For more information, please see full course syllabus of Probability

For more information, please see full course syllabus of Probability

### Experiments, Outcomes, Samples, Spaces, Events

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
- Terminology 0:19
- Experiment
- Outcome
- Sample Space
- Event
- Key Formula 2:47
- Formula for Finding the Probability of an Event
- Example: Drawing a Card
- Example I 5:01
- Experiment
- Outcomes
- Probability of the Event
- Example II 12:00
- Experiment
- Outcomes
- Probability of the Event
- Example III 16:33
- Experiment
- Outcomes
- Probability of the Event
- Example IV 21:20
- Experiment
- Outcomes
- Probability of the Event
- Example V 31:41
- Experiment
- Outcomes
- Probability of the Event
- Alternate Solution
- Example VI 43:33
- Experiment
- Outcomes
- Probability of the Event

### Introduction to Probability Online Course

### Transcription: Experiments, Outcomes, Samples, Spaces, Events

*Hi, welcome to www.eudcator.com, my name is Will Murray and*0000

*we are going to be starting the probability lectures today.*0004

*Thank you very much for joining us, we are going to jump right on in.*0007

*We are going to learn some terminology in this first lesson.*0010

*We are going to learn about experiments and outcomes and sample spaces and events.*0012

*Let us get right started with that.*0018

*The words that we are throwing at you are all used to describe probability experiments.*0020

*Let me first got introduce the word experiment itself.*0027

*An experiment is a process leading to exactly one of various possible outcomes.*0030

*We will see lots of examples of this as we go through the lectures.*0037

*But some example to keep in mind, a very simple example would be flipping a coin, or you roll a dice, *0040

*and you draw a card from a deck.*0048

*Those are the kinds of things that we can run in an experiment in exactly one of the possible outcomes will occur.*0050

*Having said that, I should say what an outcome is one of the things that can happen in an experiment.*0057

*Outcomes are also known as simple events or sample points.*0066

*All of those words are synonymous, we use them interchangeably.*0070

*A simple event, a sample point, or outcome.*0073

*If you put all the possible outcomes together that is called the sample space.*0077

*The sample space is essentially the set of all things that can happen in an experiment.*0083

*If you are going to flip one coin, for example the sample space is head and tail*0087

*because those are the 2 things that can happen.*0093

*You can get a head or a tail.*0094

*The sample space is also known as probability space.*0097

*Those words were used interchangeably, probability space or sample space.*0101

*If you are a studying probability, with your own textbook they might use the word sample space, *0106

*they might use the word probability space, those mean the same thing.*0111

*Finally, an event is a subset of the sample space that is a set of some of the outcomes.*0115

*An event means for example, if we are going to pick a card from a deck of cards, *0122

*the event could be that we get a spade because there are 52 cards in a standard deck and 13 of them are spades.*0128

*We collect all those 30 outcomes together and call them an event.*0137

*We often use the variable S to describe the set of all possible outcomes so the sample space.*0142

*We often use the variable A or sometimes A and B when we have multiple ones to describe different events in that sample space.*0149

*We will talk about the probability of an event A occurring when we run an experiment.*0158

*Let us figure out how we compute the probabilities.*0164

*When we do that, if we have an event is assuming that these are things that we can count and everything is nice and finite, *0168

*first of all we count the number of outcomes in the sample space S.*0179

*We count all the possible things that can happen in an experiment.*0184

*If you are flipping a coin, the only 2 things that can happen are head and tail.*0188

*You just count the head and the tail, you get 2 possible outcomes.*0193

*If you are drawing a card from a deck, there are 52 different things that can happen*0196

*because you can get the ace of spades or the 2 spades, up to the king of spades and so on.*0200

*There are 52 different cards that you might draw.*0205

*Those are all the outcomes.*0208

*To calculate the probability of an event, you want to count the number of outcomes in that event.*0210

*For example, if you are drawing a card from a 52 card deck and *0217

*you want to calculate the probability of getting a spade, P of getting a spade when you draw a single card from a deck.*0223

*There are 52 possible outcomes, there are 52 possible cards that you can draw, and 13 of them qualify as being in that event.*0237

