For more information, please see full course syllabus of Probability

For more information, please see full course syllabus of Probability

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### Bayes' Rule

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- When to Use Bayes' Rule
- Bayes' Rule for Two Choices
- Bayes' Rule for Multiple Choices
- Example I: What is the Chance that She is Diabetic?
- Example II: What is the chance that It Belongs to a Woman?
- Example III: What is the Probability that She is a Democrat?
- Example IV: What is the chance that the Fruit is an Apple?
- Example V: What is the Probability that the Oldest Child is a Girl?

- Intro 0:00
- When to Use Bayes' Rule 0:08
- When to Use Bayes' Rule: Disjoint Union of Events
- Bayes' Rule for Two Choices 2:50
- Bayes' Rule for Two Choices
- Bayes' Rule for Multiple Choices 5:03
- Bayes' Rule for Multiple Choices
- Example I: What is the Chance that She is Diabetic? 6:55
- Example I: Setting up the Events
- Example I: Solution
- Example II: What is the chance that It Belongs to a Woman? 19:28
- Example II: Setting up the Events
- Example II: Solution
- Example III: What is the Probability that She is a Democrat? 27:31
- Example III: Setting up the Events
- Example III: Solution
- Example IV: What is the chance that the Fruit is an Apple? 39:11
- Example IV: Setting up the Events
- Example IV: Solution
- Example V: What is the Probability that the Oldest Child is a Girl? 51:16
- Example V: Setting up the Events
- Example V: Solution

### Introduction to Probability Online Course

### Transcription: Bayes' Rule

*Hi, this is the probability lectures here on www.educator.com.*0000

*My name is Will Murray and today, we are going to talk about Bayes’ rule.*0004

*Bayes’ rule is one of the most interesting rules in probability but it is also the easiest to make mistakes with.*0009

*You get these very counterintuitive results with base rule*0016

*and you can get some very tricky problems that can really mess with your head.*0021

*It is worth being very careful when you are dealing with Bayes’ rule.*0027

*And it is also worth following the formula very carefully.*0031

*If you follow the formula, you would not go wrong but you can also get some very strange stuff, if you are not careful.*0035

*With that said, let me show you when you use Bayes’ rule.*0042

*What kind of problem you apply it for.*0047

*The idea is that your sample space, the set of all possible outcomes must be a disjoint union of events.*0050

*Let me show you what that would look like.*0057

*All the thing possible things that can happen have to be divided up into disjoint unions.*0062

*You have to be covering all possibilities and they cannot overlap at all.*0073

*Something like this, where you have an event B1, B2, and so on, up to Bn.*0078

*You have a disjoint union, very common example of Bayes’ rule is when you have a group of people*0086

*and some of them are women and some of them are men.*0092

*That is a disjoint union, something like that.*0096

*It could be a common example where you are going to see Bayes’ rule coming into play.*0098

*After you have a disjoint union, you got one more event that overlaps all of them.*0104

*You got one more event and I’m going to call it A, like this.*0110

*This sort of overlaps into these different categories, this is A.*0117

*The way Bayes’ rule is phrased or kind of question that you are going to get that*0122

*for which Bayes’ rule gives you an answer is you know that A occur.*0128

*You are given that A is true and then what is the probability that you are in one of the other of these B boxes.*0134

*That is the kind of problem that you will see Bayes’ rule being applied to.*0143

*I think after you work through some examples with me, you get the hang of it and*0147

*you will get a feel for that this is the kind of problem that is asking me for Bayes’ rule.*0151

*It always says, given that something happen, what is the probability that we are in one of these boxes?*0155

*Let me give you the actual formula that you can use to solve Bayes’ rule problems and*0162

*then we will see how it plays out in some examples.*0166

*Let me start out wit Bayes’ rule for two choices and then I will move on to the general rule for N choices.*0171

*Even the two choices, it looks a little daunting.*0177

*The idea here is that we got our sample space divided into 2 disjoint events, which I will call B1 and B2.*0180

*For example, it might be a man and women, and a group of people, B1 and B2.*0191

*We got this overlapping event that kind of laps over both of them.*0198

*We want a problems, we are going to study later on, we are going to have tennis players at a college.*0202

*There are some of that everybody at the college is either a man or woman and*0208

*some of the men play tennis and some of the women play tennis.*0212

*We will have this event that overlaps into both categories.*0217

*And then the question is, if you know that that event is true.*0221

*In the problem later on, if you know that a person plays tennis,*0226

*what is the probability that you are in one of the other of the categories?*0230

*I just solved it for B1 here.*0234

*The probability of B1 given A.*0237

*You have this fairly complicated formula to calculate it.*0241

*What it does is it reverses the roles of the A and B1.*0245

*We see we got B1 given A here and then you reverse that to have A given B1 × the probability of B1.*0249

*In the bottom, we have the same thing again, probability of A given B1 × the probability of B1.*0257

*And the same thing would be too, the probability of A given B2 × the probability of B2.*0263

*That is a fairly complicated formula.*0270

*After we do some examples, I think you will get more used to it.*0273

*Bear with me, I’m going to give you the formula for N possible choices.*0276

*Instead of B1 and B2, we will have B1 through BN.*0280

*I will give you that formula and then we will jump into some examples.*0285

*But if you are already feeling overwhelmed, if you want to try an example right away,*0288

*you might want to go ahead and skip to example 1 because that is one where you have two choices.*0292

*That is where we will use this formula right away.*0298

*Let me show you Bayes’ rule for multiple choices.*0304

*It is the same general rule as Bayes’ rule for two choices but it applies to a more complicated setting here.*0306

*We have a big sample space and it is divided up as a disjoint union of several different possibilities.*0315

*We are calling the different possibilities B1, B2, and so on, up to Bn.*0326

*The idea there is that at some point, you are interested in the probability of one of these choices which I’m going to call B sub J.*0336

