For more information, please see full course syllabus of Probability

For more information, please see full course syllabus of Probability

### Independence

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 0:00
- Formula and Intuition 0:12
- Definition of Independence
- Intuition
- Common Misinterpretations 1:37
- Myth & Truth 1
- Myth & Truth 2
- Combining Independent Events 3:56
- Recall: Formula for Conditional Probability
- Combining Independent Events
- Example I: Independence 5:36
- Example II: Independence 14:14
- Example III: Independence 21:10
- Example IV: Independence 32:45
- Example V: Independence 41:13

### Introduction to Probability Online Course

### Transcription: Independence

*Hi, this is Will Murray for www.educator.com.*0000

*We are working through the probability lectures.*0002

*Today, we are going to talk about independence.*0005

*We have already seen the definition of independence in the introductory lectures.*0009

*Today, we are going to really study it more closely in a bunch of examples and try to get you comfortable with the idea of independence.*0013

*First of all, let us take a look at the definition.*0020

*We will talk about events A and B being independent, *0023

*if the following formula holds P of A the probability of A is the same as the probability of A given B.*0026

*Let me remind you that this is conditional probability right here.*0037

*That is the probability that A is true, if you already know that B is true.*0041

*The idea of independence is that if you evaluate the probability of A without thinking about B.*0050

*And then, if you evaluated the probability of A, already knowing that B is true, you get the same answer either way.*0060

*If somebody tells you that B is true, it gives you no new information about whether A is likely to be true.*0068

*That is quite a tricky idea, we are going to practice that with lots of examples.*0076

*It is an idea that is very commonly misinterpreted by students.*0081

*I want to address some of the common misinterpretations first and just go ahead *0085

*and confront those head on and try to dispel those myths.*0090

*Then, we will see where we can go with this formula.*0094

*The common misinterpretations that I want to mention for you, one is that people think that for events to be dependent, *0098

*there must be some sort of physical connection between the events.*0107

*If they are dependent then they must be affecting each other in some physical way.*0111

*That is something you want to get away from, if you are under that impression.*0116

*The truth of it is that, the 2 events can be physically related or unrelated and they can still be independent or not.*0122

*You really want to rely on the formula.*0130

*You do not necessarily want to try to think about one event affecting the other one.*0132

*I have seen a lot of my students think about it most terms and it definitely leads them astray.*0136

*Another one that is very common is they think that independence means the 2 events are mutually disjoint.*0144

*Let me show you what I mean by that with a Venn diagram.*0151

*If 2 events are mutually disjoint then they do not overlap at all.*0155

*There would be A right there and there would be B, that is completely separate from A.*0159

*People think all those events are completely separate.*0165

*Therefore, they are independent of each other.*0169

*That is very much the wrong interpretation to put on independence.*0171

*That is not what independence means at all.*0176

*In fact, if they are disjoint like that, they may not certainly dependent because*0179

*remember that independence means that if you know the B is true, it does not change at all what you think about A.*0185

*But in this case, let me color these events in.*0192

*If you are told that one of these events is true then you know for sure that the other event is false.*0196

*If we are in this situation and I tell you that B is true, then immediately you know the probability of A *0204

*just drop to 0 because you know that there is no overlap between the two.*0212

*That is dependent.*0216

*If I tell you information about one, you just gain a lot of information about the other one.*0218

*Mutually disjoint actually implies dependent not independent.*0222

*A very easy mistake that a lot of my probability students have fallen into.*0225

*I hope you would not, now that we have tried to sort it out here.*0231

*Let me show you something you can do with the calculations on independence.*0235

*We need the formula for conditional probability.*0239

*It says that the probability of A given B is the probability of the intersection A and B divided by the probability for B.*0242

*If A and B are independent, remember what independence means.*0248

*Independence means the probability of A given B is the same as the probability of A.*0256

*If that is true, we can plug in P of A into P of A given B here.*0261

*We just get P of A is equal to the probability of A intersect B/P of B.*0268

*If we clear the denominator there, if we plug P of B over to the other side then we get P of A × P of B is equal to P of A intersect B.*0275

*That is a very useful formula for calculating probability.*0286

*Also, this works in the other direction.*0289

*You get that if this formula holds then the 2 events are independent.*0293

*You really do have to check because it only works when they are independent.*0299

*Some people like to use this formula even though it is not true, even when the events are not independent.*0303