*There are 13 possible outcomes which will give you spades.*0248

*Of course, you can reduce the fraction 13/52 and you find that the probability of getting a spade is ¼.*0253

*That is not very surprising.*0261

*Because of this of nature a probability, you are counting things all over the place.*0264

*There is a lot about counting outcomes, counting events.*0269

*A great deal of early focus in probability course is it is going to have can have a common notarial flavor, *0274

*when we try to count different events and count different sample spaces.*0281

*You will see that some of the problems that we study today and in some of the other early lectures*0285

*are all about counting different things, different complicated counting techniques.*0290

*You get the hang of it as we get into it, as we start practice it a little more.*0296

*Our first example is to roll 2 dice and to see if the sum showing on the 2 dice is 10.*0302

*Remember, these are standard dice so they are labeled 1 through 6 on the 6 sides.*0308

*I want to see if the sum is going to be a 10 between the 2 dice.*0313

*We want to identify the experiment.*0318

*We want to list all the possible outcomes and the event that we are interested in,*0321

*which means the subset of outcomes that would give us 10.*0325

*We want to find the probability of the event.*0330

*The probability that we are going to get a 10 when we roll 2 dice.*0331

*This is a pretty standard probability problem.*0336

*The experiment here is just roll the 2 dice.*0339

*We are rolling the dice, roll the 2 dice.*0347

*The outcomes are all the different possible ways that the 2 dice can call up.*0355

*And it is very helpful here if you have a mental image of the dice being distinct.*0365

*Maybe think of one of them as being red and one of them as being blue.*0372

*Think of you are rolling a red dice and a blue dice. *0376

*Let us think of all the different combinations that can arise there.*0379

*The red dye could be 1, 2, 3, 4, 5, and 6.*0384

*The blue dice could be 1, 2, 3, 4, 5, and 6.*0391

*They are all in the different possible combinations of the 2 dice there.*0401

*The red could be 1, the blue could be 1.*0405

*The red could be 1 and the blue could be 2.*0409

*The red could be 1 and the blue could be 3, and so on.*0414

*There are 6 possible combinations in that first roll there.*0419

*There are going to be 6 possible combinations in the 2nd roll, 3rd roll, 4th roll, 5th roll, all the way up to*0423

*you could get 6 on both sides.*0429

*You can get red dice being 6 and the blue dice being 6.*0431

*There are 36 possible outcomes in the set there.*0436

*Our sample space here is all those possible outcomes.*0443

*We can have a red 1 and a blue 1.*0447

*We can have a red 1 and a blue 2.*0451

*Let me see if I can make it look more blue there.*0461

*All the way up to a red 6 and a blue 6.*0465

*We have 36 possible outcomes in this experiment, 36 possible combinations of numbers on the 2 dice.*0474

*That is our sample space, all the possible outcomes.*0485

*We want to identify the event we are interested in.*0488

*The event here is that the 2 dice sum up to 10.*0492

*Let us think about which of those possible outcomes will give us a sum of 10.*0497

*You can get a red 4 and a blue 6.*0503

*You can get a red 5 and a blue 5, because those that up to 10.*0508

*Or you can get a red 6 and a blue 4, because those also add up to 10.*0515

*3 of those combinations between red and blue add up to 10.*0522

*That is the event we are interested in.*0529

*We now know what our event is and we now know what our sample space is.*0533

*Let us figure out what the probability is.*0538

*The probability of event A is the number of outcomes in A ÷ the total number of outcomes in our sample space.*0541

*The total number of things that can happen, outcomes in S.*0560

*And we calculated both of those, we looked at event A and we saw that there were 3 outcomes that will give you a sum of 10.*0568

*In S, there are 36 possible outcomes and that simplifies down into 1/12.*0576

*That is the probability of rolling a 10, when you roll a 2 dice together.*0585

*That is our final probability there.*0594

*Just to recap here, what we did was we identified our experiment which was rolling the 2 dice.*0600

*We listed all the possible outcomes, all the possible outcomes was this chart right here.*0607

*There are 36 possible outcomes, possible combinations of what might be showing on the red dice,*0613

