For more information, please see full course syllabus of Probability

For more information, please see full course syllabus of Probability

### Binomial Distribution (Bernoulli Trials)

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
- Binomial Distribution
- Binomial Distribution (Bernoulli Trials) Overview
- Prototypical Examples: Flipping a Coin n Times
- Process with Two Outcomes: Games Between Teams
- Process with Two Outcomes: Rolling a Die to Get a 6
- Formula for the Binomial Distribution
- Key Properties of the Binomial Distribution
- Example I: Games Between Teams
- Example II: Exam Score
- Example III: Expected Grade & Standard Deviation
- Example IV: Pogo-sticking Championship, Part A
- Example IV: Pogo-sticking Championship, Part B
- Example V: Expected Championships Winning & Standard Deviation

- Intro 0:00
- Binomial Distribution 0:29
- Binomial Distribution (Bernoulli Trials) Overview
- Prototypical Examples: Flipping a Coin n Times
- Process with Two Outcomes: Games Between Teams
- Process with Two Outcomes: Rolling a Die to Get a 6
- Formula for the Binomial Distribution 3:45
- Fixed Parameters
- Formula for the Binomial Distribution
- Key Properties of the Binomial Distribution 9:54
- Mean
- Variance
- Standard Deviation
- Example I: Games Between Teams 11:36
- Example II: Exam Score 17:01
- Example III: Expected Grade & Standard Deviation 25:59
- Example IV: Pogo-sticking Championship, Part A 33:25
- Example IV: Pogo-sticking Championship, Part B 38:24
- Example V: Expected Championships Winning & Standard Deviation 45:22

### Introduction to Probability Online Course

### Transcription: Binomial Distribution (Bernoulli Trials)

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

*We are starting a chapter now of working through our discrete distributions.*0006

*The first we are going to study is the binomial distribution.*0012

*This is also called the Bernoulli trials.*0016

*If you are studying Bernoulli trials in your probability course then you are using the binomial distribution.*0020

*These are synonymous terms for the same idea, let us learn what that idea is.*0026

*Before I give you the formulas for the binomial distribution, I want to tell you the general setting.*0029

*It is very important to be able to recognize this setting.*0037

*When you get some random problem, you have to figure out is this a binomial distribution, is this a geometric distribution?*0040

*Let me tell you the setting for the binomial distribution.*0046

*It describes a sequence of N independent tests, each one of which can have 2 outcomes.*0049

*You can think of running a test N times and each time you can either succeed or fail.*0056

*Every time there is 2 outcomes, that is why it is called binomial, success or failure.*0061

*It is also known as Bernoulli trials, as I mentioned.*0066

*The technical example of the binomial distribution or a Bernoulli trial is flipping a coin.*0069

*You want to think of flipping a coin exactly N times in a row.*0075

*By the way, N is always constant for the binomial distribution or for Bernoulli trials.*0079

*You always talk about N being a fixed number.*0085

*That is different from some of the distributions we are going to encounter later like the geometric distribution.*0089

*For the binomial distribution, N is always constant.*0094

*We want to think about it, like I said, flipping a coin is the prototypical example.*0097

*Although, that is somewhat limiting because people often think of the coin as being fair,*0103

*meaning it got a 50-50 chance of coming up heads or tails.*0108

*That certainly is a binomial distribution but you can also use a binomial distribution,*0113

*even when the probabilities are not even like that.*0120

*For example, if your coin is loaded, it is more likely to come up heads than tails, that is still a binomial distribution.*0123

*We will see how we adjust the formulas to account for that.*0130

*You can also think about any other kind of situation where you either have success or failure.*0133

*For example, one sports team is going to play another sports team and each time your home team will either win or lose.*0139

*That is a binomial distribution.*0147

*If you say we are going to play the other team 15 × and each time we will win or lose.*0149

*At the end, we will have a string of wins and losses.*0155

*That is a binomial distribution, that is a set of Bernoulli trials.*0158

*You can think of, for example, rolling a dice.*0162

*You think of, wait a second, there are 6 different things that can happen when I roll a dice not just 2.*0165

*Suppose you are only interested in whether the dice comes up 6 or not.*0171

*If you roll a 6, it is a success and somebody pays you some money.*0176

*If you do not roll a 6, it is a failure.*0180

*1 through 5 essentially can sort of want those altogether and count those all as a single category of failure and rolling a 6 is a success.*0182

*Essentially, rolling a dice just boils down to you roll it, do I get a 6, do I not get a 6?*0193

*That is again, a binomial distribution.*0198

*You can think of that as a set of Bernoulli trials.*0201

*The probability of success if it is a fair dice, there is just 1/6 and the probability of failure is 5/6.*0204

*There is all these different situations, they all come down to studying the binomial distribution.*0213

*They already come down to the same formulas.*0220

*Let us go ahead and take a look at the formulas that you get for the binomial distribution.*0222

