Probability Combining Events: Multiplication & Addition

Hi and welcome back to the probability lectures here on www.educator.com, my name is Will Murray.0000

The last time we learned some of the basic terminology for probability and experiments0005

about sample spaces and events.0010

Today, we are going to keep going with that with the combining events.0012

We are going to learn how to build up more complicated events.0016

We are going to learn about unions and intersections.0019

We are going to learn how to calculate the number of outcomes in events, in intersections, and unions of events.0022

We are going to learn some rules of multiplication and addition.0030

We are also going to learn about conditional probability and independence.0033

There is a lot that is going to be happening in this lesson today.0038

We are going to start out with unions of events.0041

The union of two events A and B means the set of outcomes in A or B.0044

This is one of the most loaded words in the English language, this word or.0051

Let us talk about what that means when we say or.0056

It means that at least one of A or B is true.0061

Or often mean something different in English.0065

In English, people some× talk about super salad.0071

For example at a restaurant, with your entree you get a super salad but that does not mean you get both,0074

that means you get either a soup or a salad.0079

That is not what or means in mathematics.0082

The super salad is the exclusive or.0085

In mathematics, we mean the inclusive or.0088

We mean we get one or the other, or you get both.0091

Let me draw a little diagram to indicate what I’m talking about when I talk about the inclusive or,0095

or the union of events.0102

If we have our sample space here, it is the set of all the outcomes, that is our sample space.0104

And then we have two events A and B.0110

There is A and there is B.0116

The union of those two events is all the stuff that is in A or B, or both of them.0121

We get all three regions there.0130

The stuff that is just A, the stuff that is just in B.0132

You also get the stuff that is in A intersect B.0136

Everything in there gets counted when we talk about the union.0141

When people use the word the or in mathematical settings or in computer science settings,0145

they do mean that inclusive or.0150

The exclusive or is just not something that is used as often.0153

When it is used, people are very explicit to specify it as the exclusive or.0157

We definitely, in mathematics, we are talking about the inclusive or.0164

When we want to calculate the probability of A union B, we have a formula for it.0171

What you do is you calculate the probability of A and then you add the probability of B.0177

The problem if you just stop there, is that you have over counted everything that is in both A and B.0186

You over counted this region right here in the middle, the intersection of A and B0193

because you count that once for A and once for B.0200

What you have to do is subtract off that over counting of the intersection.0204

You have already counted it twice so you subtracted off so that you have counted it exactly once.0209

You subtract off the probability of a intersect b, when you are calculating the probability of the union.0214

The easier case there is when they are disjointed events.0223

Disjoint means they do not overlap at all.0227

Let me draw a little diagram to show disjoint events.0229

Here is our sample space and here would be A and here would be B.0233

When they do not overlap at all, when there is no intersection between the two events,0240

it is much easier to calculate the probability because you just calculate the probability of A and0245

then the probability of B and then you add them up because there is no intersection to worry about.0250

That is the union of events, let us move on and talk about the intersection of events.0256

Again, I will illustrate this with a diagram.0261

A intersect B is the set of outcomes that are in both A and B.0263

I’m going to setup my little diagram, my sample space here.0272

I will draw in two events, A and B.0279

That is the whole sample space.0285

There is my first event A, there is my second event B.0287

I want all the outcomes that are in both A and B.0293

That is just this little region in the middle there, that is the intersection.0298

A intersect B there.0303

It is often helpful when you are calculating things.0307

We will see some examples of this, as we get into some of the problems.0309

If you have N possible outcomes for one experiment and N possible outcomes for a second independent experiment,0312

and you are trying to figure out all the possible outcomes for the combined experiments,0321

what you do is you multiply N × N.0326

Where N and N possible combined outcomes.0329

You will see how that comes in as we do some examples.0334

We got a couple more concepts that I want to introduce to you before we get into some examples.0338

We have conditional probability and independence.0343

Conditional probability, you read this, it is this vertical line notation here, P of A given B is how you read that.0348

The conditional probability is written P of A vertical line B but you read that as P of A given B.0358

And what it means is, it is the probability that event A is true given that you already know that B is true.0366

There is a nice way of calculating it and I can show you this again with the diagram.0375

I think it will help you understand where this formula comes from.0380

The idea here, let me setup two events A and B.0386

There is A and there is B.0392

A and B, this is all taking place in a sample space S.0398

What we are assuming here is that we already know that B is true.0403

We already know that B is true.0408

It is restricting our world down to the set of outcomes where B is true.0411

We are saying that we are already restricting our self to this red area down here in the sample space.0417