*what might be showing on the blue dice.*0619

*I did not actually list all 36 because that will be too cumbersome but we know there will be 36 of them.*0622

*And then we have to identify the event which is that 2 dice add up to 10.*0628

*There are 3 possible outcomes, 3 of those combinations add up to 10.*0636

*We mention right here a place that often confuses students when I’m first starting to study probability, *0640

*people wonder that it looks like 4 or 6 got listed twice and 5 -5 only got listed once.*0646

*That is true, that is because there are 2 different ways you can get a 4 – 6.*0654

*You get a red 4 and blue 6 or you can get a red 6 and blue 4.*0659

*That will get listed twice.*0664

*5 and 5, there is only one way you can get it which is to get 5 on both of the dice.*0666

*There really are 2 ways to get a 4 and 6.*0673

*There is only one way to get a 5 and 5.*0676

*In fact, this is even reflected when people roll dice in casinos in the game craps,*0679

*people talk about getting 10 the hard way which means getting a 5 and 5 *0685

*because that is harder than getting a 4 and 6 because there are 2 ways to get a 4 and 6.*0690

*We listed all of them, we found 3 outcomes there.*0697

*The probability of our event is just the number of outcomes in the event and the number of outcomes ÷ *0700

*the number of outcomes in our sample space.*0707

*There are 3 outcomes in our events, we counted those.*0710

*There are 36 in our sample space, 3/36 reduces down to 1/12.*0714

*Let us keep going.*0721

*In our second example, we are going to flip a coin 3 times and see if we get exactly 2 heads.*0723

*And again, we want to identify the experiment, all the possible outcomes in the event we are interested in.*0729

*We want to find the probability of the event.*0735

*The experiment here is flipping the coin.*0739

*The experiment is to flip the coin 3 times, all the possible outcomes *0742

*or all the possible combinations of heads and tails that we can get.*0755

*The possible outcomes here, I’m just going to try to list them systematically.*0761

*Let me call this S because this is the sample space that we are listing right now.*0768

*S is we can get head, head, head.*0772

*We can get a head, head, and a tail.*0778

*We can get a head, a tail, and then a head.*0781

*A head, then a tail, then another tail.*0785

*We can get a tail-head-head.*0789

*We can get tail-head-tail.*0793

*Tail-tail-head or a tail-tail-tail. *0796

*There are 8 possible outcomes here and that is really coming from the fact that 2 × 2 × 2 is 8.*0802

*Because each time you flip the coin, there are 2 different things that can happen.*0812

*Then you multiply those possibilities as you go along, so you get 8 possible combinations of flips there.*0816

*Those are the outcomes that we are interested in, that is the entire sample space.*0823

*The event that we are interested in is that we get exactly 2 heads.*0830

*I want to figure out which of those outcomes satisfies that criterion.*0844

*Exactly 2 heads would mean certainly HHH, it does not but HHT does and HTH does, head- tail- head.*0849

*Down there I see a tail-head-head*0860

*Those are the only ones there are that have exactly 2 heads.*0866

*Our event just has those 3 possible outcomes that satisfy the requirement.*0872

*And then finally, we want to find the probability of the event.*0878

*The probability of A is the number of outcomes in A ÷ the total number of outcomes*0881

*in our sample space, in our probability space, outcomes in S.*0894

*In A, we had 3 possible outcomes and in S we had 8. *0904

*That does not reduce so our final answer there is 3/8.*0910

*That is the probability of getting exactly 2 heads, when we flip a coin 3 times.*0914

*Let we make sure that all the steps there are clear, we want to identify the experiments.*0920

*Identifying the experiment is just flipping the coin 3 times.*0925

*We wanted to list all the possible outcomes.*0929

*I listed them in order here, I kept track of the first flip, the second flip, and the third flip.*0933

*There are 8 possible outcomes.*0938

*That came from 2³ because there are two things that can happen on the first flip × 2 things*0943

*that can happen on the second flip × 2 things that can happen on the third flip.*0951

*Exactly 2 heads, I just scan through that list of outcomes and I have identified the ones that have exactly 2 heads.*0955