*You have several parameters, they go into this.*0227

*You start out with the number of trials.*0232

*I want to emphasize that that N is fixed.*0235

*That is different from some of the other distributions that we are going to study later.*0239

*In particular, the geometric distribution that we will study in the very next video, that it is different from this,*0243

*and you keep flipping a coin until you get a head and that could take indefinitely long.*0251

*The binomial distribution is not like that.*0258

*You say ahead of time that I’m guaranteed I'm going to flip this coin N times,*0261

*or we are going to play the other team N times.*0265

*It is fixed, it stays constant throughout the experiment and you know that ahead of time.*0268

*P is the probability of success on any given, looks like we got cutoff a little bit there.*0276

*Let me just fill that in here.*0283

*On any given trial, *0285

*If you are dealing with a fair coin then P would be ½.*0288

*If you are dealing with a sports team playing another sports team, it depends on the relative strength of the teams.*0295

*Maybe if your team is the underdog, maybe they only win 1/3 of the games then P would be 1/3.*0302

*That is your chance of winning any particular game with the other team.*0311

*Maybe, if you are rolling a dice and you are trying to get a 6 then your probability of getting a 6 would be 1 and 6.*0316

*Your probability of failure, now, we are going to call that Q but Q is not really very difficult to figure out *0325

*because that is the probability of failing on any given trial, that is just 1 – P.*0334

*The probability of failure is, we call it Q but sometimes we will swap back and forth and alternate between Q and 1 – P.*0341

*They really mean the same thing.*0349

*If you are flipping a coin and it is a fair coin then your probability of failing to get a head or getting a tail would be ½.*0351

*If you have a sports team and we said the probability of winning any given match is 1/3 because we are the underdogs here,*0359

*that means our probability of losing is 2/3.*0368

*If we are rolling a dice and we are trying to get a 6, anything else is considered failure.*0372

*The probability of getting anything other than a 6 is 5 out of 6 there.*0378

*It is very easy to fill in Q, that is just 1 – P.*0384

*The formula for probability distribution, there are some terrible notation here and I do not like to use it *0388

*but it is kind of universal in the probability textbooks.*0395

*We are forced to deal with it.*0399

*This P of Y here, what that represents is the probability of Y successes.*0401

*Let me write that in.*0407

*That is the probability of exactly Y successes.*0409

*If you are flipping a coin N ×, this is the probability that you will get exactly Y heads.*0424

*If your sports team is going to play the other team N ×, this is the probability that you will win exactly Y games.*0430

*If you are going to roll a dice N ×, this is the probability that you will get exactly Y successes.*0439

*The formula that we have here is N choose Y, that is not a fraction there.*0447

*That is really N choose Y.*0452

*The other notation that we have for that is the binomial coefficient notation C of NY.*0454

*The actual way you calculate that is as N! ÷ Y! × N – Y!.*0460

*It is not just a fraction N/Y, that is really the formula for combinations.*0470

*N!/Y! × N – Y!.*0477

*The rest of this formula here, we have P ⁺Y and here is the really unpleasant part here.*0483

*This P right here is not the same as the P on the left hand side.*0488

*I said the P on the left hand side is the probability of exactly Y successes.*0493

*This P right here is this P right here, it is the probability of getting a success on any given trial.*0499

*It is this P up here, whatever the probability of successes on any given trial, that is what you fill in for this P.*0510

*That is the really unfortunate notation that you see with the binomial distribution, *0518

*is that they use P for 2 different things in the same formula.*0523

*I think that is really a high crime there but I was not given the choice to make up the notation myself.*0527

*I do not want to mislead you by using different notation from everybody else in the world.*0536

*We are kind of stuck with that.*0541

*Just be careful there that that P and that P are 2 different uses of the word P.*0543

*The problem with probability is everything starts with P.*0548

*We end up using the variable P in many places.*0552

*We have the exponent Y, we have Q.*0556

*Remember, we said Q is just 1 – P, that is easy to fill in and N – Y,*0559

*that is our formula, and then let us think about the range of values that Y could be.*0564

*If you are going to flip a coin N ×, how many heads could you get?*0569

*The fewest heads you can get would be 0, if you do not get any heads at all.*0573

*The most heads you can get would be N heads.*0577

*Our range of possibilities for Y is from 0 to N.*0580

*That is a formula we will be using over and over again.*0585

*There are another couple of issues we need to straighten out, before we jump into the examples.*0589

*Let us take a look at those.*0594

*The key properties of a binomial distribution and we will need to know these properties for every distribution we encounter.*0596

*The binomial distribution is just the first one, but we will be getting into the geometric distribution and the Poisson distribution,*0603

*all these other distributions later on.*0613

*For every single one, we want to know the mean, variance, and the standard deviation.*0615

*Here they are for the binomial distribution.*0619

*The mean, which is also known as the expected value means the exact same thing, mean and expected value.*0622