What is the probability that A is true given that we are restricting our world to this set of outcomes in B?0425

If you are going to be inside B and you want A to be true,0435

that means you can only be looking at this region right here because this is the only region inside B that is also in A.0439

That region that I just colored in there was exactly A intersect B.0447

To find the probability of A given B, we want to find the probability of being in that region,0455

the probability of A intersect B.0461

The probability of A intersect B but we are restricting herself to a world inside B.0467

We divide this by the probability of B.0477

We are essentially looking at the probability of the blue region ÷ the probability of the red region.0480

That is how we calculate conditional probability.0488

We will see some examples, you get some practice with that as we get into some of the problems.0491

But there is one more concept I want introduce to quickly introduce to you first which is independence.0495

Independence of two events means that suppose you are wondering about whether A is true.0501

You are wondering about whether A is going to happen and someone tells you that B is true.0508

Someone tells you the B has happened.0515

Do you now have any more information about whether A is true?0518

If that new information about B being true, that does that make you change your prediction0525

about whether A might be true?0534

If it does not make you change your prediction then we call those events independent.0538

In terms of calculations, this is watching your prediction for A given that B is true.0543

That is P of A given B.0551

That is if somebody has told you that B is true and you decide to make a prediction about the probability of A.0558

This is if you have no information at all.0563

You just provide a prediction about P of A with no external information about B.0566

If those probabilities would be the same either way, can we say that A and B are independent?0580

We say that knowing that B is true does not cause you to change any predictions you might make about A.0586

Now, independence is a very dangerous concept.0594

It is very misleading and is very frequently mistaken by students.0597

Most commonly, students get independence mixed up with being disjointness, what was being disjoint.0603

Here is what disjoint means.0612

Disjoint means that the probability of A intersect B is 0.0613

It means that A and B are completely separate from each other.0618

That is not independence.0622

Just because they are separate, it does not mean they are independent.0624

In fact, it would mean that they are dependent.0627

Because if they are disjoint and I tell you that B is true, now you know that A is false.0632

If I tell you that B is true, you will get some new information about A.0640

Disjointness is not at all the same as independence but people tend to get those mixed up.0644

People tend to think that independent means disjoint and disjoint means independence.0650

It is not all true.0654

We will try to do some examples to highlight the differences between those two.0656

To reiterate, if they are disjoint, they are not actually dependent, not independent0663

because if you know that one is true then you know that the other one is false.0668

Independence means that knowing one is true gives you no new information about the other one.0673

Let me show you how this plays out in equations.0681

If they are independent, remember independent means P of A is equal to P of A given B.0683

P of A given B, here is our formula for P of A given B.0690

That is our formula for P of A given B.0697

The P of A intersect B/ P of B.0699

If they are independent then P of A given B is the same as P of A.0703

P of A would be equal to P of A intersect B/ P of B.0709

We get this nice formula, if you multiply both sides by P of B, you will get P of A intersect B0714

is equal to P of A × P of B.0721

That is a really useful multiplication rule except you have to be careful to check that A and B are independent.0724

That is the only time you can use this multiplication rule, as if you have independent events.0733

Like I said, because students often made mistakes about what independent means,0741

you can often get into a lot of trouble by misusing this multiplication rule.0745

We will go through some examples to train your intuition to know when an event is independent and when not.0751

Let us jump into some examples here.0760

First example, we are going to choose a whole number at random, from 1 to 100.0762

We want to know, what is the probability that it will be even or greater than 80?0767

Let us set up some events here.0773

Let me setup a sample space.0776

The sample space S is all numbers from 1 to 100, it is a whole number.0779

It is 1, 2, 3, up to 100, that is my sample space, all possible outcomes there.0784

The events we are interested in are being even and being greater than 80.0792

A, I will say is the even numbers.0797

That means we have 2, 4, and so on, up to 100 is our last even number there.0806

B is that it is greater than 80.0820

Everything greater than 80, it says greater than 80 so it is not including 80.0824

That is 81 through 100 there, 81, 82, up to 100.0830

The key word that we want to focus on here is the or.0836

What we want to calculate there is a union, the probability of a union.0843

However, what we saw was that to calculate the probability of a union, A union B,0851

what we want to do is calculate the probability of A + the probability of B - the probability of A intersect B.0857

