For more information, please see full course syllabus of Probability

For more information, please see full course syllabus of Probability

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### Combining Events: Multiplication & Addition

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
- Unions of Events 0:40
- Unions of Events
- Disjoint Events
- Intersections of Events 4:18
- Intersections of Events
- Conditional Probability 5:47
- Conditional Probability
- Independence 8:20
- Independence
- Warning: Independent Does Not Mean Disjoint
- If A and B are Independent
- Example I: Choosing a Number at Random 12:41
- Solving by Counting
- Solving by Probability
- Example II: Combination 22:07
- Combination Deal at a Restaurant
- Example III: Rolling Two Dice 24:18
- Define the Events
- Solving by Counting
- Solving by Probability
- Example IV: Flipping a Coin 35:07
- Flipping a Coin Four Times
- Example V: Conditional Probabilities 41:22
- Define the Events
- Calculate the Conditional Probabilities
- Example VI: Independent Events 53:42
- Define the Events
- Are Events Independent?

### Introduction to Probability Online Course

### Transcription: 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 experiments*0005

*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 B*0193

*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 and*0245

*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 prediction*0525

*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 independent*0663

*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 B*0714

*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 this*1190

*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 separately*1230

*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 combination*1329

*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 it*1671

*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 4*1817

*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 6*1919

*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 ½*2022

*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 112*2095

*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 that*2110

*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 and*2120

*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 and*2150

*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 3*2560

*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 and*2573

*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 and*2915

*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 is*2975

*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 into*3014

*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 red*3024

*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 and*3151

*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 whether*3198

*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 here*3276

*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 0*3310

*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 or*3344

*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

*The probability of B is 1/6.*3486

*Is that equal to the probability of B given C is 0.*3489

*Is 1/6 = 0?*3493

*No, it is not equal to 0.*3494

*That tells us that B and C are not independent, they are dependent.*3497

*We did that all using calculations.*3509

*Using the calculations that we got from example 5.*3512

*If you do not know where these numbers came from, just go back and watch example 5.*3515

*The one just before this one and you will see where I got all these numbers, the 1/6, 1/6, and 5 out of 36 for A, B, and C.*3519

*We calculated the conditional probabilities 1/6, 1/5, and 0.*3527

*We calculated some others in example 5 but I just extracted the ones I would need here.*3532

*To check independence using calculations, you just check whether the conditional probability*3537

*is equal to the probability without making the assumption that the other event is true.*3543

*That is what I check, I just check the numbers each time and if they were equal, I said they are independent.*3551

*If they are not equal, I said they are dependent.*3557

*That kind of answers the problem but it would be nice to have some intuitive reason for why these might be true.*3563

*For A and B, it is not so obvious.*3571

*If the red dice is 3, what is that tell you about the total being 7?*3572

*And the answer is it does not really tell you anything because the total being 7 was a 1/6 chance anyway.*3580

*If the red dice is 3, then I know to get the total being 7, I need the blue dice to be a 4 and there is a 1/6 chance of that.*3589

*It is telling me that the red dice is 3 gives me no new information about whether the total was 7.*3597

*That is why A and B are independent.*3603

*How about A and C?*3608

*The red dice is 3 and the total is 8?*3609

*You might think that that is similar to the previous one, to the red dice is 3 and the total is 7, but it is not.*3612

*Here is why.*3618

*To get the total being 8, let us think about how that can happen.*3620

*How can the total be 8?*3627

*You can get 2-6, you can get a 3-5, you can get a 4-4.*3629

*These are always to get a total being 8.*3636

*You can get a 5-3, or you can get a 6-2.*3638

*In particular, for the total to be 8, you cannot roll a 1 on either dice.*3644

*If I tell you that A is true, if I tell you that the red dice is 1, what I really told you is in particular that the red dice is not a 1.*3651

*It is slightly more likely for the total to be 8 because we know now that the red dice is not a 1.*3667