*The first one has 3 heads, that does not count.*0962

*The second one and the third one both have exactly 2 heads.*0964

*The 5th one here has exactly 2 heads.*0969

*And that is why I took those and put them into now the set, describing the event that we are interested in.*0972

*The probability of that event is just the number of outcomes in A which is 3 ÷ the number of outcomes in S which is 8.*0980

*That is where we got the probability being 3/8 there.*0989

*For our next example, we are going to be drawing cards from a deck.*0995

*We are going to take a standard 52 card deck and we are going to keep drawing cards until we find the ace of spades.*1000

*The question is what is the probability that it will take us between 20 and 30 cards, *1011

*including 20 and 30, that is what inclusive means.*1015

*When it says inclusive, it means you include those starting and ending values.*1018

*It says identify the experiment, all the outcomes, and the event that we are interested in.*1024

*The experiment here is drawing the cards.*1031

*We are drawing the cards, do not know how long it is going to last because we might get the ace of spades on the very first pick. *1037

*Or if we get unlucky, we might have to draw 51 cards and then get the ace of spades on the very last pick.*1047

*The set of possible outcomes really depends on where we get the ace of spades.*1054

*We could get the ace of spades on the first pick, if we are very lucky.*1061

*We could get it on the second pick, that would still be pretty lucky if we get the ace of spades on the second pick, *1070

*anywhere up to the 52nd pick.*1076

*It would be our worst case scenario, if we got the ace of spades as the very last card that we draw.*1080

*Remember, we said without replacement which means after we draw a card, we are not putting it back in the deck.*1087

*We know for sure that eventually I’m going to find the ace of spades.*1094

*If we are putting the cards back in the deck, I might potentially never find the ace of spades*1098

*because I could keep drawing forever and just keep drawing new cards.*1102

*The event here is that we are going to draw between 20 and 30 cards.*1106

*Our event would be that we get the ace of spades on the 20th pick or possibly on the 21st pick, *1112

*or all the way up to the 30th pick.*1130

*That is our event there, that is our A.*1135

*Generally, use A and B for events.*1140

*We use the letter S for the sample space.*1143

*The probability of A, that is what we are really trying to calculate here, *1146

*is the number of outcomes in A ÷ the number of outcomes in S.*1150

*We just need to calculate that.*1169

*The number of outcomes in S, we already said there is 1 pick up to 2 picks up to the 52nd pick.*1171

*There are 52 possible outcomes in S.*1177

*If we look at A, 20, 21, up to 30, there are 11 of those outcomes because we are counting the 20 and 30 on both ends.*1181

*It is 20, 21, 22,23, 24, 25, 26, 27, 28, 29, 30.*1190

*It is 11 outcomes in the event A.*1195

*We actually have 11 outcomes not 10.*1200

*It seems like there are 10 but there is extra 1 on the end.*1203

*That is our probability of drawing the ace of spades somewhere between the 20th card and the 30th card.*1207

*Let me go back over that and make sure everything is clear.*1218

*The experiment here is drawing the cards.*1221

*We are drawing the cards one by one.*1223

*We might just draw a single card.*1225

*We might end up drawing all the way to the 52nd card because we stop whenever we get the ace of spades.*1226

*And we are not putting cards back in the deck, that is the very key point to mention here.*1233

*The outcomes are that we could get the ace of spades on the first pick, the second pick, all the way up the 52nd pick.*1239

*That is 52 possible outcomes in our sample space.*1245

*The event that we are interested in, according to the problem is that we might get the ace of spades*1250

*on the 20th pick, the 21st pick, or all the way up through the 30th pick.*1257

*There are 11 of those possibilities that we are interested in to satisfy the requirements of the problem.*1263

*When we are calculating the probability, we divide those numbers together.*1269

*The number of outcomes in the event we are interested in ÷ the total number of outcomes just gives us 11/52.*1273

*In example 4, it is the same experiment as before.*1283

*We are going to roll 2 dice, I think we saw that in example 1.*1288

*This time want to see if the sum is a prime number.*1292

*Again, we are going to identify the experiment, all the outcomes in the event we are interested in and *1295