*There are 2 different notations for it.*0629

*We use the Greek letter μ for mean and we also say E of Y for expected value.*0634

*Those are really the same thing.*0641

*People also some× say the average.*0643

*All those things are essentially should be used synonymously.*0644

*But it is N × P.*0649

*Remember, N is the number of trials and P is the probability of success on any given trial.*0651

*The variance, two different notations for that, V of Y and σ² is N × P × Q.*0657

*Remember, Q is just 1 – P.*0665

*Sometimes you will see that written as N × P × 1 – P but they mean the same thing.*0668

*Standard deviation is always just the square root of the variance.*0673

*If you tell me the variance, I can always calculate the standard deviation very easily.*0678

*Just take the square root of the variance and that is the square root of N × P × Q.*0685

*We will be calculating those in some of our examples.*0691

*Let us go ahead and get started with those.*0694

*In our first example, we have the Los Angeles angels are going to play the Tasmanian devils in a 5 game series.*0698

*Maybe this is football or baseball, let us say baseball.*0706

*The angels have a 1/3 chance of winning any given game.*0711

*I guess the Tasmanian devils are bit stronger than the Angels.*0714

*What is the chance that the Angels will win exactly 3 games here?*0718

*Let me write down our general formula for the binomial distribution.*0724

*The general formula for binomial distribution is P of Y is equal to N choose Y, binomial coefficient there, combinations N choose Y.*0730

*P ⁺Y × Q ⁺N – Y.*0741

*Let me fill in everything I know here.*0747

*The Y I’m interested in is 3 games because I want to find the chance that they are going to win exactly 3 games.*0750

*My Y will be 3, my P is the probability that the Angels will win any particular game and that is 1/3 there.*0757

*My N is the number of games that they are going to play in total.*0769

*It is a 5 game series, that is where I’m getting that from, N = 5 and Q is just 1 – P.*0773

*Q is 1 - P which is 2/3, 1 -1/3 there.*0782

*The probability of 3, probability of winning 3 games is 5 choose 3 N choose Y 5 choose 3 × 1/3³ × 2/3⁵ – Y.*0788

*Be careful here, it is rather seductive to get your binomial coefficients and your fractions mix up here *0812

*because we are mixing them both on the same formula.*0820

*The fractions are 1/3 and 2/3, that 5/3 is a binomial coefficient, it is not a fraction.*0823

*We want to expand that out as a binomial coefficient, 5!/3! × 2! × 1/3.*0830

*I got 1/3³ and then 1/3³ × 2/3⁵ – 3, that is 2/3².*0843

*5!/3! that means the 1, 2, 3, will cancel out.*0856

*We would just get 5 × 4/2 ×,*0861

*Let us see, I’m going to have 3²/5 in the denominator here because I got 3 copies of 1/3 *0866

*and then 2 more here and 2² in the numerator.*0874

*5 × 4/2 that is 20/2 is 10 × 2²/3⁵.*0880

*10 × 2², 2² is 4, 10 × 4 is 40.*0891

*3⁵, 3 × 3 is 9 × 3 is 27 × 3 is 81 × 3 is 243.*0896

*That is my exact probability of winning exactly 3 games.*0918

*443 is just about 6 × 40 because 240 is 6 × 40.*0923

*This is very close to 1/6, if you want to make an approximation there.*0929

*The exact value would be 40/243.*0934

*The wraps that one up, let us just see how we solve that.*0944

*I used my generic formula for the binomial distribution.*0948

*The probability of exactly Y successes is N choose Y × P ⁺Y × Q ⁺N – Y.*0952

*I’m going to fill in all the numbers that I know here.*0961

*My N came from the fact that it was a 5 game series.*0963

*My Y is the number of game that I want to win, that was the 3 here.*0967

*The 1/3 is the little P, that is the probability that I will win any particular game.*0974

*My Q is just 1 – P, that is 2/3.*0983

*I drop all those numbers in here, very careful, the 5 choose 3 is a binomial coefficient.*0986

*It is a combination, it is not a fraction.*0992

*I simplify these fractions down while I’m simplifying down the binomial coefficient there.*0999

*Doing the arithmetic, it simplifies down to 40/243 which I noticed is approximately equal to 1/6.*1007

*That is my probability of winning exactly 3 games out of this 5 game series.*1014

*In example 2 here, we got a big exam coming up and we studied most of the material but not all of it.*1023

*In fact any given problem, we have a ¾ chance of getting that problem right.*1031

*Most likely, we will get a problem right but not guaranteed.*1036

*The exam that we are going to take is 10 problems and I guess we are really hoping to get an 80% score or better.*1039

*I would like to score 80% on this exam but I really only studied ¾ of the problems.*1049

*This is really a binomial distribution problem.*1055

*Remember that you use the binomial distribution, when you have a sequence of trials and each trial ends in success or failure.*1060

*How does this correspond to that?*1069

*We have 10 problems here, each problem we will do our best to solve it and will either succeed or fail.*1071

