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Lecture Comments (6)

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(x≤5)—P(x≤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?

Binomial Distribution (Bernoulli Trials)

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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 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

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 down2679

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 deviation2913

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 0.8943100

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