*we are also going to find the probability of the event.*1300

*The experiment, just like before, is to roll 2 dice.*1303

*I think this was in example 1 before.*1309

*Yes, that was example 1.*1312

*Our experiment is the same thing, we are going to roll 2 dice but now the outcomes, those are also the same as before.*1314

*Those are all the possible combinations you can get when you roll 2 dice.*1329

*And again, it is very helpful to think of the dice as being red and blue.*1333

*You can get a 1, 2, 3, 4, 5, and 6 on the blue dice.*1337

*You can also at the same time get 1, 2, 3, 4, 5, and 6, on the red dice.*1345

*In terms of combinations, you can have red 1 blue 1.*1352

*You can have red 1 blue 2, and so on.*1356

*There are 36 possible combinations all the way down to red 6 blue 6.*1361

*That is my set of outcomes, that is my sample space there, 36 possible combinations there.*1375

*This is S and we want to identify the event here.*1382

*The event here is the sum showing on the 2 dice is a prime number.*1386

*Let us figure out which of those combinations could give us a prime number.*1393

*I’m going to use a new screen for this because it takes a little bit of calculation.*1397

*The event here is, I’m going to call it A.*1404

*The set of combinations for which the sum is a prime number.*1410

*Let us remember our prime numbers.*1421

*The prime numbers between 2 and 12, because those are the only possible combinations you can get by rolling 2 dice.*1424

*You can get a 2, 2 is prime, 3 is prime, 5 is prime, 7 is prime, and 11 is prime.*1431

*Those are all the possible totals we are looking for.*1440

*I want to figure out which combinations will give me those possible totals.*1444

*To get a 2, I could roll a 1 on the red dice and 1 on the blue dice.*1450

*That will give me a total of 2.*1456

*That is the only possible combination that does give me a total of 2.*1457

*To get a 3, I can get 1 on the red dice, 2 on the blue dice.*1461

*Or I can get a 2 on the red dice and a 1 on the blue dice.*1468

*To get a 5, I could roll a red 1 blue 4.*1474

*I could roll a red 2 blue 3.*1482

*That was supposed to be blue. *1487

*I could roll a red 3 blue 2.*1492

*Or I could roll a red 4 blue 1.*1498

*There are 4 possible combinations that would add up to 5.*1505

*For 7, I could roll a red 1 blue 6.*1509

*I think I’m going to stop switching colors back and forth because it is slowing me down here.*1516

*I can roll a red 1 blue 6, red 2 blue 5, red 3 blue 4, red 4 blue 3, red 5 blue 2, or red 6 blue 1.*1522

*There are 1, 2, 3, 4, 5, 6 combinations that can give me a total of 7.*1535

*To get 11, I could roll a red 5 and a blue 6, or a red 6 and a blue 5.*1539

*Those are the ways that I can get 11.*1552

*There is a very nice way to keep track of these.*1553

*Instead of trying to list all of these and not being really sure if you got them all,*1556

*let me remind you of what the sample space look quite.*1560

*For blue, we can get 1, 2, 3, 4, 5, 6.*1564

*For red, we can get 1, 2, 3, 4, 5, and 6.*1570

*For red, we can get a red 1 blue 1.*1578

*We can get a red 2, blue 1.*1582

*We can get a red 1 blue 2, and so on.*1587

*We can fill in the rest of this chart but when we are trying to list totals, *1592

*if we look at the total of 2, let me circle that in green here.*1598

*The only way to do it is by getting a 1-1, that is a total of 2.*1604

*If we want a total of 3, we can get 2-1 or a 1-2.*1608

*Those are the ways that you get a total of 3.*1613

*A total of 5 would be, you ca get a red 1 blue 4.*1617

*I’m sorry, I wrote that in the wrong place there.*1628

*That would be a red 4 blue 1.*1631

*You can get a red 3 blue 2.*1636

*You can get red 2 blue 3, or a red 1 blue 4.*1641

*What you get here to get a total of 5 is the diagonal stripe on the table.*1650

*It is 2 stripes below the stripe for 3 because 5 is 2 bigger than 3.*1660

*To get a 7, you will get a diagonal stripe on the table.*1666

*2 places below, you are going to get 6 outcomes across this diagonal stripe.*1670