*What you are given is some information about one more event which will the overlap all of these choices.*0348

*There is an A that overlaps all of these choices.*0356

*You are told that A is true.*0360

*You are given that A is true, this is conditional probability here.*0363

*The probability of B sub J given A is equal to, this is the same as the Bayes’ rule for two choices in the numerator.*0367

*We switched the rules of the B sub J and A.*0375

*The probability of A given B sub J × the probability of B sub J.*0378

*And then in the denominator, we have the same kind of thing except we are adding up all those possibilities over all those events.*0384

*The probability of A given B sub I × the probability of B sub I.*0392

*You add that up over all the I's from 1 to N.*0398

*Let us check that out in the context of some examples.*0402

*The first one is just only two possibilities but I think it is already fairly challenging.*0405

*We will get the hang of it, probably out after we do several examples together.*0411

*Example 1, it is a problem having to do with a home test for diabetes.*0416

*Imagine, you buy this little kit at the pharmacy and you take it home and you test yourself for diabetes.*0422

*The setup here is that among all the people who take these tests, 20% of them actually are diabetic.*0431

*Maybe the people who take these has some reason to suspect that they might be diabetic.*0437

*20% of them actually are diabetic.*0442

*Now, these tests are not 100% accurate.*0444

*Sometimes they have false positives and sometimes they have false negatives.*0446

*The way it works is, if the person taking the test, if you are diabetic then there is a 90% chance that the test will show positive.*0451

*If you are not diabetic then there is an 80% chance that the test will show negative.*0460

*It looks like the test is sort of 90% accurate, if you diabetic.*0467

*It is 80% accurate, if you are not diabetic.*0470

*The idea here is that a woman takes the test and it shows positive.*0474

*It shows positive and then the question is what is the chance that she actually is diabetic?*0480

*It is a fairly complicated problem.*0487

*Let me try to graph out the possibilities here, I’m going to label some events.*0489

*Our disjoint events are the fact that when you take the test, you either are diabetic or not diabetic.*0494

*I'm going to call those the B1 and B2.*0505

*B1 is the event that you are diabetic, that you have the disease.*0507

*B2 is the event that you are not diabetic.*0515

*What we are given in this scenario is that a woman takes the test and it shows positive.*0524

*The test came a positive for this particular woman.*0531

*And that is what I'm going to call the event A, is that the test is positive.*0534

*Let me draw out how these events all fit together.*0545

*Because as I said, it is a little complicated.*0547

*Here is our total sample space and there are two things that can happen.*0553

*Either the woman is diabetic or not diabetic.*0562

*That gives us a disjoint union.*0565

*B1 is diabetic and B2 is that she is not diabetic.*0567

*She takes this test, we do not know which of the category she is in.*0584

*She takes the test and she shows up positive.*0592

*The test registers that she is positive.*0597

*The question is given the she shows positive on the test, what is the probability that she is actually diabetic.*0600

*In other words, we want to find the probability that she is actually diabetic.*0609

*The probability that B1 is true given that the test shows her as being positive.*0618

*We want to find the probability of B1 given A.*0624

*That is a classic Bayes’ rule scenario.*0628

*Once we phrased it in this form, it is fairly easy to apply the formula.*0631

*I will do that on the next slide.*0636

*Let me make sure that you understand the setup, before we move on to the next slide and apply the formula.*0638

*I think the setup is the trickiest part of this formula.*0644

*The idea here is we are setting up two events, diabetic and not diabetic.*0649

*Those gives us a disjoint union of all the people in the world.*0655

*Everybody in the world is either diabetic or not diabetic.*0658

*At least according to this simplified scenario.*0661

*We have this event that kind of overlaps both of them.*0666

*The test is positive, we are given that the woman has taken the test and the test is registered positive.*0668

*because the test has registered positive, she could still be diabetic or not diabetic.*0676

*I want to find the probability that she actually is diabetic given that the test has registers positive.*0683

*we are going to calculate that using Bayes’ rule on the next slide.*0689

*Let me expand this formula on this side.*0693

*We want to find the probability that she is diabetic given that the test has registered positive.*0697

*I'm just going to expand that out using Bayes’ rule for two events.*0706

*I gave you this formula a couple slides ago.*0710

*You can check out this formula, if it does not look familiar.*0715

*It is the probability, we will switch the rules of A and B1.*0718

*A given B1 × the probability of B1 divided by, the denominator is the same thing,*0722

*A given B1 × the probability of B1 + the same thing for B2, the probability of A given B2 × the probability of B2.*0731

*Let us try to fill in all of these probabilities because we can read them off from the stem of the problem.*0749

*If you remember, first of all, let me remind you of what these are.*0756

*B1 is the event that somebody is diabetic.*0763

*B2 is the event that somebody is not diabetic.*0769

*A is the event that somebody who takes the test shows up positive.*0777

*And we were told in the stem of the problem, you can go back and check,*0783

*that 20% of the people who take this test actually are diabetic.*0789

*That is the probability of B1, that is 20%.*0797

*I will just write that as 0.2.*0799

*You can fill that in 0.2.*0803

*That means 80% of them are not diabetics.*0807

*I will fill in a 0.8 there.*0809

*We are also given something about the accuracy of the test.*0812

*The test is 90% accurate, when a person is diabetic.*0818

*That means that if the person is diabetic, the probability that the test will be positive is 90%.*0824

*That tells us the probability of A given B1.*0833

*Given that the person is diabetic, the probability that the test will show up positive is 90%.*0836

*I can put that in for A given B1.*0841

*The last part is a little confusing here because we want to find the probability of A given B2.*0847

*That is the probability that the test is positive given that the person is not diabetic.*0855

*What the problem told us was that if the person is not diabetic, the test is accurate 80% of the time.*0862

*What that means is the test will show negative because that is the accurate result for a non diabetic person.*0872

