means that if you want to repeat the sampling process many times then the true pop figure would fall within the samples margin of error 95% of the time. pollsters often aim for margin of error of 3 or 5 points at 95% confidence. for a fixed level of confidence you'll need a larger sample if you want to narrow your margin of error It was stated that the true population percentage should fall within my confidence interval. But how sure can I be that it really does? Think of the confidence level as measuring this. If you want to be more confident that the true figure for the population will fall within your confidence interval, then you'll need a bigger sample. Pollsters, for example, often aim for 95% confidence. They might tell you that the estimated percentage of people who will vote for a certain candidate, based on their sample, is 40%, with a +/- 5 percentage point margin of error, at 95% confidence. As the text notes, a 95% confidence level really means that if you draw 100 samples from the population, the true percentage figure will fall within the confidence interval in 95 out of 100 samples. But we can think of it crudely as a measure of how sure we are that the real figure falls within our confidence interval.

simple random sampling

in which each object in the target population has the same probability of being included in our sample

file-drawer effect

it has been suggested that research scientists attempting to find a correlation between two variables are less likely to publish their results when they fail to find any correlation the so called fire drawer effect- they put their results away in file drawer and forget about them

Vague questions

its hard to find even one clear meaning for the question ex) how do you conceptualize existence?

Margin of error

margin of error for a stat is half the length of what is called the confidence interval, often expressed as for example +/- 3 points or +/- 5 points. if you want a small margin of error you need a larger sample size

Framing effects

the effects that tendentious wording, question order, physical circumstances etc can all have on survey response are collectively called framing effects -a collective term for the various ways in which the wording of questions or choices can affect the responses people give or the choices they make 3 important points: 1) a considerable amount of careful thought must go into the design of any survey in order to avoid any framing effects 2) dishonest pollsters can easily influence the results they get by deliberately incorporating the right framing effects into their surveys 3) framing effects are not confined to surveys. in ordinary convos or debate, person can shape answers. doctors know more people will have surgery if they say 95% success rate vs 5% death rate Even if you draw a large random sample, you still have to test the people or objects in your sample in order to find out whether they have the trait you're interested in. For example, if you want to know what percentage of people in a state carry a certain gene, you have to select your sample. But even if you have gathered a large random sample of people in the state, you still have to test them for the gene, and you could flub this test, which will affect any conclusions you draw about the population. In this chapter, we are concerned with testing people for their opinions, which we do by asking them questions in surveys. The way you ask the questions can heavily influence the answers you get, and if the questions are poorly framed, then any conclusions you draw about the population based on this sample could be wrong. The chapter describes several ways in which questions can be poorly worded, with the effect that we will get misleading impressions about the opinions people actually have. Read the examples involving tendentious, complex, ambiguous, and vague questions (and false alternatives) on pp.219-220. Note also the way in which the order in which questions are asked, and the physical circumstances under which they are asked, can also influence the responses you get (pp.220-221).

Fallacy of hasty generalization

the fallacy of drawing conclusions about a pop based on a too small sample Statisticians have formulas that can be used to compute the needed sample size, once we have decided on the precision we desire and so on. We will not learn any of these formulas here, and we will not be computing any sample sizes. These are exercises for a course on statistics. What is important for our purposes is to realize that if you draw conclusions about a whole population based on an absurdly small sample, then you must realize that there is a huge risk of error associated with such an inference: i.e., there will be a high probability that your estimate is wrong. In such cases, we say that you are committing the fallacy of hasty generalization. You must use your common sense in deciding what sample size is absurdly small, but there will be some clear-cut cases (e.g., generalizing about millions of people based on a dozen in your sample). mom smoked all her life and hasn't gotten cancer, smoking can't be bad for you

Ambiguous questions

there are at least two fairly definite but different things the question could be asking ex) do you believe in evolution this is ambiguous because evolution could mean multiple things

variance

variance of the population in respect to the characteristic you're interested in. to the extent that they are all alike, you don't need to sample as many of them

selection effects

when we reason from a sample that was not authored randomly we increase the risk of selection effect selection effect- a misleading impression created by a sample that was not randomly selected When you draw conclusions about an entire population based on a sample that is biased, you run the risk of a selection effect, which is a misleading impression about the population based on a biased sample - often the result of not selecting the sample randomly. If your target population is the entire population of Illinois, but you only draw your sample from those going into hospitals around the state, then your sample will be biased toward senior citizens and sick people, because they go to the hospital more often than young and healthy people. So, not everyone in Illinois has an equal chance of getting into your sample. If you draw conclusions about the population of Illinois based on this sample (say, about the average age of IL residents), there is a good chance you'll draw a wrong conclusion, and this will be a selection effect - i.e., an effect of the non-random way you selected your sample.

Tendentious questions

worded so as clearly to favor one response over the other and a survey that uses this question is likely to overstate the prevalence of pro death penalty attitudes in our sample and the population find a way to word the question as not to prod respondents in any particular direction by making one response seem more reasonable than the other, and clearly the second question is better in this respect

if A is the outcome of getting a 1 when we roll a die, what is ~A?

~A is P(2 v 3 v 4 v 5 v 6)

12) P[(Av~A) & (A&~A)] =?

12. 0%, since the event on the right side of the central & is impossible; if one of the events can't happen, then it's impossible that both happen.

10) P(A v ~A) =?

10. 100%; for any event, it's certain that it will either happen or not happen.

If i randomly draw a single card from a deck then the outcome spade and diamond are a) mutually exclusive b)jointly exhaustive c) independent d) none

(a) You can't get both spade and diamond on one draw. But they clearly do not exhaust the possibilities, and they are not independent, since (for example) P(S) = ¼, but P(S|D) = 0.

if i roll a die then (123456) are a) mutually exclusive b)jointly exhaustive c) both d) neither

(c) These are the only outcomes, and no two of them can both happen on a single roll, so they are all mutually exclusive.

if i randomly draw a single card from a deck, then the outcome (jack) and outcome (diamond) are: a) mutually exclusive b)jointly exhaustive c) independent d) neither a or b

(c) They are not mutually exclusive because you can get a jack and a diamond on one draw: you just have to draw the jack of diamonds. They are not jointly exhaustive, because there are plenty of other possibilities, such as spade and queen. But they are independent. Why? Remember that two events are independent if neither changes the other's probability. Now consider: the probability that it's a jack is 4/52 = 1/13. The probability that it's a jack if it's a diamond = 1/13 also, since 1 of the 13 diamonds is a jack. P(D) = ¼, since one fourth of all the cards are diamonds. And P(D|J) = ¼, since one of the four jacks is a diamond. So, neither outcome changes the other's probability at all, making them independent.