*That is a very big mistake in probability.*0310

*It is important to realize that this formula P of A intersect B is equal to P of A × P of B is, if and only if, the events are independent.*0312

*If they are not independent then that formula will not hold true.*0327

*Let us see how independence plays out in some examples here.*0332

*The first example is, we are going to roll 2 dice.*0336

*I want to think of 1 dice being red and 1 being blue.*0341

*We got 3 events given to us.*0345

*One is that the red dice is a 3, one is that blue dice is a 4, C is that the total is 7.*0347

*We are going to look at these combinations of events two at a time and ask whether each pair is independent.*0354

*Remember, our formula for independence that was on the first slide, just a couple slides back.*0362

*Our basic formula for independence is that the probability of A is equal to the probability of A given B, *0367

*the conditional probability of A given B.*0377

*That is our criterion to check whether 2 events are independent.*0380

*In order to evaluate this, we really need to calculate a lot of probabilities.*0385

*Let us calculate the probabilities of each one of these things.*0389

*The probability that the red dice is a 3, P of A is 1/6 because that is the chance.*0392

*If you forget that you are rolling blue dice, you just think of the roll of the red dice, *0400

*what is the probability that it comes out to be a 3?*0406

*It is 1/6.*0408

*The blue dice is 4, the probability of B is 1/6.*0410

*Similarly to how we got the first answer.*0417

*The probability of C is that the total is 7.*0420

*Remember, when you are rolling dice, let me just list all the different ways you can get each total.*0423

*You can get a 2, a 3, you can get a 4, a 5, a 6, a 7, 8, 9, 10, 11, or 12.*0431

*It is pretty useful to remember the number of different ways you can get each one of these totals.*0442

*The way you can get a 2, the only way you can get that is with 1-1.*0448

*There is one way to get that.*0451

*You can get a 3 two ways because you can get a 1-2 or 2-1.*0454

*4, there is 3 ways because you can get 1-3, 2-2, or 31.*0460

*This pattern continues, there are 4 ways to get a 5, 5 ways to get a 6, 6 ways to get a 7.*0466

*It drops off after that.*0474

*There is 5 ways to get an 8 because you can get 2-6, 3-5, 4-4, 5-3, and 6-2.*0476

*4 ways to get a 9, 3 ways to get a 10, 2 ways to get 11 because you can get 5-6 or 6-5.*0486

*There is just one way to get a 12.*0493

*If you add all those up, you get 36 which is the total number of possible outcomes when you roll 2 dice.*0495

*In particular, how many ways can you get the total being 7?*0505

*There are 6 ways to get a total of 7 because you can get 1-6, 2-5, 3-4, 4-3, 5-2, 6-1.*0508

*That probability of C there is 6/36 total outcome is 1/6.*0520

*In order to check independence, let us calculate some conditional probability.*0528

*Let us check for PMA given B, that means if the blue dice is a 4, what is the probability that the red dice is a 3?*0534

*The blue dice being a 4 certainly does not change the probability that the red dice will be a 3.*0546

*That is 1 out of 6.*0552

*We notice that, that is the same as the probability of A.*0556

*That was the criterion that you check for independence.*0564

*A and B, they are independent.*0567

*Let us check the probability of A given C because we need that to check whether A and C are independent.*0580

*The probability of A given C says we are given that the total is 7.*0590

*Now, what is the probability that the red dice is 3?*0595

*If the total is 7, remember there were 6 ways that the total can be a 7.*0599

*How many of those have the red dice being 3?*0605

*One of those ways has the red dice being 3 because one of those ways is a red dice being 3 and the blue dice being 4.*0609

*That is one of the ways of being 7.*0622

*That is the only way to have a red dice be 3 and get the total to be 7.*0624

*There are 6 ways total to get 7.*0628

*One way has the red dice being 3 and the blue dice being 4.*0630

*The probability of A given C is 1/6.*0635

*If we look back at the probability of A by itself, that is also 1/6.*0638

*We have met our criterion, A and C are independent.*0645

*Now, B and C, what is the probability that the blue dice is a 4, that was 1/6.*0660

*The probability of B given C, there are 6 ways you can get a 7 and only one of those has the blue dice being 4.*0666

*That is 1 out of 6 which is the same as the probability of B, even when we have not assumed that C was true.*0678

