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### 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

### 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 ahead0085

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 because0179

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 A0204

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 probability0715

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 01169

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 teacher1273

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 B1290

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 information1305

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 or1348

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

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 explain1394

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 around1425

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 calculation1504

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 students1516

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 student1564

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 1/101651

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 11927

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 and1941

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 total1971

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 combinations1991

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 because2277

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

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 A2439

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 true2626

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