*There will be 6 outcomes in there, dividing the blue can add up to 6.*1681

*If I need to get 11, you can get a red 6 blue 5 or a red 5 blue 6.*1687

*There is a shorter diagonal stripe there, this is to get 11 and this one is to get 7.*1699

*We can look at these diagonal stripes and the lengths of the stripes to calculate the total probability of this event.*1708

*We just add up all the lengths of those stripes.*1715

*That is kind of a good way to check that we have not overlooked any possibilities here.*1718

*That is our event, let us calculate the probability now.*1724

*The probability of A is equal to the number of outcomes in A ÷ the number of outcomes in S.*1727

*I need to count out the total number of outcomes in A.*1748

*If I look at 2, I see 1 possible outcome.*1751

*If I look at 3, I see 2 possible outcomes.*1754

*5, I see 4 possible outcomes.*1759

*7, I see 6 possible outcomes.*1762

*11, I see 2 possible outcomes.*1764

*Again, I can check these numbers by looking at the lengths of the stripes over here.*1767

*7 has a stripe of 6.*1773

*5 has a stripe of 4.*1775

*3 has a stripe of 2.*1777

*2 has a stripe of 1.*1779

*11 has a stripe of 2.*1780

*Those correspond to these numbers here.*1783

*And if I just add all of those numbers up, I see 2 + 6 is 8 + 4 is 12 + 2 is 14 + 1 is 15.*1787

*I get 15.*1801

*And then remember, the total number of outcomes in S, that is the total size of this chart which was 36.*1803

*It looks like 15 and 36, I can reduce that if I take a 3 out of each one. *1809

*It reduces down to 5/12, that is my probability.*1813

*If I roll 2 dice, what is the probability of getting a prime number?*1818

*The answer is 5 out of 12.*1822

*Let me remind you of everything we did there.*1826

*First, I made a chart of all the possible outcomes.*1830

*I did not fill them all in because it would have taken too long and I got the idea of what the shape would be.*1833

*There are 36 possible outcomes depending on the 6 possibilities for the red dice and the 6 possibilities for the blue dice.*1841

*I identified my event which is that the sum was prime, which means the sum has to be 2, 3, 5, 7.*1850

*I have looked at my chart and I try to list all the outcomes that would lead to the sum being 2, 3, 5, or 7.*1857

*That is where I got 1 -1 for 2.*1864

*2 possibilities adding up to 3, 4 possibilities adding up to 5, 6 possibilities adding up to 7*1867

*and then just 2 possibilities adding up to 11.*1874

*To find the probability, I need to count up all those possibilities.*1877

*I have added up all those numbers, that is what I was doing over here.*1881

*Add them all up to get to 15, that is where this 15 comes from,*1884

*and then divide by the total number of outcomes in the sample space which is 36.*1889

*15/36 reduces down to 5/12.*1894

*That is how we got that probability.*1898

*For example 5, it is a little bit different from some of the others.*1903

*We are going to flip a coin repeatedly until we get a head and*1906

*I want to calculate the probability that we are going to flip at least 3 times.*1910

*The idea is we keep flipping this coin over and over again.*1915

*As soon as we see a head, we stop, and then that is the number of flips that we done in total.*1918

*What is the probability that that number will be 3 or greater?*1923

*Let us identify the experiment and all the outcomes and then find the probability of that event.*1930

*The experiment here is just flipping a coin over and over again.*1937

*And then as soon as we see the first head, we stop.*1951

*The outcomes, this is a little different from all of our example so far *1958

*because what can happen here is we could get a head right away on the first flip.*1965

*Our first possible outcome is we get a head right away or we might get a tail on our first flip*1973

*and then get a head and then would stop.*1980

*Or we might get a tail and then another tail and then get a head and then we would stop.*1982