*That is from that 6 right there.*0686

*B and C are independent as well.*0690

*All 3 of these pairs of events are independent.*0695

*We have answered our question kind of just by doing the calculations.*0706

*Let me remind you how those all went.*0709

*Assume that I check this formula on all 3 pairs of events here.*0712

*Is the original probability equal to the new probability, the conditional probability *0715

*when you are assuming that one of the other events is true?*0722

*For the red dice being 3, without making any other assumptions that is 1/6.*0725

*There is a 1/6 chance that you are going to roll a 3 on the red dice.*0732

*Suppose you know that the blue dice is 4.*0736

*There is still a 1/6 chance that the red dice is 3.*0738

*Since, those probabilities equal each other, A given B = the probability of A, we say that they are independent.*0742

*For the probability of A given C, that is assuming that our total is 7.*0750

*What is the chance now that the red dice is 3?*0755

*Of the 6 ways to get a total of 7, that is where that is 6 is coming from, one of them is with a red 3 and a blue 4.*0757

*One out of those 6 ways gives you a red 3.*0771

*That is our probability of A given C is 1/6.*0776

*That matches the original probability of A.*0779

*Again, we have independent events.*0781

*And finally, for BGC, it is actually very similar to A and C.*0784

*If the total is 7, there are 6 ways to get that.*0789

*That 6 is being invoked right there as well.*0792

*One of them is with a blue 4.*0796

*We have 1 and 6 is the conditional probability.*0802

*1 and 6 was the original probability of B as well.*0804

*Since, those match, we do say that B and C are independent.*0808

*In the next example, we are going to return back to the same experiment.*0813

*We are still going to roll two dice, one red and one blue.*0818

*We have the same 3 events, A, B, and C except we are going to introduce a 4th event D.*0820

*We are going to calculate some independence with B.*0826

*If you are about to look at example 2, hang on to this experiment and hang onto these numbers P of A = 1/6, P of B = 1/6, P of C = 1/6.*0829

*We are going to throw in a 4th event D and we are going to see if each of these events A, B, and C are independent from D.*0840

*It will be all the same numbers and even all the same numbers.*0845

*You want to remember these numbers as well for the next example, example 2.*0850

*Let us jump right into it.*0854

*Example 2 here is a follow-up from example 1.*0856

*If you have not just watched example 1, go back and watch that first because I want to be using the same experiment.*0860

*We are rolling 2 dice and I'm going to use the same numbers that we just calculated in example 1.*0867

*In order to understand this example too, you really need to check back and understand example 1 first.*0873

*The idea is that we are rolling two dice, that is our experiment.*0881

*Think of them as a red one and a blue one and we got 4 events.*0884

*Some of these events, we have already calculated the probability of.*0888

*The red dice being 3, we calculated that probability was 1/6.*0891

*That was back in example 1.*0895

*Blue dice being 4, that was also a probability of 1/6.*0897

*Again, from example 1.*0901

*Finally, the total is 7.*0903

*That also from example 1 was probability 1/6.*0905

*What we have not seen yet is this new event D, is that the total is 8.*0909

*Let us figure out the probability of that.*0914

*Let me just remind you that back in example 1, we figured out that there are 5 ways to get an 8, if you roll two dice.*0916

*Let me just remind you of what they are.*0924

*If I list the red dice first each time, you can get a red dice 2 and a blue dice 6.*0930

*You can get a 3-5, red 3 blue 5.*0940

*You can get a 4-4, you can get a 5-3 and you can get a 6-2.*0944

*There are 5 ways you can get an 8, if you roll two dice.*0951

*And there are 36 total outcomes.*0957

* The probability of D here is 5 out of 36.*0959

* It is the first one that does not match all those 1/6.*0966

*We are being asked if A and D are independent, B and D, and C and D?*0969

*Let us calculate the necessary conditional probabilities.*0974

*I will remind you the formula for independence is that PMA is equal to, since we are going to use D, *0978

*I will give you the formula in terms of D, PMA given D.*0986

*The conditional probability there.*0989

*The probability of A given D that means we are assuming that D is true.*0993

*We are assuming that we are in one of these 5 scenarios.*0999

*The probability that the red dice is 3.*1004

*There is just one of those 5 scenarios that give me a red dice of 3.*1006

*It is 1 out of 5.*1011

*Now, that is not equal to the original probability of A which was 1/6.*1013

*That means A and D are not independent, they are dependent events.*1021

*They are not independent, they are dependent.*1031

*Let us calculate B and D, the probability of B given D.*1039

*Again, if you know that D is true, you know that you are in one of these 5 events here.*1046