*The test will show negative 80% of the time.*0879

*That means the test will show positive 20% of the time.*0883

*The probability of A given B2 is 20 percent.*0888

*That is really because the probability that the test is positive given that the subject is a not diabetic.*0893

*Given that the subject is not diabetic, that is the trickiest one there.*0908

*Remember, when we said if it is not diabetic, the test is 80% going to show negative.*0913

*That means it is 20% going to show positive.*0918

*That is filling in all the numbers, that is the hard part.*0922

*Now, it is just a matter of simplifying all these numbers down.*0925

*I’m going to write them on the next line, just to get rid of all the jargon.*0929

*0.9 × 0.20/0.9 × 0.20 + 0.8 × 0.20.*0935

*I will switch the roles there.*0946

*I see I got a 0.2 multiplied by everything and I can just cancel out the 0.2 from everything,*0947

*dividing by 0.2 from the top and bottom.*0953

*I still see that I got these decimals everywhere.*0955

*You will multiply by 10/10 and I will just get rid of some of my decimals.*0958

*That will give you 9/9 + 8.*0962

*That is pretty easy to figure out, it is 9/17.*0966

*That is our final answer, that is the probability that a woman actually has diabetes if this test shows that she is positive,*0972

*if the test shows positive for diabetes.*0984

*That is a little surprising because if you think about it 9/17 is pretty close to ½.*0986

*We said that this test seems to be accurate, 80% of the time if she is not diabetic.*0994

*It is accurate 90% of the time if she is diabetic.*1000

*And yet, when she gets a positive reading, it is only about a ½ chance that she does in fact have diabetes.*1003

*That is a little surprising and that is really why Bayes’ rule can lead you into some of these very counterintuitive results.*1011

*Essentially, that is a result of the fact that not many of the subjects in the world are diabetic.*1021

*It skews the numbers to one side but it is a little surprising because*1028

*if you think the test is 80% accurate on one side of the ledger and it is 90% accurate on another side of the ledger,*1032

*how come it turns out to be accurate about ½ the time, when it shows a positive result.*1041

*That is the way it works out and that is why you really have to trust the math in Bayes’ rule*1050

*and the intuition is sometimes completely wrong.*1055

*Let me just remind you where we got everything here.*1061

*I just wrote the formula for Bayes’ rule here for two events.*1064

*When we have a union of two events, that is the formula for Bayes’ rule.*1070

*I got that from the second slide of this lecture series.*1073

*You can just play it back and to find that formula earlier on in the video.*1078

*And then I was filling in all these probabilities.*1084

*B1 and B2 are the probabilities that someone is diabetic and not diabetic.*1088

*For this population, that is a 0.2 and 0.8.*1093

*That is why I got those numbers.*1096

*The 0.9 comes from the probability that the test will be positive given that the person is diabetic.*1099

*That came from the problem stem.*1107

*It said that if a person is diabetic then there is a 90% chance that the test will show positive.*1109

*This 0.2 is the most confusing part, it says that what is the probability of A given B2,*1115

*the probability that the test shows positive given that the person is not diabetic.*1121

*That is the probability that the test is inaccurate.*1127

*We said that the person is not diabetic, the test has an 80% chance of being accurate.*1132

*It has a 20% chance of being inaccurate.*1138

*That is where that 0.2 comes from.*1141

*Once you drop all the numbers in, it is a pretty simple matter to simplify the fractions down to 9/17,*1143

*which of course is approximately ½.*1151

*You get this surprisingly low number, when you look at all the numbers in the original problem.*1154

*That is Bayes’ rule for two outcomes there.*1161

*Let us go ahead and try another example.*1166

*In this example, we are visiting a small college and it has 220 women and 115 men.*1169

*Tennis seems to be very popular at this college.*1176

*40% of the women play tennis and 30% of the men play tennis.*1178

*The idea here is that you find an extra tennis racket.*1183

*You know it belongs to somebody who left a tennis racket lying around.*1186

*It must be one of these tennis players.*1190

*You want to find out what is the chance that this tennis racket belongs to a woman?*1193

*Let me setup some events here and we will use Bayes’ rule to calculate things out.*1199

*I'm going to say A is the event that we have a tennis player because that is the event that we are given.*1207

*We are given that we found a tennis racket.*1219

*We know somebody plays tennis.*1221

*B1 and B2 are going to be the sets of women and men respectively.*1224

*And those give us a disjoint union of all the students at this college.*1234

*Let me draw a map of all the students at this college.*1242

*You are all the students at this college and I guess there are few more women than men.*1248

*I will try to draw this somewhat to scale.*1253

*There is B1 which is women and there is B2, the set of all the men at this college.*1258

*Overlapping both of those is the set of tennis players at this college.*1272

*The woman and man of course do not overlap each other but the tennis players overlap both*1279

*because some of the women play tennis and some of the men play tennis.*1286

*Here is the set of tennis players and what we are given is that we have found the tennis rackets.*1292

*We found something belonging to a tennis player and we have to figure out what is the chance that it belongs to a woman.*1299

*We are trying to find the probability that we are looking for a woman given that the person we are looking for is a tennis player.*1306

*That is the probability of B1 given A.*1318

*We have a nice formula for that, straight from Bayes’ rule.*1321

*I'm just going to copy the formula directly from the second slide of this lecture.*1324

*Remember, that you switched the A and B1 and then you multiply that × the probability of B1.*1332

*You divide that by the same thing P of A given B1 × P of B1.*1341

*Then, you add on the same thing with P of B2.*1349

*We want to fill in all those probabilities.*1358

*All those probabilities had been given to us in the problem.*1362

*If we fill them in right, this will be fairly fast.*1365

*The probability of A given B1, that means the probability that somebody plays tennis given that they are a woman.*1368