In order to calculate the union, we need to calculate the intersection.0871

We need to go back and calculate the intersection first.0878

Let us calculate the sizes of all these sets.0882

First of all, how many even numbers are there from 1 to 100?0887

There are 50 of these.0892

How many numbers are there greater than 80?0896

81 through a 100, that is 20 numbers.0903

And how many even numbers are there greater than 80?0906

That is the intersection, things that are both even and greater than 80.0911

That is all the even numbers greater than 80, 82, 84, up to 100.0915

There are 10 numbers in there, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100.0927

There are 10 of those.0935

The size of the union is equal to the size of A.0937

I’m really doing this by counting now.0949

Let me go back after we work this out by counting and we will do it by probability.0951

And you will see a slightly different argument, they will get us to the same answer.0956

A + B - the size of the intersection, A intersect B, which we said A was 50, B is 20, the intersection we said is 10.0960

That simplifies down to 60, 70 -10 is 60.0975

The probability of A union B, that is what we are looking for because it was not or.0980

A union B is equal to the size of A union B ÷ the size of the sample space.0997

That is 60 ÷ 100 of which reduces down to 3/5.1007

That is our probability of getting an even number or a number greater than 80.1015

This is a good example of solving a probability problem by counting.1024

By counting the sizes of all the sets there are.1031

We have counted the size of A, we have counted the size of B.1035

We have counted the size of A intersect B.1038

We counted the size of A union B and then we divide them all together.1040

We could also solve this by probability.1046

I like to show you how that goes.1050

The arguments are very similar but they will end up at the same answer.1052

But sometimes students get mixed up between the two approaches.1057

Sometimes you will start out with one approach and then switch over to the other approach,1063

or mix the two approaches together and you get confused.1066

You end up getting tangled up in getting the wrong answer.1069

Let me show you how the two different approaches look and show you what they have in common and how they are different.1072

If we do this by probability, if we use this formula up here, the probability of A is the number of even numbers over 100.1079

There is 50 even numbers over 100 which is ½.1093

The probability of B, B was the event that our pick is greater than 80.1098

The number of numbers greater than 80 is 20/ 100 possible choices, that is 1/5.1106

The probability of A intersect B is the probability of getting a number that is greater than 80 and is even.1117

We said that, where is our A intersect B?1130

There it is right there.1132

We said that there is 10 numbers that are greater than 80/100 possibilities that is 10/ 100, which is 1/ 10.1134

The probability of A union B is the probability of A + the probability of B - the probability of the intersection A intersect B.1144

I’m using this formula right here.1161

That is ½ + 1/5 – 1/10.1164

If I put those all over common denominator which would be 10, I will get 5 + 2 - 1/10 which is 7 -1/6/ 10, which is 3/5.1171

Of course, that agrees with the answer we got by counting but it looks a little different in flavor.1184

I want to emphasize that a lot of probability problems have a breakdown like this1190

where you can solve them strictly by counting, kind of following this technique.1195

Or you can solve them by probability, following this technique on the right.1201

Each way can work but if you mix up the 2 then you are liable to make mistakes.1206

If you kind of mix the two together, it is a recipe for disaster, let me say.1214

You want to decide, am I going to do this by counting or am I going to do this by probability?1221

Then, stick to your technique and pursue it all the way through.1227

Sometimes you can do it by both techniques but if you do, you want to do it completely separately1230

and make sure that your answers agree with each other at the end.1235

Let me go back and show you quickly the steps we went through here.1239

We want to solve it by counting, we first listed our sample space.1244

We have listed our two events, the even numbers and the ones greater than 80.1247

Those are our two events A and B.1251

Count the numbers of outcomes in each one.1254

A have 50 outcomes, B had 20 outcomes, and then the intersection had 10 outcomes.1258

They are numbers that are both even and greater than 80.1264

To get the size of the union, we add up the sizes of A and B and subtract the intersection that turned out to be 60.1268

And then the probability, finally we divide 60 ÷ S, that sample space.1276

60 ÷ 100 reduced down to 3/5.1282

That is solving the whole problem by counting the sizes of the set.1286

Instead, if we do it by probability, we calculate the probability of everything as we go along.1291

The probability of getting an even number is ½.1296

The probability of getting a number greater than 80 is 1/5.1300

The probability of getting both even and greater than 80 is 1/10.1303

And then we use this generic formula that I gave you in one of the introductory slides, P of A + P of B - P of A intersect B.1309

Reduce those fractions down and it reduces down to that same answer 3/5.1319

In example 2, we have a combination deal with restaurants where they are offering a combination1329

of a main dish and a drink and the dessert.1334

We want to figure out how many different types of combination meals we can make.1338