*Only one of those is the blue dice of 4.*1052

*It is that one right there, the probability of 4-4, 1 out of 5.*1056

*Again, that is not equal to the original probability of B because the original probability of B was 1/6.*1060

*B and B, since those numbers did not come out the same, B and D are not independent.*1069

*Those are dependent events.*1076

*Finally, let us calculate the probability of C given D.*1084

*We are assuming that D is true.*1091

*We are assuming that the total is 8 and then we are asking what is the probability that the total is 7?*1093

*If we know the total is 8 then it is not possible for the total to be 7.*1099

*The probability of C given D is 0.*1106

*Since, the original probability of C was 1/6 and certainly not equal.*1110

*C and D are also dependent, since the probability has changed, once we assume that D was true.*1118

*This illustrates one of those misconceptions that students are some× prone to, when they are studying probability.*1133

*People think that the total being 7 and a total being 8, those could never happen again together.*1140

*Those must be independent of each other.*1146

*That is not the way it works.*1148

*Remember, independence means if you know one of them is true,*1150

*it does not change your calculations or your probabilities that the other one is true.*1152

*If you tell me that the total is 8 then I do change my probability for 7.*1159

*I change it to 0 because I know that it cannot be 7.*1164

*If you have told me that it is 8, I will change my probability quickly and change it down to 0*1169

*which means that those are dependent events.*1174

*As soon as I know one is true, I change my calculations about the other one.*1176

*Just to recap there, what we did was calculated these probabilities back in example 1.*1182

*We calculated the probabilities of A, B, and C so we can find where those numbers come from back in example 1.*1188

*We also calculated the probability of D and that was 5 out of 36 because there are 5 ways to get a total of 8.*1194

*To calculate independence, I wanted to check whether this formula holds for each combination of events here.*1202

*I calculated probability of A given D, that means there are 5 possibilities in the D world.*1211

*There are 5 ways to get an 8 and only one of them , that right there, the 3 and 5 makes A true.*1218

*It is 1 out of 5, that is not equal to the 1 out of 6 that we had before.*1225

*That is why we say those are dependent.*1230

*Similarly with B, there is one way out of those 5 to get the blue dice being a 4.*1232

*That is not equal to the probability that we had before.*1240

*B and D are dependent.*1243

*For C given D, that is even less likely because as soon as you know the D is true, *1246

*the probability of C is 0 which does not match the original probability of C.*1251

*C and D must be dependent events.*1257

*Knowing that one is true, affects your impression of the probability of the other one.*1260

*In example 3, we got a class of 10 students here and the teacher*1273

*is going to randomly choose students to present two problems at the blackboard.*1278

*We are going to call it problem 1 and problem 2.*1281

*I guess Sally and Tom are two of the students in this class and we got two events here.*1284

*A is the event that the teacher picks Sally to present problem number 1 and B *1290

*is the event that the teacher picks Tom to present problem number 2.*1295

*The question we have to answer is whether A and B are independent.*1300

*I threw this one in and it is a little bit of a trick question because there is a little more information *1305

*that we need to know for sure, before we can really answer this.*1313

*Let me break that down for you.*1317

*The answer depends on the way that the teacher is going to choose the students.*1322

*Here is the big question.*1346

*Can the teacher choose the same student to present both problems or *1348

*is the teacher committed to choosing different students to present the 2 different problems?*1353

*The answer depends on whether those teachers must choose different students.*1359

*I will show you why that really makes a difference, different students.*1367

*This is a matter of choosing with replacement or choosing without replacement.*1376

*That is a key concept in probability and it is a very subtle one.*1381

*I spent some time talking about this in one of the earlier lectures.*1385

*If you go back and look at the earlier lectures, I think the lecture is called making choices.*1389

*If you look back at that lecture on making choices, I really tried to explain*1394

*the difference between with replacement and without replacement.*1399

*And that difference is very important in this problem.*1402

*We show you how that affects it.*1407

*With replacement, suppose we are choosing with replacement.*1411

*What that means is after the teacher chooses a student to present problem number 1, *1419