*We find that right here, 40% of the women play tennis, that is 0.4.*1376

*What is the probability that someone is a woman?*1382

*Here, we have 220 women and 115 men, there are as 370 students total at this college and 220 of them are women.*1387

*The probability of getting a woman is 22 out of 37.*1403

*It is actually 220 out of 370 but I’m just simplifying that down.*1408

*In the denominator here, the first two terms are the same.*1412

*0.4 × 22 out of 37.*1416

*Now, the probability of A given B2 of the probability that we have a tennis player given that we have a man.*1421

*Only 30% of the men at this college play tennis, 0.3.*1428

*If you have a man, there is a 30% chance that he is a tennis player.*1436

*The probability that you have a man in the first place is 150/370.*1440

*I will just go ahead and reduce that to 15/37.*1445

*That is probably the trickiest part now.*1450

*We just have to simplify these fractions here.*1451

*I can multiply by 37/37 and I will simplify things a bit.*1457

*We get 0.4 × 22/0.4 × 22 × 22 + 0.3 × 15.*1462

*I have canceled all my denominators of 37.*1479

*And this is not one I think that simplifies in any terribly nice way.*1482

*I just went ahead and calculated it out to a decimal.*1488

*What I calculated was 0.662 is my probability that this tennis racket that I found, it belongs to a woman.*1493

*About a 66% chance that when we go to the lost and found, that will be a woman coming to claim that tennis racket.*1507

*Let me remind you the setup here.*1517

*I think the hardest part about Bayes’ rule problems is kinda setting everything up, identifying the events,*1520

*and getting yourself ready to go with that formula.*1526

*Once you get the formula, you fill in all the probabilities and then you reduce the fractions.*1530

*The setup here is to, first of all divide our world into women and men.*1536

*We got two disjoint events that cover the whole world.*1541

*We got women here and men.*1544

*And then, we have this third event, the group of tennis players.*1547

*This overlap into women and men.*1555

*There are some women tennis players and there are some men tennis players.*1557

*We set up our events here and the problem is to tell us given that we have a tennis player,*1561

*because we found a tennis racket and it must belong to a tennis player.*1568

*Given that we have a tennis player, what is the probability that it belongs to a woman?*1571

*What is the probability of B1 given that we are in A?*1576

*Once you identify it that way, it is fairly easy after that to expand it out using Bayes’ rule*1580

*because now you are just putting a formula.*1587

*That expands out into this formula, I just copy that straight off the second slide of this lecture.*1590

*Then, I filled in all the probabilities.*1598

*The probability of A given B1 means that if you are a woman, what is the probability that you play tennis?*1601

*That comes from this 40% right here.*1607

*That is where that 0.4 came from and that 0.4.*1610

*Then, probability that a man plays tennis is 0.3.*1614

*Just the overall probabilities of B1 and B2, that comes from the total number of women and men*1620

*at the university not worrying about tennis right now is 220/370 for women and 15/370 for men.*1628

*I just simplify down the calculations and I came up with the decimal which was equivalent*1638

*to about 66% chance that this tennis racket must belong to a woman.*1644

*In example 3 here, we got the state of New Alkenya.*1653

*We are looking at the voting trends in this state.*1659

*Apparently, 40% of the voters there are democrats and 50% are republicans, and the leftover 10% are all freemasons.*1662

*This new party of freemasons.*1670

*We are surveying how many people support freeway lanes for pogo sticks?*1673

*How many supports giving us a dedicated lane on the freeway for pogo sticks?*1680

*24% of the democrats are in favor of that idea.*1684

*30% of republicans think that is a good idea.*1687

*50% of the freemasons are in favor of that.*1690

*What we found is that a voter, we have done a telephone survey.*1694

*We grab a voter at random.*1698

*It turns out that this voter is strongly in favor of the freeway lanes for the pogo sticks.*1699

*The question is what is the probability that this particular voter is a democrat?*1704

*Let me setup some events here.*1710

*We are going to divide the world now into democrats and republicans, and freemasons.*1714

*Those are going to be, my B1 and B2 and B3.*1722

*B1, I will call it BD is the set of democrats.*1730

*B2 will be the set of republicans, I will call that BR, republicans.*1742

*B3 will be the set of freemasons, I will call that B sub F.*1749

*The event that overlaps into all three of those categories is the fact that at random,*1759

*likes the idea of dedicated freeway lanes for pogo sticks.*1769

*Event A would be a pogo stick supporter there.*1776

*Let me indicate how these events all fit together.*1781

*Let me draw you a map of all of these voters.*1786

*There is no overlap between democrats and republicans and freemasons.*1790

*Those really are three completely separate categories there.*1796

*I put the democrats on the left.*1803

*It is appropriate to their political persuasion.*1807

*There is B, the democrats.*1810

*I will put the republicans over here, the republicans.*1813

*And then we have the freemasons.*1820

*Overlapping all three of those is the set of people who support freeway lanes for pogo sticks.*1823

*There is the set of people who like freeway lanes for pogo sticks and some democrats like that idea and*1836

*some republicans think it is a good idea and some freemasons think it is a good idea.*1842

*What we are really trying to calculate here is the probability that a random voter is a democrat.*1847

*The probability of BD, but we are given that that person supports the freeway lanes for the pogo sticks.*1856

*Given that A is true, that is what we want to calculate.*1865

*I want to expand that out on the next page.*1872

*We are going to use our formula for Bayes’ rule but*1875

*I think probably the trickiest part of this problem has already been done on this page.*1878

*On the next page it is just expanding a formula and then doing some arithmetic.*1883

*The key steps are really here on this page.*1887

*It is identifying the three events here, democrats, republicans, and freemasons.*1890

*Those give us a disjoint union of the space of all voters.*1894

*And then, we have this one overlapping event which is that a voter could support freeway lanes for pogo sticks.*1898