*It is exactly 10 problems, each one is N success or failure.*1077

*That is definitely a binomial distribution.*1081

*Let me go ahead and set up the generic formula for binomial distribution.*1085

*The probability of getting exactly Y successes is N choose Y × P ⁺Y × Q ⁺N – Y.*1088

*In this case, let me fill in here, while my N is the number of trials here.*1102

*That is the number of problems I will be struggling with, N is 10.*1107

*P is my probability of getting any particular problem right.*1114

*We were told that that is ¾.*1121

*My Y is the number of successes that I would like to have.*1124

*In this case, I want to score 80% or better which means out of 10 problems, I got to get 8 of them, or 9 of them, or 10 of them right.*1129

*Our Y, in turn be 8, 9, and 10.*1137

*We have several calculations here.*1143

*The Q, remember is always 1 – P, that is our chance of failure on any given problem.*1148

*1 – P, 1 -3/4 is ¼, if you give me a single problem, that is the chance I will not get it right.*1155

*Since, I need to find the probability of getting exactly 8, 9, or exactly 10 problems.*1166

*I will be adding up 3 different quantities here, P of 8.*1173

*I will give myself some space because I do not get a little bit messy.*1177

*+ P of 9 + P of getting exactly 10 problems right.*1181

*I will workout each one of those.*1188

*P of 8, just dropping Y=8 into this formula, is 10 choose 8, N was 10 × P ⁺Y.*1190

*P is ¾ ⁺Y is 8 × Q was ¼ ⁺N- Y.*1200

*N was 10 so N – 8 is 2 + P of 9 that is 10 choose 9 × ¾⁹ × ¼¹ + P of 10 is 10 choose 10 × 3/3 ⁺10 × ¼ ⁺10 – 10 which is 0.*1208

*I want to simplify that, these numbers are going to get a bit messy.*1239

*At some point, I’m going to throw out my hands in despair and just go to the calculator.*1242

*Let me simplify a bit on paper first.*1246

*In particular, these binomial coefficients, I know how to simplify those.*1250

*Remember, you are not going to mix up the fractions.*1254

*This 10 choose 8, that is 10!/8!/10 -8! = 2!.*1257

*The 10! And 8! Cancel each other just leaving 10 × 9/2.*1267

*That is 45 there, 45 × 3⁸/4 ⁺10 because we have 8 factors of 3 and 8 factors of 4 and then 2 more factors of 4.*1273

*10 choose 9 here is 10!/9! × 1!.*1293

*That is just 10!/9! which is all the factors are cancel except the 10 there.*1301

*10 × 3⁹/4 ⁺10.*1310

*Finally, we had 10 choose 10.*1319

*There is only one way to choose 10 things out of 10 possibilities.*1323

*In case you want to confirm that to the formula, it is 10! ÷ 10! × 10 - 10 is 0!.*1327

*But 0! Is just 1, remember, so that is 1.*1335

*That is 1 × ¾ ⁺10, 3 ⁺10/4 ⁺10, that ¼⁰ is just 1, that does not do anything.*1341

*At this point, I do not think the numbers are going to get any nicer by trying to simplify them as fractions.*1351

*I’m going to go ahead and threw these numbers in, all these numbers to my calculator.*1359

*Let me show you what I got for each one of those.*1363

*For the first one, I got the 0.2816 +, in the second one I got 0.1877 +, in the last part I got 0.0563.*1365

*What these really represent these 3 numbers right now, represent your probabilities *1383

*of getting exactly 8 problems right, that is P of 8 right there.*1388

*The probability that you get exactly 8 problems right, you score exactly 80% on the test.*1394

*This is probability of getting 90% on the test so you got exactly 9 problems right.*1399

*This is your probability of getting 100%, getting all 10 problems right.*1405

*Not very likely, you got 5% chance every single test, if you are only ready with ¾ of the material going in.*1408

*If we add those up, 28 + 18 + 5 turns out to be, I did this on my calculator, 52.56 is approximately,*1419

*52.5%, I will round that up to 53%, that is your probability of getting 80% or more on this exam.*1433

*You studied ¾ of the material, your probability of getting 80% on the exam is 53%.*1445

*Let me show you how I derived that.*1453

*I started with the basic formula for the binomial distribution, here it is.*1456

*And then, I filled in all the quantities I know.*1461

*The N = 10 that come from the stem of the problem.*1464

*The P, the probability of getting any problem right is ¾, that also comes from the stem of the problem.*1469

*The Y that we are interested in, we want to get 80% or better, that means that we want to get 8 problems, *1480

*or 9 problems, or 10 problems right.*1488

*Because if you are shooting for 80% and if you end up getting 90 or 100, that certainly is acceptable.*1491