The restaurant is offering us 4 different types of drinks, 12 different main dishes, and 3 possible desserts.1342

This is an illustration of the multiplication rule.1348

The solution here is, how many different types of drinks I can have?1354

I can have 4 types of drinks and then independent of that is my choice of main dish.1360

For each drink combination, I can have a main dish.1381

There are 12 different kinds of main dishes I can have, for each type of drink I might pick.1387

For my dessert, I could have 3 different kinds of desserts and I multiply those together.1395

This is the multiplication rule, since these are all independent choices.1402

What I get for my drink does not affect my main dish.1406

What I get for my main dish does not affect my dessert.1408

I just multiply those together and get 4 × 3 is 12 × 12 is 144 combinations.1412

If you think about how many different possible types of combination meals that restaurants can serve,1429

the answer is 3 × 4 × 12 is 144 different kinds of combination meals.1434

You can go that restaurant every day, a 144 days in row and order a different combination each day.1442

On the 145 day, you have to repeat some combination that you have seen before.1448

It is the multiplication rule in action there.1455

For example 3, we are going to roll 2 dice.1460

And it is useful often when you are rolling dice, it is very useful to think of them as being a red one and a blue one.1462

We are being asked what is the probability that the red one shows a 4 and the blue one is odd?1469

We are going to set up some events here.1477

S is my sample space of all possible outcomes.1483

Remember, for rolling two dice, we have this example in the previous lecture.1488

You can go back and check it out, if you do not number exactly how it works out.1493

But rolling two dice, the red one can be 1 through 6.1496

And then, for each one of those, the blue one can be 1 through 6.1500

You can get red 1, blue 1.1504

You can get red 1, blue 2.1510

All the other combinations where the red is 1 and all the combinations where the red is 2,1516

all the combinations where red is 3, all the way on up to if the red is 6 and the blue is 6.1520

There are 36 possible combinations there, 36 possible outcomes in our sample space.1528

Our two events here, our A is the event that the red one is a 4.1535

Let us list out those possibilities there, at least the first few.1551

The red would be a 4, the blue one could be 1.1555

The red one could be a 4, the blue one could be a 2, and so on, up to the red one could be a 4 and the blue one could be a 6.1560

Those are all the possibilities there, for the red dice to be a 4.1575

B is going to be the event that the blue dice is odd.1579

The blue dice is odd.1584

Let us think about the number of possibilities there, about how you can have the blue dice being odd.1591

You can have a red one being 1 and the blue one is a 1.1601

You can have the red one being a 1 and the blue one being not 2 but 3, and so on, up to the last possibility there.1606

Be the red one is 6 and the blue one could not be 6, or the last one would be 5, if we list them all out.1617

Those are all the possibilities for the blue dice to be odd.1623

Remember I said a lot of these probability problems can be solved two different ways.1627

You can solve them by counting or you can solve them thinking about probabilities all the way through.1631

I would like to try and do this problem both ways so that you can see the distinction.1637

And you can get the same answer either way but you can see the different kinds of thinking.1643

And in particular, I do not want you to kind of mix up the counting arguments and the probability arguments.1649

Let me show you first the counting arguments.1656

By counting, we want to find the probability of A intersect B.1659

Let us think about how we can count A intersect B and then we will divide it1671

by the number of outcomes in our sample space.1681

A intersect B means the red dice is 4 and the blue dice is odd.1684

The reason I'm using intersection there is because this key word here and, we want A intersect B.1691

To be in A intersect B, the red dice has to be 4 and the blue dice has to be odd.1705

Red dice can be 4, blue could be 1.1712

The red dice could be 4 and the blue could be 3.1716

The red dice could be 4 and the blue could be 5.1725

Those are the only ways that the red can be 4 and the blue could be odd.1733

S is the sample space of all possible outcomes.1739

We already figure out that there are 36 of those corresponding to 6 choices for the red dice, 6 choices for the blue dice.1743

If we work that out, we got 3/36 which of course simplifies down to 1/12.1753

By simply counting things up, we work out that our probability is 1/12.1765

That is a counting based argument.1771

Let me give you another argument that is based on thinking about probability directly.1773

This is a little more subtle but the arithmetic is pretty nice at the end of it.1783

It think about these two events, the red dice being a 4 and the blue dice being odd.1788

If I roll two dice and I tell you that the red dice is 4, does I tell you anything about the blue dice?1793