*That one event which we are calling A, overlaps all three of those.*1906

*We are given that we found the voter who does support those lanes.*1912

*The question is, what is the probability that our voter is a democrat?*1916

*We are going to expand this out on the next page using the Bayes’ rule for multiple events.*1921

*Let us work that out on this page.*1929

*Let me remind you what we are trying to calculate.*1933

*We are trying to calculate the probability that a voter is a democrat given that she supports freeway lanes for pogo sticks.*1936

*I'm going to write-down the Bayes’ rule formula which was that, let me write that generically over here in the corner.*1946

*The probability of b sub J given A is the numerator, you switch them.*1954

*Probability of A given B sub J × the probability of B sub J.*1963

*I’m just copying this from that third slide back in the game.*1968

*You can see where this formula comes from.*1973

*Divided by the sum from, as I goes from 1 to N, the same things probability of A given B sub I × the probability of B sub I.*1975

*We are going to expand that out.*1997

*Here, we have three events.*1998

*We have someone being a democrat, we have someone being a republican,*2000

*and the probability of someone being a freemason.*2007

*And then of course, we have the pogo sticks overlapping all of them, that is the A.*2011

*A is the pogo stick crowd.*2016

*But according to our formula here, in the numerator, we have the probability and*2019

*we switched them A given B sub D × the probability of B sub D.*2030

*The D is like the J in our formula before, B sub D.*2038

*Our numerator is going to be quite long because we are adding up three different things.*2043

*The probability of A given B sub D × the probability of B sub D + the probability of A given B sub R ×*2047

*the probability of B sub R + the probability of A given B sub F × the probability of B sub F.*2064

*That is probably the hard work done now.*2078

*We are just going to fill in all of these probabilities.*2081

*The probability of A given B sub B that means if somebody is a democrat, what is the probability that they support the pogo sticks.*2086

*From the original stem of the problem, we were given that if somebody is a democrat*2098

*then there is a 20% chance, I will put 0.20 that they support the lanes.*2108

*The probability of being a democrat in the first place is 40%, 0.4 is the probability of B sub D.*2114

*The first set of terms on the denominator is just the same, 0.2 × 0.4.*2123

*Now, the probability that a republicans supports these freeway lanes is, let us see.*2130

*The survey said that 30% of the republican support these pogo sticks.*2138

*The probability of being a republican is 0.5, in the first place.*2144

*That is because 50% of the voters are republicans, in the first place.*2149

*The probability that a freemasons supports the lanes is 0.5.*2154

*Given that you are a freemason, there is a 50% chance that you support the lanes.*2158

*The probability that you are a freemason in the first place is 0.1 because 10% of the voters are freemasons.*2162

*It is a fairly quick process to simplify this.*2171

*In fact, I think I see I have a lot of decimals.*2177

*You know what I'm going to do is multiply top and bottom by 100 and that will get rid of all my decimals for me,*2180

*because I have decimals everywhere here.*2188

*I have 2 × 4, that is not 2.4 that is 2 × 4.*2190

*2 × 4 + 3 × 5 + 5 × 1.*2195

*That is going to simplify pretty easily now.*2204

*That gives me 8/8 + 15 + 5 which is 8/28, which is 2/7.*2205

*That is our probability that a random pogo stick supporting voter is a democrat.*2220

*That is our answer there.*2228

*Let me remind you how we got to each of the steps there.*2230

*First of all, I was recalling the Bayes’ rule formula for multiple events.*2234

*This formula is copied straight off the third slide of this lecture, if you go back and check, you will find it.*2238

*That is the Bayes’ rule formula for multiple events.*2245

*In this case, my disjoint events are being a democrat, being a republican, and being a freemason.*2248

*The one that overlaps them all is supporting these pogo sticks.*2256

*That was our A here.*2259

*The question is what is the chance that you are a democrat given that you support these pogo sticks?*2261

*According to Bayes’ rule, we can expand that out and we sort of switch the A and B sub D here × the probability of B sub D.*2267

*We have this big sum in the denominator.*2280

*A given B sub D × B sub D, A given B sub R × B sub R, and A given B sub F × B sub F.*2283

*If we plug in all the probabilities, I got all of these probabilities from the stem of the problem.*2293

*It tells me, first of all that 40% of people are democrats, 50% of people are republicans and 10% of people are freemasons.*2298

*It also told me that given that you are a democrat, there is a 20% chance that you support the lanes, the pogo sticks.*2309

*Given that you are a republican, there is a 30% chance that you support them.*2317

*Given that you are a freemason, there is a 50% chance that you support them.*2321

*And then, I just did a little arithmetic.*2326

*I did not like these decimal so I multiply top and bottom by 100 which gave me a 10 here to turn 0.21 to 2.*2328

*A 10 to spare here to turn that 0.4 to a 4.*2336

*I got 2 × 4, 2 × 4, 3 × 5, 5 × 1.*2339

*That simplify down nicely into the fraction 2/7.*2343

*Example 4 is a real classic of probability.*2354

*This is a famous paradox in probability.*2357

*It is one that a lot of people get very confused about.*2361

*It is going to be fun to work through it.*2365

*It is a Bayes’ rule problem, the idea is that we have three bags.*2368

*One bag has 2 apples, one with 2 oranges, and one with an apple and an orange.*2372

*Of course, these bags are opaque, you cannot see into the bags and you cannot tell which fruit are in which bag.*2378

*You pick one bag at random and you draw out one fruit.*2384

*You look at the fruit and it turns out to be an apple.*2388

*The question is, what is the chance that the other fruit in that same bag is also an apple?*2391

*In other words, what is the chance that you got the bag with two apples?*2398

*A lot of people think that the answer to this problem is ½ because you are obviously in the 2 apple bag or the apple orange bag.*2401