1)old joke: guy brings own bomb on plane bc he's afraid that there will be a bomber and says the chance of two bombs being on a plane are so small... whys this ex of gamblers fallacy?

1. Suppose that the probability of somebody showing up on a plane with a bomb is 1/1,000,000. Then, BEFORE either person has decided to bring a bomb, the probability of two people independently showing up on the same plane with bombs is 1/m × 1/m = 1/trillion. But that's the wrong probability for this guy to think about. What he needs to consider is the conditional probability: P(other guy brings bomb | I bring a bomb). This probability is 1/m, just as it was before, since the events are independent. Consider an analogy from the text. BEFORE I start tossing a coin, the probability of getting tails every time in three tosses (T1 & T2 & T3) is ½ × ½ × ½ = 1/8. But AFTER I get two tails in a row, what is the conditional probability P(T3|T1 & T2)? I.e., what is the probability I'll get the third tail, after getting the first two? It's 1/2, not 1/8. The trials are independent. And so long as the two guys boarding the plane are independent (they're not planning anything together, and don't know each other), the same lesson applies here.

12)if A is certain and B is impossible, then what is P (A&B)

12. Zero, because if one even can't happen, then it's impossible for both to happen together.

14) if A is certain and B is the event of rolling snake eyes with 2 die, then what is P (A&B)?

14. 1/36, since you'll multiply this probability by 1 (snake eyes = 1 & 1)

you roll a die once and then draw one card. what is prob that you roll a 5 and draw a spade?

2. These outcomes are independent: what happens with the die and what happens with the cards have nothing to do with one another. So, we multiply the 1/6 probability of getting 5 on the die by the 1/4 probability of getting a spade, and get 1/24.

what is the prob of drawing either a king or a club from a standard deck on a single draw?

2. This is similar to the problem we did for illustration (diamond or 10). The outcomes are not mutually exclusive, since you could get the king of clubs. P(K & C) = 1/52, since there is one king of clubs in the deck. So we have: P(K ∨ C) = P(K) + P(C) - P(K & C) = (4/52 + 13/52) - 1/52 = 16/52 = 4/13

TRANSLATION

As we saw in the last chapter, it's often helpful before doing any calculations to translate the problem into the language of probability first. In the example we're now using, the disjunctive probability of diamond or 10 can be written: P(D ∨ 10). According to GDR, P(D ∨ 10) = P(D) + P(10) - P(D & 10). The last term usually is the one most confusing for students. Why do we subtract P(D & 10)? This is what the Venn diagram is meant to help illustrate. If you add the area of circle A to the area of circle C, you'll count area B twice, because the circles overlap. In probability, if two events can both happen (i.e., if they're not mutually exclusive), then when you add their probabilities together, you'll count their "overlap" (the case where they both happen) twice, and get the wrong figure. So, for events that are not mutually exclusive, you have to subtract the "overlap" P(A & B) from their sum.

Odds

Example: If you buy one ticket in a fair lottery with 1,000,000 tickets, then the probability of winning is 1/1,000,000. But the odds against winning are 999,999:1, because there are 999,999 outcomes in which you don't win, and one in which you do win.

in a lotto 100 tickets sold and only 1 ticket is a winner. jack and jill each buy tickets, the outcome "jack wins" and the outcome "jill wins" are (-?-) but we cannot determine whether they are (?) because (?)

First blank: mutually exclusive, since they can't both win if they don't share tickets. Second blank: jointly exhaustive. Third blank: since we don't know whether they bought all the tickets. If they bought all the tickets, one of them has to win. If they didn't, then some third player might win.

what is the prob of drawing either the ace of spades or a club from a standard deck on a single draw?

Letting As = ace of spades, and C = club, we want P(As ∨ C). Are these outcomes mutually exclusive? Yes; you can't get both on a single draw. So, P(As & C) = 0. There is one ace of spades in the deck (1/52), and 13 clubs (13/52). So, P(As ∨ C) = P(As) + P(C) - P(As & C) = (1/52 + 13/52) - 0 = 14/52 = 7/26.

Understand the concept of random selection on p.186

Random selection: when an object is being selected from a set of random objects, the selection is random if , and only if, every particular object in the set has the same probability of being selected Drawing card, fact is that each card is as likely to be drawn as the other P(A)=3/4 four things can happen and none are more likely than the other, and A happens in 3 of them

if someone claims that the odds against some event are "a thousand to one", then strictly speaking how many outcome must he believe are possible?

Strictly speaking, 1,000:1 odds are only possible if there are 1,001 outcomes that are possible. But in most cases people don't mean this literally.

if the odds against outcome A are 7:2 then what is the probability of A? If the probability of A is 3/5 what are the odds of A?

The answer to the first question is 2/9. Since the odds are 7:2, there are nine outcomes possible, and two of them are favorable. The answer to the second question is 2:3. If five outcome are possible, and there are three ways our event can happen, then there are two ways it can fail to happen. In this case, the odds favor our event.

JANE AND THE BANK

The example in the chapter involving Jane and the bank (p.193) is meant to illustrate the point that when you're multiplying two contingent probabilities together (i.e., each is greater than zero and less than one), the product always has to be smaller than either one by itself. This is simple arithmetic, because you're really taking a fraction of a fraction when you multiply them. This means that whenever A and B are independent events (strictly speaking, whenever both P(A) and P(B|A) are contingent probabilities), the joint probability P(A & B) must be smaller than either probability alone. If you think about this, it's just common sense: if you bet that Jane will joint a rights group and work at a bank, then Jane has to do both of two things for you to win that bet. If you only bet that she'll work at the bank, then she only has to do one of those two things in order for you to win. It's more likely that she'll work in a bank than it is that she'll work in a bank and do something else on top of that.