*We have to add up all those different possibilities.*1497

*The Q is always 1 – P, since P was ¾, Q was ¼.*1500

*We just drop those in for the different values of Y, the 8, 9, and 10.*1506

*We get these binomial coefficients and some fairly nasty fractions which I did not want to simplify by hand.*1512

*We sorted out the binomial coefficients into 45 and 10 and 1.*1520

*Each one of those multiplied by some fractions gave me some percentages,*1525

*the probabilities of getting 8 problems, 9 problems, 10 problems right.*1530

*We would put those all together and we get a total probability of 53%.*1534

*If you are shooting for an 80% on an exam and you study 3/4 of the material, your chance of getting that 80% is 53%.*1540

*You are likely get 80% but it is definitely not a sure thing.*1552

*You might want to study a little more there.*1556

*Example 3, we are going to keep going with that same exam from example 2.*1561

*It is telling us that each problem is worth 10 points, what is your expected grade and your standard deviation?*1565

*Remember, expected grade is a technical term.*1573

*As soon as you see the word expected in a probability problem, that does not mean the English meaning of the word expected.*1575

*That does not mean what grade you are going to get but that means on average, what your grade be if you take this exam many times.*1586

*What is your average going to be?*1596

*It is asking for the expected value of your exam score.*1599

*We have learned the formula for the expected value of the binomial distribution.*1605

*That is the same as the mean of the binomial distribution.*1610

*The formula we learned this back on the third slide of this lecture, it is N × P.*1613

*In this case, the N was 10 and the P is the probability of you getting any particular problem right, that was ¾.*1620

*The expected value of Y, I’m using Y here to mean the number of problems that you get right.*1629

*I should probably clarify that a little earlier.*1639

*Problems you get right.*1642

*We just figured out down below that that is 10 × ¾ is 7 ½.*1651

*That 7 ½ problems but each problem is 10 points.*1658

*That is 75 points on the exam, that is your expecting great.*1667

*Remember, I said that that is a technical term, that is you are expected grade.*1675

*In real life, there is no way you can get a 75 on the exam because all the problems are worth 10 points each.*1681

*In real life, when you take a single exam, you will have to get a multiple of 10.*1690

*You might get a 60% on the exam, a 70, 80, etc.*1699

*You will not get a 75 on the exam, I guarantee you because we are not talking about partial credit here.*1707

*Your actual score would be 60, 70, 80, or so on.*1715

*What I mean when I say that your expected grade is 75, what I mean is that if you take this exam many times,*1721

*or if you take many exams, your average over the long run will be 75 points.*1729

*Maybe, for example if you take 2 exams, your total on the 2 exams might be 150 which means you are averaging 75 points per exam,*1752

*even though you are not going to get exactly 75 points on any exams here.*1761

*That was your expecting grade, your standard deviation, a good steppingstone to calculating that is to find the variance first.*1766

*Let us find the variance of Y, variance of your score.*1774

*Variance, we learned the formula for that, it was also on the third slide of this video, NPQ.*1778

*Our N here is 10, our P is ¾, and our Q is ¼.*1786

*Remember, Q is always 1 – P.*1793

*If we simplify that, we get 30/16 that is not extremely revealing at this point.*1796

*But remember, that was just the variance, that is not our standard deviation.*1805

*To get the standard deviation, you take the square root of the variance.*1808

*Our σ is the square root of the variance, that is always true.*1813

*It is √ 38/16 and I can simplify that a bit into √30 on top and √16 is just 4.*1819

*It does not really do anything good after that, I just threw it into a calculator.*1833

*What it came back with was that that is approximately equal to 1.369 problems.*1839

*Our unit here is the problem because Y was measured in the number of problems that we get right, 1.369 problems. *1848

*Our standard deviation in terms of points on the exam, that is equal to 13.69 points because each problem was worth 10 points.*1857

*Our σ there is 13, it is approximately equal to 13.69 points on the exam.*1876

*You can estimate your score on the exam, your expected grade would be 75 points.*1890

*Your standard deviation as you take many exams will be 13.69 points up above and below 75 points.*1896

*Let me recap how we calculated that.*1906

*This really came back to remembering the formulas from the third slide that we had earlier on the videos.*1909

*If you do not remember those, just go back, check them out on the third slide of this video and you will see them.*1915

*The expected value is NP, the variance is NPQ.*1921

*Now, I’m just dropping in our values for N.*1927

*N is 10, P is ¾, our Q is 1 - P is ¼.*1931

*That tells us the expected value and the variance, in terms of the problems on the exam *1939

*because we define our random variable in terms of the problems that we expect to get right.*1946

*To convert into actual points on the exam, we multiplied by 10 because each problem is worth 10 points.*1953

*7 1/2 problems I expect to get 7 1/2 problems right that means I expect on average to get 75 points on the exam.*1960

*I will never get exactly 75 because with 10 point problems, my score will definitely be some multiple of 10.*1968

*But on average, if I take many exams, I will get 75 points.*1976

*The variance, drop in those numbers I get 30/16, that is the variance not the standard deviation.*1981