*that student presents problem number 1, goes back and sits down and then the teacher looks around *1425

*and possibly chooses the same student again to present problem number 2.*1430

*That would be with replacement, meaning a student gets replaced into the pool to potentially present another problem.*1436

*If we are choosing with replacement then A and B are independent.*1445

*I will try to walk you through the intuition of that and then just to be safe, I will do the calculations as well.*1457

*A and B are independent and here is why.*1465

*It is because, if you know that the teacher picks Sally for number 1,*1471

*does that make the teacher more or less likely to pick Tom for number 2?*1476

*The answer is it does not affect it all.*1481

*If the teacher has picked Sally for number 1 and then Sally went back and sat down,*1484

*the teacher is just as likely to pick Tom for number 2, whether or not the teacher had picked Sally for number 1.*1488

*That is assuming that we are choosing with replacement.*1495

*Now, if we go without replacement, we get a different answer.*1500

*Without replacement, I'm going to try to explain this in words and I will also do the calculation *1504

*so that you can see a numerical basis for what I’m saying intuitively.*1510

*Without replacement, what that means is the students going to pick 2 different students*1516

*to present the two different problems, then A and B are claiming that they are dependent.*1521

*They are not independent, they are dependent.*1534

*Here is why.*1536

*If you know that the teacher picked Sally for number 1, that means Sally does not get to present problem number 2.*1541

*That means Tom is a little more likely to get picked for problem number 2.*1552

*Knowing that the first event is true makes the second event a little more likely to be true.*1557

*That is because we are working without replacement and we know that the same student *1564

*is not going to have to present both problems.*1569

*If Sally gets picked for number 1, then Tom is a little more likely to get picked to present number 2.*1572

*Those are dependent because as soon as you know that one is true, it changes your information about the second one.*1579

*That is the intuitive way to think about these things but let me show you how you can actually calculate these.*1586

*With replacement, let me calculate the probabilities of these events.*1594

*We will do a little calculation here.*1603

*We are going to use a formula for independence that I showed you in one of the earlier slides.*1606

*Here is that formula.*1611

*It said that the probability of A intersect B was equal to the probability of A × the probability of B.*1616

*That is true if and only if, the 2 events are independent.*1624

*We are going to calculate that formula, check whether it is true, and use that to determine whether they are independent.*1628

*Suppose, we are working with replacement, that means after the 1st student presents number 1, that person sits back down, goes back into the pool.*1635

*The teacher picks again for number 2, it could be the same student again.*1645

*With replacement, the probability of A, the probability that Sally get picked for number 1 is 1/10 *1651

*because there is 10 students there.*1659

*The probability of B is the chance of picking Tom for number 2, that is 1/10, 10 students to choose from.*1661

*The probability of A intersect B, that means that Sally gets picked for number 1 and Tom gets picked for number 2.*1670

*You got to get 1/10 2 × in a row, that is 1/100.*1677

*The probability of A intersect B is equal to the probability of A × the probability of B because 1/100 is 1/10 × 1/10.*1685

*That works and they are independent because we check that that formula holds true.*1699

*That is working with replacement.*1710

*That is assuming that a student gets replaced in the pool after possibly presenting the answer to number 1.*1712

*By contrast, suppose we are working without replacement.*1720

*Without replacement that means after you picked a student to present problem number 1, that person is not replaced in the pool.*1726

*That person is off the hook now and you have to pick somebody different for number 2.*1736

*The probability of A, what is the chance that Sally got picked for number 1 is still 1/10.*1742

*The probability of B, the chance that Tom got picked for number 2 is still 1/10.*1748

*The probability of A intersect B, Sally getting picked for number 1 and Tom getting picked for number 2.*1756

*The probability of getting Sally for number 1 is 1/10.*1766

*Once we have picked Sally for number 1, there is only 9 students left because we are not replacing Sally into the pool.*1770

*There is 1/9 chance that Tom gets picked for number 2.*1777

*It is 1/10 × 9 that is 1/90.*1782

*If we check our little formula for independence, probability of A intersect B, *1789

*we check whether that is equal to the probability of A × the probability of B.*1796

*We see that on the left we have 1/90 and on the right we have 1/10 and 1/10.*1802

*Those are not equal to each other.*1806

*Those are not equal to each other because 1/90 is not equal to 1/10 × 1/10.*1808