Fair Betting Odds

The other topic mentioned in this brief chapter is fair betting odds. Fair betting odds are odds that reflect the actual probabilities involved in an experiment or wager. If the probability of winning a wager is 1/36, for example (you're betting on "snake eyes" (1-1) with a roll of two dice), then the odds are 35:1 against, and this means that you should receive $35 in winnings for every dollar you bet. If, instead, someone gives you odds of only (say) 30:1, the betting odds are not fair, because they don't reflect the 1/36 probability of winning. (As the chapter points out, odds given in sports betting are not fair, but no one expects them to be. It is understood by bettors that bookmakers adjust the odds to guarantee themselves a profit after all the losers pay and all the winners get paid. So, when you see that the odds on a horse are (say) 5-3 against, you should not assume that the bookmaker estimates the probability of this horse winning at 3/8. The applications of probability in gaming are fascinating and often highly complex (the mathematical theory of probability was invented centuries ago largely as an aid to gamblers). But we will not dwell on gaming here.)

Remember the Principle of Indifference mentioned in Exercise 6

The principle of indifference which is associated with the classical theory of probability directs us to assign each outcome of an experiment 1/n probability when n outcomes are possible and each,so far as we can tell, is equally probable. But since the outcomes are not equally probable in every situation, there are cases in which we should not assign probabilities using this principle

Counterexample

The text provides a card-drawing counterexample on p.202 to prove that the form of inference in question is invalid. Another simple example is below, using a single die roll. P(you get a 5 | you get an odd number) = 1/3, but P(you get an odd number | you get a 5) = 1 These simple examples suffice to prove that the form of inference we're examining is not valid.

t or f: if the order in which the numbers are picked did not matter in our lottery, you would have a better chance of winning with 26 9 12 45 than with 5555?

True. There is only one way to pick 5, 5, 5, 5: you have to pick 5 every time. But there are multiple ways (24, it turns out*) in which to pick 26, 9, 12, 45. So, if the order makes no difference, then you should pick the mixed number. * The calculation is 4! (4 factorial) = 4 x 3 x 2 x 1 = 24. But you don't need to know how to do this calculation. The important and obvious point is that there is more than one way to draw the mixed set of numbers.

Mutually Exclusive Outcomes

two events are mutually exclusive if the occurrence of one precludes the occurrence of the other- i.e. they cannot both happen. so if a and b are mutually exclusive, the P(A&B)=0. if you draw a single card from a standard deck, then 2 of spades and 3 of diamonds are mutually exclusive outcomes Mutually exclusive outcomes simply are those outcomes that cannot happen at the same time, or in a single trial of an experiment. When you toss a coin, for example, heads and tails are mutually exclusive because you can't get both on one toss. Thus, P(H & T) = 0. The joint probability for mutually exclusive outcomes is always zero.

An urn contains 5 red marbles, 3 green, 2 blue. a) if we randomly select a marble twice without replacement, then what is the prob of getting red both times? b) if we randomly select a marble twice without replacement, what is the prob of getting blue on the first and green on the second? c) randomly select marble twice with replacement, what is prob of getting red on first and blue on second? d) randomly select marble twice with replacement, what is prob of getting blue twice? e) randomly select two marbles with replacement, what is prob of getting red at least once? use negation rule here

1 a. There is no replacement, so the draws are dependent. Getting a red marble both times means P(R1 & R2), which is P(R1) x P(R2|R1). Since the first marble is not replaced, we get the following (fractions are left unreduced so that it's easier to see where they come from): P(R1) x P(R2|R1), with no replacement = 5/10 x 4/9 = 20/90 = 2/9 The first factor is 5/10 because 5 of the 10 marbles are red. The second factor is 4/9 because we're assuming that we draw red the first time and don't put it back, reducing the number of red marbles to 4, and the number of marbles in the urn to 9. 1b. We're calculating P(B1 & G2), on the assumption that the first marble drawn is not replaced. On the first draw, P(B) = 2/10 since there are two blue marbles. Not replacing the first marble reduces the denominator to 9 for the second draw. The numerator is 3, because we're assuming that we got a blue marble the first time, which means all three green marbles are still in the urn when we draw the second time. So, P(B1) x P(G2|B1), with no replacement = 2/10 x 3/9 = 6/90, or 1/15 1c. We're calculating P(R1 & B2), with replacement. P(R1) is 5/10. Since we're replacing the marble after drawing, P(B2|R1) really is just the probability of getting a blue marble if this were your first draw; it's just like starting over. So, P(B2|R1) = 2/10, since there are two blue marbles. So, P(R1) x P(B2|R1), with replacement = 5/10 x 2/10 = 10/100, or 1/10 1d. Remember: when you see the phrase "both times," you should think of joint probability and multiplication. We want P(B1 & B2), and since there is replacement, the probability will be exactly the same on both draws. We only need to multiply 2/10 x 2/10 and get 4/100, or 1/25. 1e. You're instructed to use the Negation Rule (NR) to calculate the probability of getting a red marble at least once on two draws with replacement. Students often find this sort of problem perplexing, but it really is easy to solve with the NR if you just think about what the complementary event is in this case. The NR states that P(A) = 1 - P(~A). So, P(red at least once) is just 1 - P(we don't get red at least once). Not getting red at least once means we don't get red on draw 1, and we don't get red on draw 2, either; and this is just P(~R1 & ~R2), with replacement. So, what is P(~R) on any given draw? It's 5/10, because 5 of the ten marbles are not red. Replacement means the probability will be the same both times. So we have: P(red at least once) = 1 - P(~R1 & ~R2), with replacement = 1 - (5/10 x 5/10) = 1 - 25/100 = 75/100, or 3/4

Example

1. The unconditional probability of getting 4 when rolling one fair die is 1/6. 2. I've rolled the die many times, and I haven't gotten any 4s yet. 3. So, 4 must be "due" - i.e. the probability of getting a 4 on the next roll must now be higher than the normal 1/6 probability. This is fallacious reasoning. The die has no memory, and the world is not "keeping track" of your die rolls. The next time you roll that die, you have the same 1/6 probability of getting 4 as you did the first time.