No, it tells you nothing about the blue dice.1799

These events are independent.1804

The reason that is true is if I tell you that one of these events happen, if I tell you that the red dice is 41817

and then I say that does change what you predict about the blue dice?1826

It does not change what you predict about the blue dice at all.1829

Since knowing that one is true does not change the probability of the other.1832

Remember, independence does not mean disjoint.1856

Disjoint means that the two things cannot happen at the same time.1858

This is not what we are talking about.1862

We are talking about independence means that if I tell you that one is true,1863

it does not change your prediction that the other might be true.1867

Because these two events are independent, we have an easy way to calculate the probability of the intersection.1874

This is from that opening slide.1881

If you do not remember this, just go back a few slides and you will see it.1882

The probability of A intersect B, when they are independent is the probability of A × the probability of B.1888

That is really using independence very heavily right there.1900

This equation would not be true if they were not independent.1903

But let us think about the probability of A and B independently.1907

The probability that the red dice is 4, if you just forget that you are rolling a blue dice,1911

the probability that you roll on dice and you get a 4 is 1 out of 61919

because there are 6 possible outcomes that one of them gives you a 4.1924

What is the probability that the blue dice is odd?1928

Again, for get that you are rolling a red dice, the probability that the blue dice is odd,1931

there are 6 possible outcomes.1937

3 of them will give you an odd number, 3 out of 6 is ½.1939

We can fill those in here, P of A × P of B is 1/6 × ½, multiply those together and you get 112.1946

That is the probability that both of those events will occur.1959

That is the probability of the intersection of those events.1963

And notice, that agrees with our earlier answer of 112.1966

We got 112 either way.1970

But using two quite different arguments, one is based on counting and one is based on the theory of independence.1972

Let me recap what we did there.1979

We first listed our sample space which is all the possible combinations of the two dice.1983

There are 36 combinations depending on what the red shows and what the blue shows.1987

A is the probability, A is the event that the red dice show a 4.1993

We have listed out the 6 possible combinations there.1999

We also calculated the probability of A is 1/6 because if you forget that you are rolling a blue dice,2002

the probability that the red dice will be 4 will be 1 out of 6.2008

The probability that the blue dice is odd, if you list out all those possible combinations,2012

I do not think I ever bother to count those.2020

There will be 18 combinations but we found the probability just to be ½2022

because if you forget that you are rolling a red dice then the probability that the blue dice will be odd,2030

there are 3 ways for it to be odd.2036

1, 3, and 5 ÷ 6 possible combinations.2037

3/6 or about ½ came from.2041

If we do counting argument here, we just try to count the size of A intersect B.2045

Here the outcomes in A intersect B meaning the red dice is 4 and the blue dice is odd.2054

There are three ways for that to happen, 4-1, 4-3, and 4-5.2059

Out of 36 total possible outcomes, we get t3 out of 36 is 112.2062

If we use a probability theory though, if we use the theory of independence,2070

knowing that the red dice is 4 does not tell you anything about the blue dice.2074

And knowing that the blue dice is odd, does not tell you anything about the red dice.2079

That means that these are independent.2084

And we have another way to calculate probabilities of intersections when the events are independent.2086

You can just multiply their probabilities.2092

We already calculated the probabilities to be 1/6 to ½, multiply those together and get 1122095

which agrees with the answer we got using the other argument.2101

In example 4, we are going to flip a coin 4 × and we want to find the probability that2110

there would be at least 3 heads given that the first two flips are heads.2115

This is an example where we are given some partial information about a problem and2120

we want to calculate the probability of an event based on knowing that another event is true,2125

based on that partial information that we have.2134

Let me set up some events here, our sample space.2139

S is all the possible things that can happen when you flip a coin 4 ×.2144

When you think about it, each time you flip there is two possible things that can happen and2150

if you string those together, it is 2 × 2 × 2 × 2.2155

You can get head-head-head-head, head-head-head-tail, and so on.2162

There is two choices for each one of those spot, each one of those 1st, 2nd, 3rd, and 4th spots.2172

There are 16 possible outcomes in our sample space here.2178

We want to identify the events that the problem is asking us about, 16 possible outcomes.2185

I see this we are not given that means we are going to use conditional probability.2197