*You think that it got to be one of the other, it must be ½.*2410

*It turns out that that answer is not correct.*2413

*This problem is a lot more subtle than that.*2416

*What we want do is analyze this using Bayes’ rule.*2421

*Let me set up some events here.*2425

*My three events are going to be the three bags.*2428

*My B1 will be the apple-apple bag.*2432

*That represents the bag with 2 apples.*2436

*My B2 will be the apple-orange bag.*2439

*My B3 will be the bag with 2 oranges.*2444

*My overlapping event, the event that I'm given is true is the event A which is that we draw an apple.*2449

*We are given that we drew an apple and then the question is what is the probability that we are in the 2 apple bag?*2461

*Let me just make a diagram of how everything here fits together.*2469

*We got these three bags here.*2479

*The three bags do not overlap each other.*2484

*There is the apple-apple bag.*2487

*There is the apple-orange bag.*2490

*There is the orange-orange bag.*2493

*We have this other event that overlaps them all.*2498

*Actually, this is not quite drawn accurately because this is the event that we drew an apple.*2504

*If we drew an apple then we could not have been drawing from the orange-orange bag.*2514

*This region right here is actually does not exist.*2519

*It does not overlap into orange-orange category here.*2524

*But that is okay, the calculations that we do will take that into account.*2528

*We do not really need to worry about that.*2533

*We do not need to make a special case for that.*2535

*That will be built into the calculations later on.*2537

*What we are really trying to find out is the probability that assuming that we drew an apple,*2540

*what is the chance that the other fruit in the same bag is an apple?*2547

*That means that we must have been drawing from the bag that has 2 apples.*2551

*What we are asking that, what is the chance, let me make that a little more clear.*2559

*The probability that we are in the apple-apple bag given that we drew an apple as the first fruit.*2564

*That is what we are trying to calculate.*2573

*I'm going to expand this out using Bayes’ rule on the next slide.*2576

*Let me just remind you what these events are.*2583

*We got 2 apples, we got an apple-orange bag, we got a 2 orange bag.*2587

*Those fit together as a disjoint union to cover all the possible outcomes.*2602

*We know we are going to draw one fruit.*2607

*It is going to come from one of the bags and bags do not overlap.*2609

*There is this event that we are given that we draw an apple.*2613

*That does overlap into the different categories.*2617

*And we are given that that is true and we want to find the probability that*2620

*we are in the apple-apple bag given that we drew an apple.*2624

*Let us calculate that out.*2630

*I’m going to use the Bayes’ rule formula for multiple possibilities.*2632

*That is the formula that I gave you on the third slide of this lecture.*2638

*You can go back in the video and you can find that.*2643

*I will remind you what the formula is now.*2646

*The probability of B sub J given A is equal to, you flip the probability of A given B sub J ×*2648

*the probability of B sub J divided by the sum from I equals 1 to N of the probability of A given B sub I*2659

*× the probability of B sub I.*2681

*And that is the formula we are going to be using.*2684

*What we are trying to calculate in this particular problem is the probability that*2687

*we are in the apple-apple bag given that we drew an apple initially.*2694

*I'm going to expand that out using the Bayes’ formula for multiple events.*2700

*It is according to the formula, you switch and you have the probability of the apple given B sub AA ×*2709

*the probability of B sub AA.*2717

*The denominator is going to be a little long and messy because we have three different events we need to calculate here.*2722

*The probability of the apple given B sub AA × the probability of B sub AA +*2730

*the probability of the apple given B sub AO, that was our second event there.*2739

*× the probability b sub AO.*2746

*And one more event here is the probability of the apple given B sub OO × the probability of B sub OO.*2749

*I can fill in a lot of these probabilities.*2771

*Remember, we are going to pick one of these bags at random.*2775

*Let me fill in those probabilities first.*2778

*The probability that we are in the 2 apple bag is 1/3.*2780

*The probability here that we are in the 2 apple bag is 1/3.*2783

*The probability that we are in the apple-orange bag is 1/3.*2787

*The probability that we are in the orange-orange bag is 1/3.*2789

*We are calculating those before we know that we drew out an apple.*2792

*That is why those are each 1/3 because we do not yet have any information that we drew out an apple.*2796

*But now, let us calculate these other conditional probabilities.*2802

*If we are in the apple-apple bag, what is the probability that we would get an apple?*2806

*If we know we are in the bag with 2 apples, then the probability that we get an apple is 100% or 1.*2811

*The probability of apple given 2 apples is 1.*2818

*The probability of an apple given that we are in the apple-orange bag.*2822

*For the apple-orange bag, what is the probability of getting apple?*2828

*There are 2 fruits in there, one of them is an orange.*2832

*It is a ½.*2836

*The probability of getting an apple given that we are in the orange-orange bag is 0.*2838

*Because if you are in the orange-orange bag, there is no way you can get an apple.*2844

*When you combine all these numbers together, I think that is the hard part already done now.*2849

*You just have to do a little arithmetic.*2853

*We get 1 × 1/3 in the top.*2855

*In the bottom, we get 1 × 1/3 + ½ × 1/3 + 0.*2859

*That 0 drop right out.*2869

*You know I got a 1/3 everywhere.*2872

*If I multiply the top and bottom × 3 then I will get a 1/1 + ½ which is 1/ 3/2.*2874

*If I can fit that around, I get 2/3.*2885

*That is a little bit surprising because you remember, when we first looked at that problem,*2891

*we said if we got an apple, we know we must be in the apple-apple bag or we must be in the orange-orange bag.*2898

*You would think at that point, it is a 50-50 chance that you are in the 2 apple bag or the apple-orange bag.*2903

*But in fact, that is not true.*2912

*Essentially, if you got an apple it is telling you that that is more likely to have occurred*2914

*by pulling it out of the apple-apple bag.*2920

*Bayes’ rule kind of gets you out of a very sticky and counterintuitive situation.*2924