10) t ot f: if i roll two fair die, then the prob that the sum on the two die will be 12 is the same prob that the sum will be 3

10. False (there is only one way to get a sum of 12: you have to get 6 on each die; but there are two ways to get a sum of 3: you can get 1 & 2, or 2 & 1 (imagine that the two dice are differently colored). So, P(12) = 1/36, and P(3) = 2/36, or 1/18.

11) P[(Av~A) v (A&~A)] =?

11. 100%, because the event on the left side of the central wedge is certain (see #9).

11) t or f: P(A&B) = P(B&A)

11. True (it makes no difference in what order you write the conjuncts; if you doubt this, try any of the joint probability problems we did in #1 with the terms reversed).

12) t or f: if a excludes b then b excludes a (excludes is a symmetric relation) 13) t or f: for any possible event a, P(AIA)=1 14) t or f: if A B and C are jointly exhaustive then P(A) + P(B) +P(C) =1 15) t or f) if a b and c are jointly exhaustive of the possibilities then 1- P(AvBvC)= 0 16) t or f: of a b c are jointly exhaustive the P(~A & ~B & ~C) =0

12. True (This can be proven fairly easily, but we'll omit the proof here. Don't confuse probabilistic independence with other kinds of independence, such as financial independence, which is not symmetric, since your parents might be financially independent of you, even while you are dependent on them.) 13. True 14. True (if they're the only things that can happen, then they must account for all the probability) 15. True, because if they're exhaustive, P(A v B v C) = 1. 16. True, because if they're the only things that can happen, then it is impossible that none of them happen.

13) which has highest probability a) it will snow if alaska next year b) it will snow in alaska next year and it will snow in texas next year c) it will snow in alaska next year or it will snow if texas next year d) it will snow in texas next year

13. (c), because it's a disjunctive probability, which means we'll add the probability of (a) to the probability of (d). Since (a) and (d) are contingent (neither is absolutely certain or impossible), the disjunctive probability must be larger than (a) and larger than (d) (whenever you add fractions, you get a bigger fraction). [Incidentally, the outcome with the lowest probability is (b), because it's a joint probability, which means we'll multiply P(a) and P(d), and get a smaller probability (multiplying fractions gives us a smaller fraction).]

consider the two probabilities below and indicate whether a is more probable than b, b is more probable than a or the probabilities are the same. a) the probability that two randomly selected people have the same bday b) the probability that two randomly selected people both have january 5 for their birthday

14. The answer is (a), and you can figure this out without doing much math. How many ways can two people share a birthday? Discounting leap years, there are 365 ways. How many ways can two people share Jan 5th in particular? Just one way: the first person has her birthday on 1/5, and the second person has the same birthday. So, it's far more probable that they'll share some birthday or other, than it is that they'll share Jan 5 in particular. (The probability of (a) is 1/365. The probability of (b) is 1/365 × 1/365 = 1/133,225.]

16) t or f: if A and B are contingent and logically equivalent then P(A&B)=1

16. False (to say that they are logically equivalent just means that IF one of them happens, THEN the other one must happen - i.e. P(A|B) = 1 and P(B|A) = 1. It does not mean that they are both certain to happen. For example, the statements below are equivalent: A: Everyone will vote for me in the election. B: No one will fail to vote for me in the next election. But is P(A & B) = 1? Surely not. That would mean I'm certain to win. All we can say is that IF A comes true, THEN B will come true - i.e. P(B|A) = 1.

2) from the fact that P(AIB)=n, it does not necessarily follow that P(BIA)=n. but it might never the less be true in some cases that the inverse probability is the same. provide an example to show this.

2. To show that it's possible for P(A|B) to be equal to P(B|A), we only need two independent events, since in such cases neither event will change the probability of the other. Suppose you toss a coin, and I toss a coin. Then P(I get tails | you get tails) = ½ And P(you get tails | I get tails) = ½ In other words, each of us has a ½ probability of getting tails, no matter what happens with the other person's coin, because the tosses are independent.

2) indicate whether each statement is gambler fallacy a) i was mugged yesterday so i am safe now at least, since the probability of getting mugged two days in a row must be low b) i drew a card from a deck and did not replace the card before drawing again. the card was a 3. so i am less likely to get a 3 again on the second draw c) our house was destroyed by a tornado last year. so we feel pretty safe this year since the chances of the same house being destroyed again must be small d) I've been taking plane trips for years with no incidents. so the probability of being in a plane crash is steadily climbing for me e) the last shell fired by those soldiers fell wide of the target. so probably the next shell will land closer f) im a former lottery winner. so the chances of my winning again are less than those of any other player that has never won before

2a. This is the Gambler's Fallacy. If you don't change your routine at all, you have the same probability of getting mugged today as you did yesterday (muggers don't cross you off a list). 2b. This is not fallacious. Not replacing the first card does decrease the probability of getting a three next time, from 4/52 to 3/51. 2c. This is the Gambler's Fallacy again. Assuming the atmospheric conditions don't change dramatically, the probability of the house being destroyed next year by a tornado is no different than what it was last year (Mother Nature doesn't cross you off a list, either). Remember: never confuse the probability of it happening twice BEFORE it has happened for the first time with the conditional probability of it happening AGAIN, AFTER it has already happened once. 2d. This is the Gambler's Fallacy. Assuming the plane trips are of the same distance, that the weather conditions are similar each time, etc., every time you take a plane trip it is, from the standpoint of probability, as if you're taking one for the first time. The probability of a crash is the same each time. You don't become increasingly "due" for a crash just because you've had no crashes before, any more than you become "due" for tails after getting heads many times in successive coin tosses. 2e. Since there is learning and correction involved in shelling, there is no fallacy here. 2f. This is fallacious reasoning. The probability of winning the lottery is the same every time you play.

if you roll a die twice, what is the prob of getting an odd number at least once?