*To get the standard deviation, you take the √ of that and that simplifies down into 1.369 problems.*1988

*Converting that to points gives me a standard deviation of 13.69 points on this exam.*1996

*In example 4, we are going watch the heralded Long Beach jack rabbits play and *2008

*they are going to be playing in the world pogo sticking championship.*2015

*Apparently, they are very good at this, as you expect jack rabbits to be.*2019

*Each year they have an 80% chance of winning the world championships.*2025

*They are obviously the dominant force in the world pogo sticking championships.*2029

*The question is we want to find the probability that they will win exactly 5 × in the next 7 years.*2035

*7 years of championship, they will play every year, each year they got 80% chance.*2042

*We also want to find the probability that they will win at least 5 × in the next 7 years.*2048

*We are going to calculate both those.*2056

*We need a little more space for this.*2057

*I put an extra slide in here for us to work these out.*2059

*This is still the example of the long beach poly jack rabbits in their pogo sticking championship.*2064

*We are going to be playing 7 championships here.*2071

*Again, this is really a Bernoulli trial.*2077

*Why is this a Bernoulli trial, it is because we are playing 7 championships.*2080

*Each year we will win or we will lose.*2084

*We have a probability of winning each year or losing each year.*2087

*Let me fill in the generic formula for Bernoulli trials.*2091

*P of getting exactly Y successes is N choose Y × P ⁺Y × Q ⁺N – Y.*2095

*That is our generic binomial distribution formula.*2110

*Let me fill in whatever values I can.*2116

*The N here is the number of trials that we are doing here.*2118

*We are going to track this over 7 years so our N is 7.*2122

*P is our probability of success on any given trial.*2129

*That is the probability that the jack rabbits will win in any given year.*2134

*We said that that is 80%.*2138

*I will give that as a fraction, I will try to work this not using fractions, that is 4/5.*2139

*Q is always 1 – P so that is 1 - 4/5.*2147

*In this case, that is 1/5.*2155

*Finally, what is the Y value that we will be interested in here?*2157

*Our Y value, we want to win exactly 5 × for the first part of the problem.*2162

*We want to find the probability of exactly Y successes.*2169

*In the second part of the problem, we want to win at least 5 ×.*2173

*That means we could win 5 ×, we could win 6 ×, we could win 7 ×.*2176

*Let us calculate all of those.*2182

*P of 5, when I plug in Y = 5 here, that is 7 choose 5 × P is 4/5 ⁺Y is 5 × Q is 1/5 ⁺N – Y, that 7 – 5 is 2.*2185

*Now, I just have to expand and simplify these fractions.*2216

*7 choose 5 that is not 7 ÷ 5.*2220

*7 choose 5 is 7!/5! × 2!.*2224

*Let me cancel the factorials.*2234

*That is just 7 × 6 because the 5! takes care of all the other factors, ÷ 2!.*2236

*That is equal to, 7 × 6, 6/2 is 3, 7 × 3 is 21.*2247

*I still have 4⁵/5⁷ because there are five 5 in the first fraction and two 5 in the second fraction.*2255

*What will we get here is 21 × 4⁵ ÷ 5⁷.*2274

*That does not really turn out to be any particularly interesting numbers.*2287

*I just left that as a fraction, I did not bother to plug that into my calculator but that is our answer to part A.*2291

*That is the probability that the jack rabbits will come home with exactly 5 championships within the next 7 years.*2298

*In part B, we want to get at least 5 championships.*2306

*That means we really want to figure out the probability of getting 5 or 6, or7 championships.*2311

*We figured out the probability of 5 already, let us find the probability of 6.*2319

*We use a same formula except we put in Y equal 6, so 7 choose 6, 4/5⁶, 1/5⁷ -6 that is 1 there.*2324

*7 choose 6 is 7!/6! × 1!.*2342

*And then we have of 4⁶/5⁶ × 5 ⁺15⁷.*2349

*7!/6! Is 7, that 7 × 4⁶ or 5⁷.*2358

*Not a particularly interesting number by itself.*2368

*Let me go ahead and figure out P of 7, the probability of winning all 7 matches.*2372

*I’m going to use the binomial distribution formula.*2377

*Although, it might be a little easier if you think about this directly but I want to practice the formula.*2382

*It is 7 choose 7 × 4/5⁷ × 1/5⁷ -7 which is 0.*2386

*7 choose 7 is 7!/7! × 0!.*2401

*We just have a 4/5⁷ because the 1/5⁰ is just 1.*2407

*7!/7! Is just 1, we get 4⁷/5⁷.*2415

*It would have been easier to think about that as saying we have a 4/4 chance of winning.*2421

*In order to win 7 matches, we have to win all 7 × in a row.*2428

*It is 4/5 × 4/4, 4/5⁷.*2433

*That is probably an easier way to get there more directly but I just want to practice using the probability distribution formula.*2438