*Intuition might tell you that there is a 50-50 chance but the calculations actually tell you, and they are correct,*2930

*that there is a 2/3 chance that you were in the apple-apple bag.*2937

*There is a 2/3 chance of the other fruit in the bag is indeed an apple.*2941

*Let me remind you how that went.*2947

*We were calculating from the previous page that we are assuming that we have an apple and*2950

*we want to find the probability that we are in an apple-apple bag.*2957

*I use the Bayes’ rule formula for multiple events.*2963

*I got that from one of the earlier slides in the lecture.*2967

*It says that the probability of any particular event given this overlapping event is the probability,*2971

*the conditional probability we kind of switch them.*2980

*A given B sub J × the probability of B sub J and then we moved the denominator.*2982

*You add up all the events, A given B sub I × the probability of B sub I.*2988

*In this case, our events were the B sub 1 was the apple-apple bag.*2994

*The B sub 2 was the apple-orange bag.*3002

*B sub 3 was the orange-orange bag.*3007

*That is how I expanded out the denominator there.*3011

*The apple-apple, apple-orange, and orange-orange.*3015

*The numerator is the one we are interested in, that is the apple-apple.*3020

*We expand all that big formula out, then we fill in all the probabilities.*3025

*The probability of each bag without knowing that we have an apple is just 1/3.*3031

*That is where I got all of those numbers being 1/3.*3036

*And then the probabilities of getting an apple added each bag is a little trickier.*3040

*If you are in the apple-apple bag, you are guaranteed to get an apple.*3045

*That is a 1.*3049

*That is where that one come from as well.*3052

*If you are in the apple-orange bag, there is a ½ chance that you get an apple.*3054

*If you are in the orange-orange bag, there is a 0 chance that you get an apple.*3058

*That is where that 0 comes from.*3062

*It is just a matter of collecting all the fractions together and doing a little arithmetic to simplify it down to 2/3.*3064

*In our last example for Bayes’ rule here, we got a family that has 3 children and we are given that there are no twins.*3078

*We are given that there are at least 2 girls and we want to find the probability that the oldest child of those 3 children is a girl.*3086

*Again, let me setup some events here.*3096

*We are asked the probability that the oldest child is a girl.*3101

*Let me set up an event B1 is the event that the oldest child is a girl.*3104

*And that oldest child could also be a boy.*3118

*I will call that B2 is a boy.*3124

*Overlapping both of those is the event that we are given.*3129

*We are given that there are at least 2 girls.*3133

*My event A will be that there are at least 2 girls.*3136

*Let me draw a map of this experiment.*3144

*This B1 and B2 are disjoint events because that oldest child is either a boy or a girl.*3155

*There is B1 and there is B2.*3164

*But overlapping both of those is the possibility that at least 2 of them could be a girl.*3168

*That is my A.*3176

*And we are given that A is true and we want to find the probability that the oldest child is a girl.*3180

*We want to find the probability of B1 given that A is true, given that at least 2 of the children are girls.*3187

*We have a formula for that, that we have two events here.*3201

*This is really using Bayes’ rule for two events.*3203

*Let me expand out that formula.*3206

*According to the second slide of this lecture, you can just play it back and find it, if you do not remember it.*3209

*It is P of A given B1.*3215

*I will switch those around, × P of B1.*3220

*In the denominator is that same expression.*3225

*Let me copy that again.*3228

*P of A given B1 × P of B1 + the same thing with B2.*3230

*A given B2 × the probability of B2.*3240

*We have to figure out all those probabilities and drop them into the formula and calculate it out.*3248

*The probability of B1 is fairly easy.*3254

*That is the probability of the oldest child is a girl, that is ½.*3257

*We can fill that in.*3262

*The probability that the oldest child is a boy is also ½.*3263

*That is at least if you have no extra information about that.*3267

*The probability of A given B1 and A given B2 is a little trickier.*3273

*We calculate those down here.*3280

*The probability of A given B1 that means that we are given that the oldest child is a girl.*3283

*We are given that the first child is a girl.*3292

*We want to figure out the probability that at least 2 children are girls.*3299

*The probability that at least 2 children are girls, if the first child is a girl means the probability that*3304

*at least 1 of the younger 2 children is a girl.*3313

*We got 2 younger children here.*3330

*The 2 younger children could be a girl – boy, girl – girl, it could be boy – girl.*3332

*It could be boy – boy.*3342

*At least one of them being a girl, there is 3 of those situations where at least 1 of them is a girl, that is ¾.*3347

*The probability of A given B2, we need to fill that in.*3357

*That is the probability of at least 2 girls given that the oldest child is a boy.*3363

*If the oldest child is a boy, then to get at least 2 girls, you got to have both of the younger 2 being girls.*3368

*Both of younger 2 must be girls.*3378

*If you look at the same listing above, there is 1 out of 4 situations.*3395

*There gives you both of the younger 2 being girls, that is 1 out of 4.*3400

*I can fill in these numbers here.*3404

*Probability of A given B1 was ¾.*3407

*This probability is 1 out of 4.*3411

*That was also ¾, just like below.*3414

*I can just write this in terms of numbers now.*3419

*¾ × ½/ ¾ × ½ + ¼ × ½.*3423

*I see I got a ½ everywhere.*3434

*I can erase those if I multiply top and bottom by 2.*3436

*I know that I got ¼ everywhere.*3440

*If I multiply top and bottom by 4 as well, that will cancel all my 4th.*3445

*In the top, that will just leave me with a 3.*3449

*In the bottom, that will leave me with 3 + 1.*3452

*I have getting rid of all my denominators there.*3456

*I get 3 out of 4 and that is my final probability.*3459

*If I know that there at least 2 girls in the family, the probability that the oldest child is a girl is ¾.*3465