3. These "at least once" problems can be solved in two ways: with the Negation Rule, or with the General Disjunction Rule. The Negation Rule method often is easier, but we'll use both for illustration. The results of each method should be the same. The Negation Rule states that P(A) = 1 - P(~A). We want to know the probability of getting an odd number at least once when rolling a fair die twice. Half the values on a die (1, 3, 5) are odd, so on each roll the probability of an odd number is 1/2. We begin this way: P(we get odd at least once) = 1 - P(we don't get odd at least once) We write it this way because not getting odd at least once is the complement (basically, the opposite) of getting odd at least once. Continued → P(WE GET ODD AT LEAST ONCE) = 1 - P(WE DON'T) Now, what does "we don't get odd at least once" mean? It means you get "not odd" (in other words, you get an even number) on BOTH rolls. We can write this as P(E1 & E2) in other words, even on roll #1, and even again on roll #2 So, P(odd at least once) = 1 - P(E1 & E2) And what is P(E1 & E2)? Since the rolls are independent, the probability is ½ × ½ = ¼. So, P(odd at least once) = 1 - (1/2 × 1/2) = 3/4. Continued → NOW WITH GDR Now we'll solve the same problem with the General Disjunction Rule. The probability of getting an odd number at least once is the probability of getting odd either on the first roll (O1), or on the second roll (O2), or on both rolls (O1 & O2). And since you could get an odd number both times, these outcomes are not mutually exclusive. So, we have: P(odd at least once) = P(O1 ∨ O2) P(O1 ∨ O2) = P(O1) + P(O2) - P(O1 & O2) = (1/2 + 1/2) - (1/2 × 1/2) = 1 - (1/4) = 3/4 [Our answer with GDR is the same as our answer with the Negation Rule, as it should be. Remember: we subtract (1/2 × 1/2), the probability of getting an odd number on both rolls, from the sum because the outcomes are not mutually exclusive: you can get odd both times. Since the rolls are independent, P(O1 & O2) = ½ × ½.]

3) suppose the prob of your dna matching dna found at the scene of a crime is1/500000 if the dna is not yours. can we infer from this info that the dna is yours?

3. This is similar to an example we used. The answer, of course, is no. How probable it is that it was your DNA at the crime scence will depend on other information in addition to the DNA. If you have an identical twin, for example, who shares all your DNA, then the probability that the DNA they found is yours, given that your DNA matched the DNA at the scene, drops significantly. And even if you don't have a twin, other information could render it virtually impossible that the DNA came from you (e.g., the information that you were in another state when the crime was committed, or that someone has a sample of your DNA they could plant at the scene, or whatever; we have to reason based on the total evidence). Notice also that we'd have to consider the probability that the DNA test itself was performed incorrectly. All we know for sure is that the test indicated your DNA was a match.

which is more probable: a) the chicago bears will win the next Super Bowl b) the chicago bears will win the next Super Bowl and it will snow in chicago at least once next year

6. (a) This is just like Jane and the bank. Both probabilities are contingent, and so (b)'s probability is P(Bears win) x P(Cubs win), and the product must be a smaller fraction than P(Bears win). (Note: These events probably are independent, but it would not really matter if they were not, so long as neither one entails the other. P(Cubs win | Bears win) is definitely less than 1 and greater than zero (one team's victory might inspire the other team, but it won't guarantee that the other team will win), so we're still going to multiply a fraction by a fraction, and we have to get a smaller fraction.)

which are independent? a) successive draws from a deck where each card is replaced after and the deck is reshuffled? b)performance of stock marker today and tomorrow c) when you get when you flip a coin 100 times and what you get on flip 101 d) one of your parents is a genius and you are a genius

7. (a) yes; (b) no; (c) yes; (d) no, assuming there is at least some genetic component to intelligence.

9) suppose i walk into a room just as someone rolls double 6 with two fair die. i now engage in the reasoning bellow and in doing so i commit a fallacy sometimes called the inverse gamblers fallacy. explain why the reasoning is fallacious and why its called the inverse gamblers fallacy 1) the probability of rolling double 6 is greater if you roll the die several times than if you only roll the die once 2) so it is likely that this person had already rolled the dice several times before rolling double 6 as i just entered the room

9. The "Inverse Gambler's Fallacy" has the following form: 1. The unconditional probability of event E is low. 2. E happened in this trial I just observed. 3. So, there must have been lots of previous trials before this one. The idea is that if there were many previous trials (many previous rolls of the dice in this case, with no double-sixes in any of them), then it would be more likely that E (double six) would happen in the trial I just observed. But when the trials are independent, this is fallacious reasoning: even if the dice had been rolled many times before I walked in, the probability of double six on the roll I saw just now would still be just 1/36. This is called the "inverse" Gambler's Fallacy because instead of making a prediction about the NEXT trial, we're inferring the existence of EARLIER trials. But it's just as fallacious either way. To use another analogy to illustrate the point in #9, suppose that someone just won the lottery. Someone says "Oh, you must have played many times before without winning, so you were due to win." They're committing the inverse Gambler's Fallacy. It would make no difference if this person had played the lottery before without winning. This person's chances of winning the lottery would still be exactly the same as everyone else's this time; each time the lottery is played, it's as if everyone is playing for the first time as far as the probabilities are concerned. So, inferring the existence of earlier "trials" for this person is fallacious reasoning because those earlier trials would not change the probability of this person winning this time around.

DIAMOND OR TEN

In our example, we want P(D ∨ 10). Notice that these events are not mutually exclusive, because you can get both outcomes on a single draw, if you get the ten of diamonds. In the entire deck of 52 cards, there is only one ten of diamonds, and so P(D & 10) = 1/52. There are 13 diamonds in the deck (there are 13 of each suit), and there are four tens (there are four of each card). So, using GDR, we get P(D ∨ 10) = P(D) + P(10) - P(D & 10) = (13/52 + 4/52) - 1/52 = 16/52 = 4/13.

DISJUNCTIONS

GDR is used to compute disjunctive probabilities. So, you should use this formula when a problem asks you to compute the probability that this or that will happen. For example, what is the probability, on one random draw from a standard deck of cards, that the card you draw will be a diamond, or 10? In these types of problems, you must add probabilities, instead of multiplying them as we do when computing joint probabilities. This should make sense: the probability that the card will be either a diamond or 10 should be greater than the probability of either of these outcomes alone. (Think of it as a wager: you win if the card is either diamond or 10; then your chances of winning are higher than they would be if the card had to be a diamond.)