*The probability that we will win at least 5 matches, you just add up those 3 numbers.*2447

*The probability of 5 + the probability of 6 + the probability of 7.*2455

*The fractions are actually fun to workout here, I did work them out.*2462

*I will work them out, 21 × 4⁵/5⁷.*2466

*P of 6 is 7 × 4⁶/5⁷.*2473

*P of 7 is just 4⁷/5⁷.*2480

*We can factor out that 5⁷ in the denominator.*2485

*Also, in the numerator we got lots of 4 and we factor out 4⁵/5⁷.*2488

*The numerator, I still have a 21 + 7 × 4 because 4⁶ is 4⁵ × 4 + 4².*2496

*7 × 4 is 28, 4² is 16.*2510

*This is 4⁵/5⁷.*2516

*21 + 28 is 49 + 16 is 65.*2521

*But 65 is 5 × 13, we cancel out one of those 5’s and we get 4⁵ × 13/5⁶ *2530

*because one of the 5 was canceled with the 5 from the 65.*2546

*We divide from the numerator there.*2551

*That is our probability of winning at least 5 games or 5 championships over the next 7 years.*2556

*Let me recap here.*2565

*This is kind of a classic binomial distribution problem, classic Bernoulli trials.*2567

*It does not have to be coin flipping even though people always talk about coin flipping.*2572

*In this case, it is a question of the Long Beach jack rabbits in the pogo sticking championships.*2576

*Every year, they either win or they lose.*2582

*They win with a probability of 4/5.*2586

*They lose with a probability of 1/5.*2590

*They will play for 7 years, that is why we have our N = 7.*2593

*The question was asking us what our chance of winning exactly 5 × is, that is where we get that 5.*2599

*And then later on, we want the probability of winning at least 5 ×.*2607

*We are going to calculate 6 and 7 as well.*2611

*What we are really doing here is dropping 5, and 6, and 7 in for Y into the binomial distribution formula.*2614

*I just dropped Y = 5, that 5 in the denominator came from the probability but that came from Y there and that 2 was 7 – Y.*2623

*And then, I simplified all the fractions and I got that is my probability of winning exactly 5 championships.*2638

*I found the probability of winning 6 championships the same way by running all the way through Y = 6, then I ran through Y=7.*2646

*There is an easier way to calculate that but I want to practice the binomial distribution formula.*2655

*Add all those together to find the probability of winning at least 5 × because it is at least 5 ×, *2663

*that is why you put a greater than or equal to.*2669

*We got to check 5, 6, and 7.*2672

*We have a common denominator 5⁷ on all of these.*2675

*These numbers, you can do some nice factoring, factor out 4⁵ and simplify everything down*2679

*and you get still bit of a cumbersome number but that tell us our probability of winning at least 5 × in 7 years.*2686

*I want you to hang onto the numbers from this example because in example 5,*2696

*which we are about to do refers back to this example.*2699

*It is the same scenario with the long beach poly jack rabbits playing in the pogo sticking championship.*2704

*I think they are going to play for a different number of years but the probability will be the same.*2711

*Let us check that out but remember the numbers from this example.*2716

*In example 5, again, each year the Long Beach jackrabbit has an 80% chance of winning the world pogo sticking championship.*2723

*We want to find the expected number of championships that they will win in the next 5 years and the standard deviation in that total.*2730

*Let us calculate that out.*2740

*I’m going to use my generic formulas for expected value and variance, and standard deviation.*2741

*You can find those formulas on the third slide in this lecture.*2748

*If you do not remember those formulas, where they come from, *2753

*just check back in the third slide of this lecture and you will see the following formulas.*2755

*The expected value of Y is N × P, this is for the binomial distribution.*2761

*The variance of Y is NPQ.*2768

*The standard deviation which is always just the square root of the variance.*2774

*That makes it in this case, the square root of NPQ.*2781

*We are going to calculate each one of those for this particular problem.*2786

*N, remember is the number of trials that we are running.*2790

*In the previous example, we are running this over 7 years but now we are just running it over 5 years.*2793

*I’m getting that from right here, that is my N, N = 5.*2799

*The P here is the probability of winning any particular year and we are given that that is 80%.*2805

*As a fraction, that is 4/5.*2811

*I’m not going to go ahead and figure out what Q is.*2814

*Remember, Q is always 1 – P.*2819

*That is easy, that is 1 - 4/5 is 1/5.*2821

*I think that is all I need to know for this one.*2827

*Let us go ahead and calculate these out.*2830

*The expected value, the expected number of championships that they will win over the next 5 years is N × P.*2833

*N is 5, P is 4/5, that is just 4 championships.*2844

*That of course should not be at all surprising because we are going to play for 5 years in a row.*2853

*We have an 80% chance of winning in any given year.*2864

*We expect to win 4/5 of the years.*2867

*If we play for 5 years, we expect to win 4 out of 5 of those years on average.*2871