*That is why P of A given C is slightly higher than P of A alone.*3674

*I guess actually that means I have told you that the total is 8,*3684

*that means that the red dice cannot be 1 which makes it slightly more likely for it to be 3.*3689

*That is why we get a 1/5 over here and a 1/6 over there.*3698

*I’m giving you a little bit of information when I tell you that the total is 8.*3702

*I’m saying, I’m really telling you that the red dice cannot be 1 which makes it slightly more likely for it to be 3.*3705

*B and C is easier to understand intuitively.*3712

*If I tell you that the total is 8, I'm giving you some information about whether the total is 7.*3716

*I told you that the total is not 7.*3723

*I changed the probability on you.*3725

*As soon as I tell you that the total is 8, you know for sure that the probability is not 7.*3727

*I have really changed your thinking.*3732

*Those are really dependent events, they do affect each other.*3734

*That is kind of an intuitive reason why we got these answers that we did.*3739

*Why A and C are dependent, B and C are dependent, but A and B are independent.*3743

*There is a lot of twist and turns into this probability and independence.*3750

*I hope you had fun playing with those.*3754

*We are going to move on to the next lesson.*3757

*I want to say thank you for joining us today for the probability lectures on www.educator.com.*3759

*My name is Will Murray, thanks for joining, bye.*3763

8 answers

Last reply by: Dr. William Murray

Fri Jan 12, 2018 9:44 AM

Post by Shahriar Arfa-zanganeh on May 27, 2017

Hi Dr. William Murray,

I was wondering if there is anything wrong with this video or if there is a bug with it, as when I get to the conditional probability part of the video, the remaining amount of it turns into a black screen that does not go away. This effect does not seem to happen when I try playing other videos of yours, so I think it may just be this one that has this issue. Would be much appreciated if the issue is resolved, as you teach the material very well.

Thanks

1 answer

Last reply by: Dr. William Murray

Tue Oct 4, 2016 11:59 AM

Post by Thuy Nguyen on October 1, 2016

Hi, about the combination meals where there are 12 main dishes, 4 drinks, and 3 desserts:

Why don't we divide out the 12*4*3 total meals by 3!? Won't some meals be counted more than once?

When do we divide out like the poker hand and when do we not like in this meal combination?

Thanks (smiles).

3 answers

Last reply by: Dr. William Murray

Mon Nov 24, 2014 9:49 PM

Post by Jim McMahon on November 19, 2014

I cannot advance the lecture to a point that I am trying to jump to by clicking on the timeline. Is there a different way to advance the lecture to the desired section?

2 answers

Last reply by: Dr. William Murray

Mon Jun 23, 2014 7:19 PM

Post by Sitora Muhamedova on June 21, 2014

In e.g # 5 calculate the conditional probability

you mentioned that the p(Aâˆ©B)=P(34)= 1/36

How come you ignore the other cases as far as I know it should be p(34,43) = 2/36.

Please explain this situation

1 answer

Last reply by: Dr. William Murray

Mon Mar 31, 2014 11:02 PM

Post by Henry Thrift on March 28, 2014

I am not sure if you answered this question. I have completed this and wanted to check my work and also see the format you use to get the answer.

P[Sâˆ©M] 0.6 0.3 0.18

P[Sâˆ©M] = P[S|M]*P[M]

P[Sâˆ©M] = .30*.60

P[Sâˆ©M] = 0.18

P[Sâˆ©F] 0.35 0.4 0.14

P[Sâˆ©F] = P[S|F]*P[F]

P[Sâˆ©F] = .35*..40

P[Sâˆ©F] = 0.14

0.6 0.18 0.42

0.4 0.14 0.26

1 answer

Last reply by: Dr. William Murray

Mon Mar 17, 2014 1:53 PM

Post by Henry Thrift on March 17, 2014

Do you use Joint and Marginal Probability Tables:

On Campus, 30% of the men smoke, and 35% of the women smoke. If our campus is 60% men and 40% women, construct the joint probability table.

S

NS

M

F