As the text mentions, the "trials" (i.e. any time when you perform the "experiment," which could be anything from tossing a coin to getting on a plane to investing in the stock market) are not always independent. There are contexts where the probabilities do change. The simplest example is drawing objects from a set without replacement: the next time you draw, there are fewer objects left to pick from, and so the probabilities are different next time. In all cases that involve at least some element of learning and memory, as well, the trials will not be fully independent. If you're playing a game that involves skill, then the learning process might increase your chances of winning the next time you play. But if it's purely random, such as lotteries or roulette wheels, then the trials will be independent.

IN SOME CASES WHEN THE TRAILS ARE NOT INDEPENDENT THIS IS NOT FALLACIOUS like if we are drawing a deck of cards without replacement... then probabilities are clearly affected and gamblers fallacy doesn't apply

Medical Testing and Inverse Probability

In addition to its appearance in legal contexts, the Inverse Probability Fallacy shows up in the context of medical testing, as well, and its effects here can also be very harmful. The text example involves a scenario in which we're told that if we have a disease (D), then a test is 99% likely to return a positive result. Someone takes the test and gets a positive (+) result, and he performs the following erroneous inference that leads him to panic: P(+|D) = 99% (the probability of getting a + result if i have the disease is 99%) So, P(D|+) = 99% (so given the positive result, it is 99% probable i have this disease) - i.e., he thinks it's 99% probable that he has the disease, given his positive result. But this does not follow. Exercise 1 asks you to explain why.

Cards

Incidentally, since some students don't play cards, or don't play with the same cards standardly used in the U.S., and since many of our examples will involve drawing cards, the composition of a "standard deck" is below. 52 cards in total 4 "suits" (club, spade, diamond, heart), 13 of each Clubs and spades are black, while diamonds and hearts are red "Face cards" are jacks, queens, and kings; there are four of each. The other cards are aces and cards numbered 1-10; there are four of each.

MUTUALLY EXCLUSIVE OUTCOMES / events P(A&B)=0

Now suppose we want to calculate the probability of getting either a king or a queen on one draw from a standard deck. On one draw, you can't get both; these are mutually exclusive outcomes. And we know that when two outcomes are mutually exclusive, their joint probability is zero. So, P(K & Q) = 0. Since there are four kings and four queens, GDR gives us: P(K ∨ Q) = P(K) + P(Q) - P(K & Q) = (4/52 + 4/52) - 0 = 8/52 = 2/13. (Of course, we could just ignore P(K & Q), since it's zero. We included it above simply for illustration.)

conditional probability

P (BIA) is read "The probability of B, given A. The probability that B will happen, given that A happens

disjunctive probability

P(AvB) The probability of A or B This is the probability that at least one of these events- or both- occurs. The v symbol is called a wedge, and it is used to write disjunctive (either or) statements

Chapter 72: Inverse Probability Fallacy

P(A|B) Inverse: P(B|A) for any conditional probability like P(AIB) there is an inverse probability P(BIA)

give an example to show that P(AIB) and P(A&B) are two different things

Suppose we draw a card from a standard deck. Let K = king, and C = club. Then P(K & C) = 1/52 (there is one king of clubs in the deck) But P(K|C) = 1/13 (there are 13 clubs, and one is a king) So, do not confuse a joint probability with a conditional probability.

Counting Rule mentioned in Exercise 7.

The Fundamental Counting Rule states that if you're going to perform several experiments (we'll use three for illustration), and if there are n1 things that can happen in the first experiment, n2 things that can happen in the second, and n3 things that can happen in the third, then the total number of possible outcomes is n1 x n2 x n3. So, to answer the question in the text, the total number of possibilities is 4 x 3 x 2 = 24.

if we are to perform 3 experiments and if in the first experiment, 4 outcomes are possible, in the second 3 are possible, and in the third 2 are possible, then how many outcomes are possible in all 3 experiments combined.

The Fundamental Counting Rule states that if you're going to perform several experiments (we'll use three for illustration), and if there are n1 things that can happen in the first experiment, n2 things that can happen in the second, and n3 things that can happen in the third, then the total number of possible outcomes is n1 x n2 x n3. So, to answer the question in the text, the total number of possibilities is 4 x 3 x 2 = 24. Since 6 things can happen on each die, the total number of possible outcomes is 6 x 6 = 36. 1/36, since there is only one way double six can happen (you gave to get 6 on each die) a) if you roll 2 fair dice, or one die twice, how many different outcomes are possible- where (3,2) and (2,3) are treated as different outcomes. Since 6 things can happen on each die, the total number of possible outcomes is 6 x 6 = 36. b) in the experiment a, what is the probability of a double 6 1/36, since there is only one way double six can happen (you gave to get 6 on each die) c) if you toss a fair coin 3 times, how many different outcomes are possible- where TT and TTH are treated as different outcomes? Since two things can happen each time, the answer is 2 x 2 x 2 = 8. d) in experiment c, are HHT and TTT equally probable? Yes; each has 1/8 probability. e) in the experiment c, is a mixed result (at least one head and one tail) as probable as a monotonous result (TTT and HHH)? No. There are two monotonous results (TTT, HHH), and six mixed results. So, the probability of a monotonous result is 2/8 = ¼, and the probability of a mixed result is 6/8 = ¾. (Of course, each specific outcome has an equal 1/8 probability. "Monotonous" and "mixed" are types of outcome.) f) when rolling two fair dice, is getting a sum of 12 as probable as getting a sum of 3? f) No. There is only one way to get a sum of 12: you have to get double six, which is 1/36 probable. But you can get a sum of three by getting 1 on the first die and 2 on the second, or vice-versa. So, the latter probability is 2/36 = 1/18.

CHAPTER 69: THE GENERAL CONJUNCTION RULE Multiplication GCR: the general formula for computing the probability that two events a and b will both occur

The General Conjunction Rule (it goes by other names as well, such as the multiplication rule) is used to compute joint probabilities: i.e., the probability that two or more events will all happen. P(A & B) = P(A) x P(B|A). or if a and b are independent : P(A & B) = P(A) x P(B). the probability that both a and b will occur is equal to the probability of a multiplied by the conditional probability of b given a We are calculating the probability that A will happen and that B will also happen, given that A has already happened. Now, if A and B are independent (for example, two independent tosses of a coin, or two draws from a deck where the first card was replaced), the probability of B is unaffected by whether A happens or not. So, in these cases, P(B|A) = P(B).