*Let me recap that problem here.*3473

*We are given that there at least 2 girls.*3476

*I will set that up as my event A.*3478

*We want to find the probability that the oldest child is a girl.*3481

*I set up one event for that, the oldest child being a girl.*3484

*I see that I called that B1, that should have been B2.*3488

*Let me change that.*3493

*I will get some good comments saying you got the wrong B1 there.*3495

*I have made that a B2.*3502

*B1 is that the oldest child is a girl.*3505

*B2 is that the oldest child is a boy.*3508

*That divides my world into two disjoint events.*3511

*The event of there are being at least 2 girls, that overlaps both of them.*3520

*We want to find the probability that the oldest child is a girl given that there are at least 2 girls.*3525

*That is P of B1 given A and then using my Bayes’ formula for two events,*3533

*this is copied off the second slide of the lecture.*3540

*You can go back and just look up this formula.*3544

*That expands into A given B1 × B1, A given B1 × B1, A given B2 × B2.*3546

*I want to fill in those probabilities and that is probably the trickiest part here.*3554

*The probability of B1 is the probability that the oldest child is a girl.*3558

*No problem, that is ½.*3562

*The probability that the oldest child is a boy, that is also ½.*3564

*A given B1, that is a little trickier.*3568

*That is what I was working out here.*3571

*That says, what is the probability of getting at least 2 girls given that the oldest child is a girl?*3573

*If the oldest child is guaranteed to be a girl, to get at least 2 girls, you have to get at least 1 girl in the next 2 children.*3581

*At least 1 of the younger 2 children must be a girl.*3592

*And then I wrote out the possibilities that these are the younger 2 children that I wrote out here, the younger 2.*3595

*The other 2 children could be girl – boy, girl – girl, boy – girl, or boy- boy.*3604

*How many of those have at least one of them being a girl?*3611

*There are 3 of those possibilities, that is 3 out of 4.*3614

*That is where this ¾ came from and that is where this ¾ came from.*3618

*A given B2, this one right here means that the probability of getting at least 2 girls given that the oldest child is a boy.*3623

*If you want 2 girls and the oldest child is a boy, then both of the younger 2 must be girls, in order to get 2 girls.*3638

*There is only one way to get that, that is right there.*3646

*That is why we have a 1 out of 4 chance.*3649

*That is where that ¼ comes from.*3652

*We have put all the fractions in.*3656

*It is just a simple matter of canceling all the denominators and multiply it by 2/2 to get rid of all the halves.*3658

*Multiplying by 4/4 to get rid of the 4th and turn into 3/3 + 1 which simplifies down to ¾.*3665

*If we know that there at least 2 girls, the probability that the oldest child is a girl is now ¾.*3673

*That wraps up the lecture in the examples on Bayes’ rule.*3681

*It is a pretty complicated rule and it is very counterintuitive.*3685

*I strongly recommend using the formula for Bayes’ rule even when it leads you to answers that might counteract your intuition.*3689

*Certainly, some of these answers seem kind of surprising to me.*3696

*The formula is guaranteed to be true.*3700

*When you set things up right, you set up your disjoint events, and you setup your overlapping event*3704

*then you can just follow the formula after that and you will get your answer.*3709

*This is the end of the lecture on Bayes’ rule.*3715

*This is part of the larger probability series here on www.educator.com.*3718

*I’m your teacher Will Murray, we will be working through some more probability lectures.*3723

*I hope you will stick around for those and thank you for joining us, bye.*3728

1 answer

Last reply by: Dr. William Murray

Thu Dec 22, 2016 11:31 AM

Post by Bharathi Viswanathan on December 20, 2016

Example 5.P(A/B1). You have mentioned BB-Boy Boy as one option. But based on problem is it an invalid option as there is only 1 boy?

3 answers

Last reply by: Dr. William Murray

Thu Aug 13, 2015 8:40 PM

Post by bo zhang on July 26, 2015

Hi,Dr.William,Thanks for the great video and it helps a lot,my teacher taught me that there are other ways to solve such conditional probability, for instance, using the tree diagram,and i have already solved example 1 with way of tree diagram, but it seems does not work when i apply it in example 2,because in tree diagram, for each step upper and lower must sum to one,and according to your problem, i cannot directly using the tree diagram to calculate the probability,and the subject is not opposite(like positive and negative when we describe a tree diagram),can you help me solve it with the way of tree diagram?

1 answer

Last reply by: Dr. William Murray

Wed Oct 8, 2014 4:53 PM

Post by Mckenzie Parent on October 7, 2014

Hi, my professor solved Bayes rule problems slightly different by using the formula P(A|B)=P(A n B)/P(B) and he takes in the concepts that P(B) is mutually exclusive. Where he does the intersection of A and B the union with A intersect A with the other variable. I wonder if it is the same methods and you derive to the same answer.

1 answer

Last reply by: Dr. William Murray

Mon Jun 23, 2014 7:17 PM

Post by Thuy Nguyen on June 21, 2014

I'm sorry, but for the Pogo Lane example, wouldn't you multiply by 10/10 in order to turn those decimals into 2, 4, etc.? In the video, you multiplied by 100/100, which would turn those decimals into 20, 40, etc.

1 answer

Last reply by: Dr. William Murray

Thu May 1, 2014 1:24 PM

Post by Michael Onizak on April 28, 2014

sorry guys *our hands not are hands

1 answer

Last reply by: Dr. William Murray

Thu May 1, 2014 1:23 PM

Post by Michael Onizak on April 28, 2014

Thanks Dr. Murray, example 4 is a great problem! After explaining this problem to a friend I have decided to post for those who are still contemplating over why it is 2/3 and not 1/2. Think of it this way. We know that the orange/orange bag is not a viable option and that we have one apple in are hand. This means out of ALL the fruit remaining to pick from (3 pieces) we have 2 apples to possibly pick and 1 orange, which we need to consider picking. This is why it is 2/3 and not 1/2.