*Since that formula failed, what we are told is that those are dependent events.*1819

*That kind of confirms our intuition from earlier.*1831

*If Sally gets picked for number 1 then Tom is a little more likely to get picked for number 2.*1835

*Assuming that we are willing to pick Sally twice in a row.*1841

*Let me recap the problem there.*1846

*We got 2 events here.*1848

*The teacher is picking Sally to do problem number 1 and picking Tom to do problem number 2.*1850

*The question is whether they are independent really hinges on whether we are willing to pick the same student twice.*1855

*That in turn, comes down to a replacement versus no replacement issue.*1863

*Remember, we discussed that issue in the second lecture in this probability series.*1867

*Just check back earlier on www.educator.com and you will see a lecture on making choices.*1875

*And I spent some time here talking about replacement and no replacement.*1881

*Choosing with replacement means it is okay to pick the same student twice.*1886

*If the teacher picks Sally for number 1, it does not make Tom any more or less likely to get picked for number 2.*1891

*That is if we are choosing with replacement, which means those are independent events.*1898

*Of course, we can confirm that using this formula for independence.*1904

*We checked the probability of one, the probability of the other, and the probability of the intersection.*1908

*Since they do multiply together correctly, that tells us that the 2 events are independent.*1913

*If we are not using replacement, that means we are not willing to pick the same student.*1919

*We are going to pick 2 different students for the 2 problems.*1924

*Then, I claim the events are dependent because as soon as Sally gets picked for number 1 *1927

*then Tom stars to get a little nervous that he is more likely to get picked for number 2.*1933

*We got some new information there.*1937

*That is working without replacement.*1940

*We can check that using the formula is calculating the probability of A and B separately and *1941

*then the probability of A intersect B is 1/90, which is not equal to 1/10 × 1/10.*1949

*The formula fails and we get that the 2 events are dependent.*1955

*They are not independent.*1961

*In example 4, we are going back to rolling two dice.*1968

*We got 2 events going on, the probability that the total is even and the probability that the total*1971

*is greater than or equal to 11, which would mean that it has to be 11 or 12.*1977

*The question is, are they independent?*1982

*Now, it is really helpful here if you remember all the different ways that you can roll two dice.*1984

*There are 36 ways that 36 outcomes in the experiment depending on the 6 combinations*1991

*of the first dice and the 6 combinations of the second dice.*1998

*We mentioned in an earlier example all the different totals you can get.*2002

*You can get a total of 2, a 3, a 4, 5, 6, 7, 8, 9, 10, 11, 12.*2006

*And then, we calculated the number of ways that you can get each of these totals.*2018

*The number of ways you can get 2, just a single way because you have to get snake eyes 1-1.*2025

*The number of ways to get 3 is 1 + 2 or 2 + 1.*2032

*There is 2 ways there.*2036

*The number of ways to get 4 is 3, 1 + 3, 2 + 2, or 3 + 1.*2038

*It continues in this pattern, it peaks at they are being 6 ways to get a 7.*2046

*Then, it drops off again 5, 4, 3, 2, where it is just one way to get 12 because you have to roll double 6.*2054

*Let us calculate the probabilities of each of these events.*2063

*The probability that the total is 7, P of A.*2069

*Let us see all the even ways here.*2074

*Remember, there are 36 total outcomes.*2076

*All the even ones, you can get 2, you can get a 4, you can get a 6, you can get an 8, you could get a 10, or you can get a 12.*2079

*I’m reading off all the even numbers here 2, 4, 6, 8, 10, and 12, then the number of ways to get each of those totals.*2094

*I add those up and it comes out to 18/36 and that is exactly ½ your probability that you are going to get an even roll.*2105

*Let us figure out the probability that it is greater than or equal to 11.*2117

*The probability of B is that it is greater than or equal to 11.*2123

*We got 36 combinations we can get in total.*2130

*To get 11, there is 2 ways you can get 11 and there is one way you can get a 12.*2134

*That is 3/36, 112 in total there.*2141

*And I think what I'm going to do to check the independents here, I’m going to check our formula.*2146

*Independence says that P of A intersect B, this is one of the formulas we have developed for independence.*2153

*It was in the original definition of independence but we developed this in one of the early slides.*2166

*If you go back and look at the first couple slides from this lecture then you will see this formula.*2172