Two things that sometimes confuse students: The Negation Rule The difference between the meaning of random selection, and the ways to achieve randomness.

The Negation Rule simply tells us that the probability of an event happening is always equal to 1 minus the probability that the event will not happen. The probability that it will snow tomorrow P(S) is just 1 - P(~S). Why? Because for any event, it is absolutely certain (probability = 1) that it, or it's complement, will occur. Either it will snow tomorrow, or it won't. Either you'll graduate, or you won't. And so on. Hence, for any event A, P(A) + P(~A) = 1; in other words, since it is necessary that one of these things happen, their respective probabilities must add up to 100%, or 1. From this it follows that P(A) = 1 - P(~A). So, if it is 30% probable that it won't snow tomorrow, it must be 70% probable that it will snow P(S) = 1 - P(~S). The Negation Rule turns out to be very useful for solving certain problems in probability, as we'll see in a subsequent chapter.

Prosecutors and Inverse Probability

The fallacy is sometimes called the "Prosecutor's Fallacy" because prosecutors are apt to commit the fallacy in criminal trials. It might be, for example, that the following is true: if you were not at the crime scene and the DNA they found is not yours, then the probability that the DNA found at the scene would still match yours is 1/5,000,000. So, if your DNA is a match for the DNA at the scene (a prosecutor might argue), the probability that it's not yours is only 1/5,000,000 - a number that would sound very damning to a jury. The inference is 1. P(DNA match | it's not your DNA) = 1/5,000,000 2. So, P(it's not your DNA | DNA match) = 1/5,000,000 But this inference is fallacious. It might be very probable, relative to all the evidence, that the DNA at the scene was not yours.

Chapter 71: The Gambler's Fallacy specifically in independent events

The fallacy we commit when we let the outcomes of previous trials for some experiment influence our estimation of the probabilities in future trails when in fact the trials are independent, like coin tosses (thinking a coin is due to land on heads since it landed on tails 4 last times... this is gamblers fallacy.) The Gambler's Fallacy is the fallacy you commit when you think the probabilities in future trials of some experiment are changing when in fact they're always the same, because the trials are independent. The probability before one has started tossing the coin of getting 3 tails in 3 tosses (1/8) The conditional probability of getting tails on the third toss given that one has already gotten tails on the first 2 tosses is STILL (1/2)

Overview

This chapter introduces the symbols we will use in subsequent chapters in this unit. There is a brief discussion about the various interpretations of probability (i.e., theories about what probability statements mean), but this is not test material. You should, however, learn The names of the various types of probability (unconditional, conditional, joint, disjunctive) on p.185 The terms "contingent" and "complement" on p.185 familiarize yourself with the basic rules of probability on p.185 Understand the concept of random selection on p.186 Remember the Principle of Indifference mentioned in Exercise 6, and the Counting Rule mentioned in Exercise 7.

Chapter 67: Odds

This chapter's main point is that we don't write odds and probabilities in the same way, although there are simple mathematical relations between them. Strictly speaking, odds are calculated in the manner described in the shaded box on p.188. But for our purposes, it is enough to remember that we write the odds against an event A by writing the number of ways we can fail to get A on the left of the colon, and the number of ways to get A on the right side. Odds against A are given by the ratio P(~A)/P(A)

Without actually computing the probability, explain how we could use the negation rule to compute the probability of getting tails at least once when we toss a fair coin 3 times

This is a case where the Negation Rule makes it easy to calculate a probability. Whenever you see a problem with "at least once," you should consider using this rule. We want to know the probability that we'll get tails at least once if we toss a fair coin three times. According to the Negation Rule: P(tails at least once) = 1 - P(we don't get tails at least once) Now, P(we don't get tails at least once) is really just the probability that we get heads on every toss. So, P(tails at least once) = 1 - P(heads on all three tosses) = 1 - P(H1 & H2 & H3) This is how the probability would be calculated. We only need to calculate the joint probability inside the parentheses, and subtract the product from one. We'll learn how to do this in a later chapter.

1) explain why the inference about disease d in last example is fallacious. what key info is needed b4 we can draw any reliable conclusion about the prob that you have the disease, given your positive result? (your answer should refer to the concept of a false positive and to the concept of a base rate)

What is most important is that you understand that the inference is fallacious, and that it is an instance of the Inverse Probability Fallacy. But it can be interesting to examine the reasons why it is fallacious. In the problem, you are given only one piece of information: the "true positive" rate for the test: i.e., the probability that the test will return a positive result if you really do have the disease. The problem is that, before you can know the probability that you really have the disease, you need two other pieces of information: 1) The "false positive" rate for this test: i.e., the probability that the test will return a positive result even if you don't have the disease. 2) The "base rate" for the disease, which basically refers to how common or rare the disease is among people like you. To give a simple example of what we mean, consider a pregnancy test that is so defective it returns a positive result for EVERYONE, whether the person is pregnant or not. Then the "true positive" rate for this test is 100%: if you are pregnant, you are 100% likely to get a positive result. The problem is that the "false positive" rate also is 100%: you are 100% likely to get a positive result even if you're not pregnant. So, when you get a positive result, it really doesn't prove anything. That's why you need to know the false positive rate before you draw any conclusions from your positive test result. The relevance of the base rate is discussed on the next slide. Suppose a man, Frank, takes our defective pregnancy test. He'll get a positive result: everyone does. Is it now 100% probable that Frank is pregnant? Hardly. The "base rate" for this condition (pregnancy) is 0% among males - the group into which Frank falls. Even if the false positive rate for this test were (say) only 1%, it still would be 0% probable that Frank is pregnant. That's why, when interpreting a positive result, you have to consider how common or rare the condition is among the population into which this person falls (the base rate for pregnancy is (obviously) higher for females, higher still for sexually active females of reproductive age, and even higher for sexually active females of reproductive age who do not use any contraception). Look at it this way. If a certain disease or condition is extremely rare, then even if you get a positive result, it might actually be more probable that the result is a false positive than it is that the result is correct. The rarity of the condition will vary by group: some are "at-risk" groups in which the condition is relatively more common. For example, people living in certain parts of Africa are at-risk for malaria; people living in Alaska are not. You have to take this fact into account when you figure out whether a positive test result is a true positive. (If you take a more advanced course on probability and statistics, you'll learn to use a formula called Bayes's Theorem, which is a straightforward consequence of the axioms of probability, and it enables you to calculate the probability of having the disease when you know the true positive rate, the false positive rate, and the base rate.)