Let me set up my events here.2201

A is the probability that there will be at least 3 heads.2202

When we try to list those out, at least 3 heads.2210

How can we get at least 3 heads?2215

List out those possibilities we can get head-head-head-tail.2220

We can get head-head-tail-head.2224

That would have at least 3 heads.2228

Head-tail-head-head, tail-head-head-head, or we could get all 4 heads.2230

That would still count as at least 3 heads.2239

Those are all the possible ways to get at least 3 heads.2243

B is the first two flips are head.2245

That was our other condition given in the problem.2250

The first two flips are heads.2254

When we list all the strings that would give us the first 2 flips are heads.2263

Head-head-head-head, head-head-head-tail, head-head-tail-head, head-head-tail-tail.2267

Those are all the strings for the first two flips being heads.2281

And now we are interested in finding the probability of A given B.2283

Let me remind you of the formula for that.2289

Probability of A given B is a conditional probability problem.2292

Probability of A given B is a probability of A intersect B ÷ the probability of B.2297

That was the formula we had in one of the introductory slide.2305

Just flip back a few slides and you will see that.2308

That means I have to figure out what the probability of A intersect B is.2312

I need to find A intersect B.2317

I just scan over those list A intersect B of A and B and see which ones are in both.2320

I see a head-head-head-tail, HHHT.2327

I lost one of my heads there.2332

HHHT, I see HHTH, HTHH does not work, THHH does not work.2335

HHHH also fixed in both of them.2347

I see three of those outcomes are in A intersect B.2352

The probability of A intersect B is 3 out of 16 possible outcomes.2358

The probability of B is, let us see, there are 1, 2, 3, 4, out of 16 possible outcomes.2364

4 out of 16 I can multiply top and bottom by 16 there to simplify things and get ¾.2374

Let me show you again where everything in there came from.2387

We get this probability problem - flip the coin 4 times.2390

Right away, I got a sample space here.2394

I got 16 possible outcomes and describing all the possible strings of heads and tails that we can get.2396

What is the probability that there will be at least 3 heads?2402

I’m going to list that out as one of my events, at least 3 heads.2406

They are all the ways that you can get at least 3 heads.2410

It looks like there is 5 of them.2413

Given that the first two flips are heads, they are all the ways to get the first two flips being heads.2416

Listed all those out and I got 4 of them.2424

The key word here is given, that tells me that I'm going to be calculating conditional probability.2427

That is what this one is all about, it is conditional probability.2433

I wrote down my conditional probability formula.2435

P of A given B is P of A intersect B/ P of B.2437

In order to calculate that, I know what P of B is because of I have already looked at B.2443

I have to find A intersect B.2449

I looked at A and B, and I try to figure out which of the strings are in both.2452

I found these 3 strings in both of them.2458

Which means the probability of A intersect B is 3/16.2460

That 16 comes from here, the t3 comes from here.2464

And the probability of B is 4/16.2469

The 4 comes from here.2472

When you simplify that down, you get the answer is ¾.2474

For example 5, we are going to roll two dice, we would think of them as red and blue.2483

We are going to define the events that the red dice is 3, that is event A, B is that the total of 7 and C is that the total is 8.2488

There is a whole bunch of possible combinations here.2498

And the problem is asking us to calculate the conditional probabilities of all these combinations.2502

B given A, A given B, C given A, A given C, and so on.2509

We need to calculate a lot of probabilities here.2515

Remember our formula for conditional probability, let me just remind you what is.2518

P of A given B is the probability of A intersect B ÷ the probability B.2522

We are being asked to find the conditional probabilities of everything in sight.2533

Which means we are going to need the probabilities of all the basic events.2538

We will also need the probabilities and of all their intersections.2542

Let us go ahead and calculate all these out.2545

The red dice is 3.2549

The probability of A, we are going to start with that.2553

The red dice is 3, that is a probability of 1/6.2556

One way to think about that is there are 6 possible outcomes where the red dice is 32560

because the blue dice can be any of 6 different possibilities ÷ 36 possible outcomes in total.2564

That is not the most efficient way to think about it.2571

The most efficient way to think about it is to just forget that you are rolling a blue dice and2573

say what is the probability that the red dice is 3?2578

It is 1/6, the total is 7.2580

We saw this one in an earlier example in the previous lecture.2583

Check back in that example, if you do not remember it because I will do it quickly now.2587

The total is 7.2592

There are 6 ways to get the total being 7.2593

You can get a 1-6, you can get a 2-5, you can get a 3-4, 4-3, 5-2, or 6-1.2596

There are 6 ways you can get a total of 7.2610

There is 36 possible rolls in all, the probability of B is 6/36 but that just simplifies down to 1/6.2612