*That is very intuitive result there but it is good that it is backed up by the formulas because probability can sometimes be counterintuitive.*2877

*Let us go and find the variance.*2886

*We are not being asked the variance in the problem but it is kind of a steppingstone to finding the standard deviation.*2888

*It was worth finding the variance.*2893

*NPQ is 5 × 4/5 × 1/5.*2895

*The 5 and 1/5 cancel, that is just 4/5.*2905

*That is the variance, that is not our full answer yet.*2910

*To get our full answer, we want to find the standard deviation*2913

*which is the square root of the quantity that we found above, √NPQ, which is √4/5.*2916

*I could distribute that square root in the numerator and get 2/√5.*2929

*Not a very enlightening number, I did go ahead and plug that into my calculator.*2934

*What my calculator told me was that that is 0.894.*2939

*That is the standard deviation in the number of championships we expect to win over a 5 year span.*2945

*I will go ahead and box that up because that is our final answer there.*2958

*Just to remind you where everything came from here.*2963

*I got these formulas for expected value of variance and standard deviation straight off the third slide of this lecture series.*2965

*You can just go back and look at those formulas.*2975

*Those do correspond to the binomial distribution.*2977

*Make sure you are working with the binomial distribution before you involve those formulas.*2980

*This one is a binomial distribution because what is happening is that,*2985

*the Long Beach jackrabbits are playing the championship year after year.*2990

*Each year they either win or they lose.*2994

*You can think of that as being almost like flipping a coin,*2997

*except that it is not a 50-50 coin because the jackrabbits are dominant *3000

*that every year the coin has an 80% chance of coming up heads.*3004

*Every year, they have an 80% chance of winning and only 20% chance of losing.*3008

*The expected value for binomial distribution we said was NP.*3015

*Variance is NPQ, the standard deviation is always the square root of the variance, square root of NPQ.*3020

*I’m just dropping in the numbers and get the numbers from the stem of the problem.*3027

*N is the number of trials that you are going to run.*3033

*In this case, that is the number of possible championships.*3036

*We are going to play for 5 years, that N = 5 come from right there from the stem of the problem.*3042

*P = 4/5 that comes from their chance of winning in any given year.*3049

*80% translated into a fraction is 4/5.*3057

*The Q is always 1 – P, that is 1 – 4/5 is 1/5.*3060

*We just plot those numbers right into our formulas, the expected value is NP.*3065

*It simplifies down to 4 championships which makes perfect intuitive sense.*3070

*If you are going to play for 5 years and you got an 80% chance of winning each year,*3075

*you expect to win about 4 championships on average.*3079

*The variance NPQ simplifies down to 4/5 but that is not what we want.*3084

*We want the standard deviation which is the square root of the variance, and that is 2 ÷ √5.*3091

*I just threw that number into my calculator and it spat out the number 0.894*3100

*is the standard deviation in the number of championships that we expect to win over any given 5 year span.*3107

*That wraps up this lecture on the binomial distribution.*3116

*This is part of the probability lecture series here on www.educator.com.*3119

*Next up, we will have the geometric distribution.*3123

*I hope you will stick around and learn about that.*3125

*It looks a bit like the binomial distribution but there are certain key issues where it is different.*3128

*In particular, you are not running a fixed number of trials anymore, you are running the trials over and over until you get a success.*3133

*That turned out to change the probability distribution and it is going to change our mean and variance, and so on.*3139

*We will look at that out in the next lecture, I hope you will stick around for that.*3145

*In the meantime, as I said, these are the probability lectures here on www.educator.com.*3149

*My name is Will Murray, thank you very much for watching, bye.*3154

3 answers

Last reply by: Dr. William Murray

Wed Aug 26, 2015 9:41 AM

Post by bo zhang on July 28, 2015

I mean when n equals to 7, i cannot using the table n equal to 10, because it is not a interval,in that case, we can only apply it with formulas, the tables only work when n equal to 5.10.15.20.25. is that right?

1 answer

Last reply by: Dr. William Murray

Thu Aug 13, 2015 8:40 PM

Post by bo zhang on July 28, 2015

Hi Dr,thanks for post the video,here is my question,sometimes,when we calculate the probabilities of binomial distribution,we can use the formulas or the tables,normally,i use both to make sure that my answers are correct, but in your example 4, the n equal to 7,and i get a correct answer when i input the formulas, but it does not work when i try the tables, for instance in example 4, p is 0.8, n is 7,if we want to find exactly 5 times,i try to use P(xâ‰¤5ï¼‰â€”P(xâ‰¤4), after i check the tables, i found out the number is 0.027 which is far from the correct number 0.27, i think the mistake occurs from n of tables, in my tables, there are N equal to 5,10,15,20,25, we can apply it only when n tends to those 5,10,15,20,25,if n equals to 7,which means we cannot directly using the tables, am i right? this n of tables mean single number or it means a interval?