*We might get 3 tails and then get a head and then we would stop.*1988

*And we do not really know how long this is going to go.*1994

*There is no assuming on how long this is going to go.*1997

*It could go on forever.*2000

*I cannot just list all these outcomes because there are infinitely many.*2002

*It could go on indefinitely.*2006

*The event that we are looking for is that we have at least 3 flips.*2008

*That is A is the set, a single flip getting us a head will be qualified.*2018

*A tail and a head will not qualify.*2026

*It would be a tail- tail-head, that would be 3 flips.*2027

*Tail-tail-tail-head, that would be more than 3 flips.*2031

*A tail-tail-tail-tail-head, that would be more than 3 flips.*2036

*It is anything that has 3 flips or more.*2041

*We want to identify the probability of that event.*2044

*Unfortunately, we cannot do this one by counting.*2049

*Because we have an infinite number of outcomes in our event, *2053

*we have an infinite number of outcomes that can happen from the experiment.*2056

*We can no longer use our counting formula from before.*2061

*Instead, let me introduce a new way of calculating it.*2065

*Our way of calculating the probability is we can add up the probabilities of each of these individual outcomes in our event.*2072

*It is the probability of tail-tail-head + the probability of tail-tail-tail-head + the probability of 4 tails*2084

*and then a head, and so on.*2096

*This is going to be an infinite sum.*2100

*We are going to have to add all these up forever.*2102

*We have to do something clever at some point.*2106

*Tail-tail-head, to get a tail-tail-head, there is a ½ chance that you get a tail right away and *2108

*then there is ½ chance that you get another tail and then there are ½ chance that you get a head.*2117

*Put those together and you get 1/8.*2123

*A tail-tail-tail-head, there are 50% chance of getting a tail, 50% chance of getting another tail, *2126

*50% chance of getting another tail, 50% chance of getting a head after that.*2132

*So it is ½ × ½ × ½ × ½, I you put all of those together you get 1/16.*2136

*Here you will be multiplying together ½⁵ which is ½⁵ is 32.*2146

*We have to add all those up, 1/8 + 1/16 + 1/32.*2158

*This is an infinite series.*2168

*Now, you have to remember something from your calculus 2.*2170

*If you do not remember how to deal with infinite series, *2174

*we got some great lectures here on www.educator.com on the second semester calculus, calculus level 2, taught by a great instructor.*2177

*And there is a lot of work in there in adding up infinite series.*2187

*Here we are going to apply that.*2191

*This is actually a geometric series.*2192

*And with a geometric series, you need to identify the common ratio.*2198

*Here the common ratio, what we are doing from each term to the next is we are multiplying by ½.*2202

*Each term is half as big as the previous term.*2209

*The important thing here is that ½ is less than 1 because that tells us when a geometric series converges, *2212

*it converges when the common ratio is less than 1.*2220

*It does converge and we learn back in those lectures on calculus 2, the sum of an infinite geometric series,*2224

*I’m going to write it in words, it is the first term or 1 - the common ratio.*2232

*In this case, our first term is that 1/8 right there.*2245

*It is 1/8/ 1 - the common ratio which is ½.*2249

*1/8/ 1 -1/2 which is 1/8/ ¼.*2255

*Do a flip here and we get 4/1 × 1/8.*2261

*I’m sorry, 1 - ½ is not ¼, it is ½.*2268

*When we do the flip there, it is 2/1 × 1/8 which is ¼.*2273

*That is our probability of having to flip this coin at least 3 times in order to see the first head.*2284

*I want to go back and calculate this in different way.*2294

*We will see a different way to solve this problem.*2296

*I want to do that on the next page.*2298

*Let me recap how we calculated it using this method first.*2299

*Our experiment was just flipping a coin here.*2305

*The possible outcomes are all the possible strings of heads and tails that we can see.*2309

*Remember, we stop as soon as we see the first head.*2316

*If we get a head right away, we stop.*2318

*If we get a tail, we keep flipping and then we get a head, we stop.*2320

*If we get 2 tails and then a head, we stop after that head.*2325

*If we get 3 tails and then head, we stop after that head, and so on.*2328

*Those are all the possible outcomes.*2332

*The event we are interested in was that we flip at least 3 times.*2334

*I listed all the outcomes that have 3 or more flips.*2339

*To find the probability of that, we cannot count these things anymore because there is infinitely many of them.*2345

*Instead, we have to find the probability of each one.*2351

*The probability of tail-tail-head is 1/8 because it is ½ × ½ × ½.*2355

*Tail-tail-head is 1/16 because you have to predict the flips 4 times in a row to get that right.*2362