C is that the total is 8.2625

Let me write down quickly the combinations that give you a total of 8.2627

You can get anything if there is a 1 on the first dice.2630

You have to get 2-6, or 3-5, or 4-4, or 5-3, or 6-2.2636

Those are all the possibilities and there is 5 of those.2647

Out of 36 possibilities, possible outcomes and all, the probability of C is 5/36.2653

We are also going to have to find the probabilities of all the intersections of these combinations of events.2661

Let us go ahead and calculate all those and then we can start to work on the conditional probabilities.2670

P of A intersect B means the red dice is 3 and the total is 7, that is the probability.2676

The only way to get the red dice in being 3 and the total being 7 is,2689

if the red dice is 3 and the blue dice would have to be 4 for that to work.2695

The only way to get that, there is only one possible outcome which would be the red dice is 3, blue dice is 4.2700

One possible outcome out of 36 things that can happen.2705

That is 1 out of 36.2710

Let us find the probability of A intersect C, that is kind of similar.2714

That means the red dice is 3 and the total is 8.2721

The only way you can get that would be if the red dice were 3 and the blue dice were 5.2726

Again, that is one possible outcome out of 36 in total.2734

1 out of 36 chance that both of those events will be true.2738

Finally, the probability of B intersect C.2744

We need this to calculate these conditional probabilities.2748

Think about what that means, B intersect C means B and C are both true.2753

B means the total is 7.2759

C means the total is 8.2762

The probability that the total is 7 and the total is 8, those cannot both occur at the same time.2763

That probability is 0.2772

There is no outcome that gives you the total being 7 and the total being 8.2775

That probability is just 0.2779

Finally, we were able to calculate our conditional probabilities.2782

We are going to use this formula over and over again here.2785

The probability of B given A is the probability of A intersect B or B intersect A, those are the same thing.2792

A intersect B over the probability of A.2801

I have calculated all these things up above so that is 1/36 ÷ the probability of A.2807

Over here we got 1/6, that flips up to 6/36 which is 1/6.2815

That is the probability of B given A.2825

Conditional probability of A given B is the probability of A intersect B again.2829

But now we divide by the probability of B.2836

But we get the same numbers, 1/36 ÷ probability of B was also 1/6.2839

Again, it simplifies down to 1/6.2847

The probability of C given A is, I’m following my formula here.2852

The probability of C intersect A which is the same as A intersects C ÷ the probability of A.2857

A intersects C, here it is up here 1/36.2865

The probability of A is 1/6, we get the same numbers again.2870

It simplifies down to the same answer 1/6.2874

Let us write that 6 a little bigger so that people can read it.2879

The probability of A given C, that is the probability of A intersect C ÷ the probability of C, this time.2884

1/36 ÷ the probability of C here is 5/36.2895

5/36, if we multiply top and bottom by 36 I will get 1/5.2903

That is kind of our first interesting surprise there because the probability of C given A and2915

the probability of A given C turned out not to be equal to each other.2923

We have to be careful from now on.2928

Which one we are calculating because they might not be equal to each other.2931

B given A and A given B did turn out to be equal but apparently that was just a fluke for those two events.2934

In general, the conditional probability does matter which order you write those events in.2941

The probability of C given B, that is the probability of B intersect C ÷ the probability of B.2949

The probability of B intersect C, we discover was 0.2960

0 ÷ 1/6 is still just going to be 0.2965

There is no probability of C given B.2972

Of course that makes sense because if I tell you that the total is 7 and then I ask what the probability is2975

if the total is 8, then of course you should say 0.2982

Because if you already know the total is 7, it cannot be 8.2987

Let us check the probability of B given C.2990

It is again, the probability of B intersect C ÷ the probability of C which again is 0 ÷ 5/36.2993

But that numerator does not really matter because we get 0 anyway.3006

First, it looked like the answers will all going to 1/6 and then we got into3014

some more interesting numbers and they started to change.3017

Let us just recap where those all came from.3020

We are rolling two dice.3023

We know that if we roll a two dice, there are 36 possible outcomes in our experiment because the red3024

and blue dice there could each be 6 possible numbers.3031

We are given three events and we have to calculate all these conditional probabilities.3035

A good thing to do at first was to calculate the probability of each of these events solo, which we did.3042

We found that there is 1/6 chance that the red dice is 3, 1/6 chance that the total is 7.3048

We got that by adding up all the total ways that we can get 7.3055

5 out of 36 chance that the total is 8.3060

We get that by adding up the 5 possible ways that you can get a total of 8 there.3064