Random Selection

When it comes to random selection, students sometimes confuse methods of achieving randomness with the definition of randomness. By definition, a draw (say, a card draw) is random if every object in the set of cards from which you're drawing has the same probability of being drawn. So, a card draw is random if every card has the same 1/52 chance of being drawn. How do you ensure that each card will have the same chance to be drawn? This is a question of method, and there might be many ways to do it, such as by drawing blindfolded, or hurling the deck into the air and just picking up the first card that lands on the floor near you. But don't confuse a method for achieving randomness with the meaning of "random." There is one meaning (equal probability for each object in the set), and lots of ways to achieve a random draw.

DEPENDENT EVENTS

When you're drawing things from a set without replacement (i.e., you don't put the first drawn object back before drawing again), the trials are no longer independent: you'll at least have one less object to pick from next time. In such cases, you must take into account the non-replacement when you find P(B|A). Consider again the case of drawing two spades on two random draws, now without replacement. On the first draw, the probability is 13/52 = 1/4. But on the assumption that you get a spade the first time and don't put it back, the probability of getting a spade next time is 12/51, because there is one less spade in the deck now, and one less card overall. So, we multiply 13/52 x 12/51 and we get 156/2,652, or around 6%. If it helps, write the problem out as P(S1) x P(S2|S1), where S1 means getting a spade on draw #1. Record above P(S2|S1) the words "no replacement," so that you remember to change the probability for draw #2.

READING THE PROBLEM

Whenever you see the word "and" or "both" (as, for example, in "what is the probability of getting this outcome in both trials?" or "on both draws?"), think joint probability, and think of multiplying probabilities together, checking to see whether the trials or outcomes are independent (when it involves drawing cards or marbles, whether the first drawn object is replaced will tell you whether the draws are independent; if the first object is replaced, the draws are independent).

IMPORTANT

You should know how to calculate probabilities with the General Conjunction Rule, and you should be able to answer conceptual questions like those in the exercises. The material in the shaded box about the Powerball Lottery is not test material. (It is possible to calculate the lottery probability in other ways, incidentally.)

Jointly Exhaustive Outcomes exhaustiveness

a set of outcomes are jointly exhaustive of the possibilities if no other outcomes are possible. there are 52 cards, and so on any single draw these 52 cards exhaust the possible outcomes of the draw, just as heads and tails exhaust all possible outcomes in a coin toss note that if a and be are jointly exhaustive then both of following are true: P(AvB)=1 P(A)+P(B)=1 if a and b are the only two things that can possibly happen, then it is certain at least one or both must happen if they are only two possibilities then together they must account for all the probability *two events A and B can be mutually exclusive without being jointly exhaustive if A and B are both mutually exclusive and jointly exhaustive then P(A&B)=0 and P(AvB)=1 -both cannot happen but it is certain one will A set of outcomes are jointly exhaustive if they represent every possibility for this experiment. For example, when drawing a card from a deck, {jack of clubs, ace of spades} are not jointly exhaustive, because there are 50 other cards you could get. Only a set that lists every one of the 52 cards would be jointly exhaustive. When you toss a coin, for all intents and purposes heads and tails are exhaustive of the possibilities. Whenever A and B exhaust the possible outcomes of an experiment, P(A) + P(B) = 100%, or 1. Why? Because they are the only things that can happen. Since nothing else is possible, A and B have to account for all 100% of the probability. And since, when A and B are jointly exhaustive, they are the only things that can happen, it is certain that one of them will happen, which means that P(A v B) = 100% as well.

state the odds against each outcome below a) getting TT on two tosses of a coin b) getting snake eyes (1&1) when rolling two fair dice c) getting the ace of spades on one draw from a standard deck d)getting a spade or diamond on one draw from a deck of cards e) winning a fair 100 ticket sweepstakes in which you bought three tickets

a) When the coin is tossed twice, with H or T possible on each toss, the counting rule dictates that 2 x 2 = 4 outcomes are possible (TT, HH, TH, HT). Each outcome is equally probable, so the odds against TT are 3:1. b) When the dice are rolled, 6 x 6 = 36 outcomes are possible. Snake eyes (1 & 1) are one such outcome, and there is only one way it happens (getting 1 on both dice); so the odds against are 35:1. c) 51:1, since there is one ace of spades out of the 52 cards. d) 26:26 = 1:1 (even odds), since half the cards in the deck are either spade or diamond. e) 97:3

imagine a lotto in which you must pick 4 numbers from 1-50 and you must pick them in the right order to win. the numbers are put on balls and the balls are drawn from a mixer.

a) how many sequences of numbers are possible? Since the balls are replaced each time, there are 50 possible outcomes each time we draw. With four draws, and using the counting rule, this means there are 50 x 50 x 50 x 50 = 6,250,000 sequences of numbers that are possible. b) is 1234 any less likely than 26 9 12 45? No. Each has 1/6.25 million probability. c) is 5555 any less likely than 26 9 12 45 No. Each has 1/6.25 million probability. d) suppose someone claims that picking a mixed number like 26 9 12 45 is better bet than monotonous number like 5555 for the reason below. evaluate the reasoning 1)there are only 50 monotonous sequences and over 6 mill mixed 2)if there are only 50 monotonous and over 6 mil mixed then it is probable that the winning sequence will be a mixed number 3) so you're more likely to win lotto by picking mixed sequence The inference is invalid. It's true that the winning sequence probably will be a mixed sequence. But no PARTICULAR sequence of numbers is any more probable than another, and you have to pick a particular sequence. So, you're no more likely to win by picking a mixed sequence.