*This formula holds if and only if, the 2 events are independent.*2176

*That will calculate the probability of A intersect B.*2180

*The probability of A intersect B means we want it to be even and bigger than 11.*2186

*That is really just the probability that we roll the 12.*2192

*That is the only way for it to be even and bigger than 11.*2195

*Which there is one way to get a 12, 6 and 6.*2198

*1 out of 36, there are 36 total possible rolls.*2201

*Now, the question is whether P of A intersect B is equal to P of A × P of B.*2206

*The probability of A intersect B is 1/36, is that equal to the probability of A is ½ × 1/12.*2219

*What is the probability of B?*2230

*It is 1/36 equal to 1/24.*2231

*Of course, no, those are not equal to each other.*2238

*A and B are not independent, those are dependent events.*2244

*They are not independent.*2253

*Let me recap how we work that out.*2262

*We are relying on this formula for independence.*2264

*The probability of A intersect B is equal to the probability of A × the probability of B.*2268

*I have calculated all those probabilities separately.*2273

*To get the probability of the total being even, that was the hardest one because*2277

*I had to look at all these even totals there and all the different kinds of rolls that can give me each 1 of those totals.*2281

*To get a total of 2, there is 1 way to do it.*2294

*To get a total of 4, there is 3 ways to do it, and so on.*2296

*I have added up all the even ones and that 18 possible rolls would give me an even total out of 36 outcomes, *2300

*36 ways you can roll in all.*2309

*18 divided by 36 is ½.*2311

*The probability of B is the probability we get more than 11.*2314

*There is 2 ways to get 11, that 2 came from there.*2319

*There is 1 way to get 12, there is 3/36 simplifies down to 1/12.*2323

*The probably of B is 1/12.*2329

*The probability of A intersect B means it has to be even and it has to be greater than or equal to 11.*2331

*The only even number greater than or equal to 11 is 12.*2339

*That is really the probability of getting a 12 which we said was 1 out of 36.*2342

*That number is coming from right there.*2348

*Now, it is a matter of checking whether this formula holds.*2352

*Is the probability of A intersect B equal to the probability of A × the probability of B?*2355

*Of course, if you work plugging in the numbers that we just figured out,*2361

*they do not work out and we get that the 2 events are not independent.*2365

*They are dependent.*2371

*Another way to think about this is that if you know that your total is greater than 11, *2374

*that means it got to be 11 or 12, which means if it is 11 or 12, the probability of A given B,*2383

*If you know that it is greater than 11, you know that it is 11 or 12.*2397

*If it is going to be even then you have to be getting a 12.*2407

*What is the probability that you are getting a 12 given that you are getting 11 or 12?*2417

*There is 3 ways to get 11 or 12 because you can get the 5-6, 6-5, or 66.*2422

*Only one of those ways gives you an even total.*2430

*The probability of A given B is 1/3.*2435

*On the other hand, the probability of A was ½ and since the probability of A *2439

*is not equal to the probability of A given B, then we must say that the 2 events are dependent.*2450

*That is kind of a quite different way of arriving at the same conclusion there.*2461

*Let us move on to our last example.*2470

*The last example, we have a teacher picking a student at random from a class of 14 girls and 12 boys.*2475

*Our 2 events here are A being the event that the teacher picks a girl.*2482

*B is the event that the teacher picks a boy.*2488

*The question is are A and B independent?*2492

*I’m setting up a test here to see whether you are falling into one of the misinterpretations of independence.*2494

*A lot of people will be tempted to say that these events are independent because you think,*2507

*if you pick a girl then you cannot pick a boy.*2511

*That is not the right interpretation of the word independent.*2514

*Let me show you quickly the computations.*2517

*If we use computations, we are going to check whether the probability of A is equal to the probability of A given B, *2519

*because that is our original definition of independent.*2535

*The probability of A was the probability of picking a girl.*2538

*There are 14 girls total, there are 14 girls and there is 26 students total.*2542

*I’m getting them by adding 14 + 12, 26.*2551

*14 out of 26 is the probability that we pick a girl.*2557

*What is the probability of A given B?*2560

*What is the probability that if you know that you pick a boy, what is the probability that you pick a girl, which is 0?*2563

*Is 14/26 equal to 0? Of course not.*2573

*That tells us that the events are not independent.*2576

*In fact, they are dependent.*2580

*Right away, just from the definition, from the original calculations, we know that the events are dependent.*2585