We got the probabilities of all the individual events and we want to find the probabilities of all the combinations.3070

The reason that I let those is because I was looking ahead to this conditional probability formula.3077

I see that it requires calculating the probability of intersections.3083

That is why I went ahead and calculated all these intersections ahead of time, A intersect B.3088

The only way to get a red dice 3 and a total of 7, as if the blue dice is 4.3093

There is only one outcome that gives you that.3099

That is why we get 1 out of 36 there.3101

Red dice is 3 and the total is 8 would have to be a 3 and a 5, 1 out of 36 again.3104

The total is 7 and the total is 8, there is no way those can both be true.3112

There is a 0 probability there.3117

We need to calculate the conditional probabilities.3119

What I was doing is just invoking this formula up here over and over again.3121

Just substituting in and out the different combinations of a's, b's, and c's.3130

I just substituted in all the combinations of a's, b's, and c's.3134

And then, I plug in all the probabilities of the intersections that I calculated above and the denominators.3137

I have plugged in the probabilities of the raw events and simplify them down into all these fractions.3144

The interesting thing that happened here was that the probability of C given A and3151

the probability of A given C turned out not to be equal.3157

You do have to be very careful about the order of these things.3160

Also, the probability of either B or C given the other one, turned out to be 0.3163

And that of course is confirmed by the calculation.3170

It also confirms with your intuition because if I tell you that the total is 7, If I say that the total is 7,3174

there is really no way that the total can 8.3180

As soon as I tell you that the total is 7, then you say the probability that it is 8 is 0.3183

I want you to hang onto all these probabilities because we are going use them again for the next example.3192

Example 6 is using this very same experiment and we are going to ask whether3198

all these events are independent of each other.3205

Instead of recalculating these probabilities, I’m just going to refer to all my answers from this example.3209

So hang on to these, make sure you understand them, and then will move on to the next example.3216

In example 6 here, this is really referring back to example 5.3224

It is the same experiment and the same events.3229

We are going to roll two dice, red and blue.3231

We are going to define the events A is the event that the red dice is 3.3234

B is the event that the total is 7.3239

C is the event that the total is 8.3241

We saw those before, we calculated the probabilities of each one.3243

In fact, let me go ahead and remind you what they were.3248

If you do not know how we got them, if you have not just looked at example 5,3252

go back and check through example 5 and you will see where all these come from.3257

The probability of A was 1/6, the probability of B was also 1/6, and the probability of C is 5 out of 36.3261

And then we also calculated several conditional probabilities that are going to be useful here3276

when we are determining independence.3282

I will just write down some of the useful ones here.3285

The probability of A given B, we calculated this last time, in the last example, in example 5.3288

It turn out to be 1/6.3295

The probability of A given C, I'm skipping ahead to the ones we are going to be using.3298

The probability of A given C turned out to be 1/5.3305

The probability of B given C, because those were disjoint events, it turned out to be 03310

because they cannot both happen at the same time.3319

We are being asked, are events A and B independent?3322

What about A and C and what about B and C?3325

For A and B, remember independence by definition means the probability of A is equal to the probability of A given B.3328

Intuitively, it means that if you calculate the probability of A with information or3344

if you calculate the probability of A with the information that B is true, you get the same answers either way.3351

Let us just check those out.3357

The probability of A, I wrote down was 1/6.3358

The probability of A given G is 1/6.3362

Those are equal so that tells us that A and B are independent.3365

That is our answer for A and B.3387

Let us look at some of the others.3389

What about A and C, we are going to look at A and C.3392

Let us calculate the probability of A and we will check whether it is equal to the probability of A given C.3398

The probability of A is 1/6, we got that up above.3409

Is that equal to the probability of A given C 1/5?3416

No, that is not equal.3421

1/6 is not equal to 1/5.3422

That tells us that A and C are not independent.3425

They are dependent.3434

I will try to give you some intuitive feel for that in a moment.3439

I want to go through and do all the calculations first.3443

Then, we will come back and see if we can make some kind of intuitive interpretation of this.3445

That was for A and C, now B and C.3453

It is tempting to say they are independent because if you are thinking like some students think,3458

they say that total is 7 and total is 8, those cannot both happen.3463

You might think all that means independent, that does not mean independent but that means disjoint.3468

Let us actually calculate it using the probabilities.3473

Is the probability of B equal to the probability of B given C?3476

That is our formal definition of independence.3482

Let us check and see if those are equal.3484