*I’m sorry, I almost said the wrong thing there.*2591

*There is a way to think about this intuitively and I hope that you kind of appreciate that as soon as you read the original problem here.*2594

*The way to think about it intuitively here is if you know that B is true, *2602

*in other words, if you know that the teacher picked a boy then you have some new information about whether A might be true.*2615

*You have a very strong new information about whether A might be true *2626

*because if you know that the teacher pick a boy then you know for sure that the teacher did not pick a girl.*2635

*You got some new information about whether the teacher might pick a girl.*2641

*The information is that that did not happen.*2645

*In fact, these events are disjoint.*2647

*There is no overlap between them which means that if you know one is true then you know the other one is false.*2655

*You do get some new information which means they are not independent, which means they are dependent.*2661

*Just to remind you, you can approach this problem sort of using your intuition.*2665

*If your intuition very well trained or using the formulas, if you want to be a little more careful.*2673

*If you use the computations, we are just checking from the original definition of independence.*2679

*We gave that as the definition in the first slide of this lecture.*2683

*Checking whether P of A is equal to P of A given B.*2689

*P of A, what is the probability that we pick a girl?*2692

*There is 14 girls out of 26 total students.*2694

*What is the probability that we pick a girl given that we pick a boy?*2699

*That would be 0 because if we pick a boy, we cannot pick a girl.*2704

*Those two numbers did not come out to be equal to each other.*2707

*That is how we know that they are dependent.*2710

*That is how you do it by the computations.*2713

*You can also do it by intuition.*2714

*If you know that B is true, that changes your estimation of whether A is true.*2717

*Because if you know that the teacher picked a boy, then you know that the teacher did not pick a girl.*2725

*That certainly gave me some new information there because now you know that A is false.*2732

*That tells you that the events are not independent, that they are dependent.*2739

*That is all the fancy ways of saying that these events are disjoint.*2746

*They do not overlap each other so you know that one is true then you know that the other is false.*2750

*That wraps up our lecture on independence.*2758

*These are the probability lectures here on www.educator.com.*2760

*My name is Will Murray, I hope you will stick around for our next lecture which is on Bayes' rule, take care, bye.*2764

1 answer

Last reply by: Dr. William Murray

Fri Aug 28, 2015 1:08 PM

Post by Hen McGibbons on August 27, 2015

i am taking intro to statistics at college starting this fall. Should i try to get through as much of the probability series before class starts? The reason i ask this is because i have read online that having an understanding of probability makes studying statistics a lot easier. I also read that in order to truly understand statistics, you must first understand probability. So am i wasting my time right now studying all these probability videos if its not preparing me for intro stats?

However, let's say for example i was thinking about majoring in statistics (which is something I am considering doing since I am undecided on a career path right now). Then i think it would be wise to start on probability first. If I'm going into the statistics field it probably makes sense that I have an intuitive understanding of statistics (which can be acquired through first studying probability) and not just an understanding based on memorizing equations (which would basically mean getting through intro stats simply by memorization).

Please let me know what your thoughts are.

1 answer

Last reply by: Dr. William Murray

Fri Aug 28, 2015 1:08 PM

Post by Hen McGibbons on August 27, 2015

excellent lecture by the way.

3 answers

Last reply by: Dr. William Murray

Sat Aug 13, 2016 10:49 AM

Post by Hen McGibbons on August 27, 2015

To see if 2 events are independent (such as event A and B) why do we only have to do P(A|B)? Wouldn't we have to check both P(A|B) and P(B|A)?

1 answer

Last reply by: Dr. William Murray

Mon Jun 23, 2014 7:35 PM

Post by Mohammed Alam on June 22, 2014

Dr. Murray,

Referring to E# II, can you please explain without using any equation or the formulas why events A and D are dependent and A and C are independent? I can find out if the events are dependent or independent by using the procedures you have shown here, but I am not getting the concept. So, please without using the procedure you have shown in the video, can you please explain why events A and D are dependent and A and C are independent? Thank you.

1 answer

Last reply by: Dr. William Murray

Mon May 19, 2014 6:11 PM

Post by Danushka Karunarathna on May 19, 2014

Hey, this is regarding example 3, how is P(B)=1/10?? isn't it supposed to be 1/9 since there is no replacement taking place?