*Here, tail-tail-tail-head, you have to predict 5 flips in a row.*2370

*You have 1/2 chance of each one, 1/32.*2374

*Adding all those up, we notice that we get a geometric series.*2378

*Our common ratio, what we are multiplying by each time is ½.*2383

*You got to remember, the sum of a geometric series which we learn back in calculus 2, *2388

*in level 2 calculus, is the first term in the series ÷ 1 - the common ratio.*2393

*That is where I got 1/8 is that first term, that is where that came from.*2400

*The common ratio is that ½.*2404

*1/8/ 1 - ½ simplifies down to ¼ is my probability.*2408

*I’m going to calculate this in a different way.*2414

*Let me show you how you could have done that, in fact without summing up the geometric series.*2417

*Here is an alternate solution for that one, the same one as before.*2423

*Alternate solution is to find the probability of A is equal to, we can calculate the complement of A which means the opposite of A.*2427

*That is the complement A bar means the complement of A, the opposite of what we are looking for.*2452

*And then do 1 - the probability of the complement of A.*2461

*Let us figure out what the complement of A is.*2466

*Remember A was the set of outcomes that have at least 3 flips.*2469

*That was tail-tail-head, tail-tail-tail-head, and so on.*2476

*The complement of A is just the outcomes that head 2 or fewer flips.*2483

*That can be if you get a head right away or you get 1 tail and then a head.*2490

*That is it, because anything else would have 3 or more flips.*2495

*The probability of a complement of A is the probability of getting a head immediately + *2499

*the probability of getting a tail and then a head.*2507

*The probability of getting a head right away on the first flip is ½.*2511

*The probability of getting a tail and then a head is ½ × ½.*2515

*We get ½ + ¼ which is ¾, that was the probability of the complement of A.*2524

*The probability of A is 1 – that, 1 -3/4 which is ¼.*2531

*Of course, that is the same answer we got before but using quite a different strategy.*2540

*It is kind of reassuring.*2544

*They ought to be the same answer but it is nice to see that they check each other.*2546

*Let me make sure that is still clear.*2553

*What we did here was we calculated the compliment of A, that means everything that is not inside A.*2557

*It is the opposite set from A.*2562

*A was tail-tail-head, tail-tail-tail-head, all the strings with 3 flips or more.*2565

*The compliment is all the shorter strings that is just getting a head or getting a tail and a head.*2571

*The probability of that is, the probably just getting a head on the first flip is ½ because you flip 1 coin.*2578

*What is the chance of getting a head is ½.*2585

*The probability of getting a tail and then a head, it would be 1/2 × ½ because you got to get 2 flips,*2587

*the way you are predicting them.*2594

*Add those up, you get ¾.*2595

*We subtract that from 1 because we are using this formula here.*2597

*1 -3/4 is ¼.*2602

*That is the probability of the experiment going for at least 3 flips.*2605

*In our last example here, we are going to draw a 5 card poker hand from a standard 52 card deck.*2615

*The question is what is the probability that we draw all spades?*2621

*We want to identify the experiment, all the outcomes in the event that we are interested in.*2626

*There is a lot of possible outcomes for this experiment.*2630

*We are going to be learning some new notation in the context of this example*2633

*because it is going to introduce something that we are going to be studying later.*2640

*We are going to learn about combinations a little bit in this example.*2644

*The experiment here is drawing the cards, the set of possible outcomes is all the different ways*2648

*to draw 5 cards from a standard 52 card deck.*2670

*This is a little tricky to count.*2675

*Let me show you how it works out.*2677

*Think about it, we are drawing from a 52 card deck, there are 52 ways we can draw the first card.*2680

*After we drawn that first card, there are 51 cards left in the deck.*2692

*There are 51 choices we can make for the second card.*2697

*There are 50 choices left for the third card.*2703

*And then, 49 choices for the 4th and 48 choices for the 5th.*2712

*That seems like there is this very large number of ways to get a particular a poker hand.*2718

*However, once you got the 5 cards in your hand, we do not care what order there in.*2727

*Which means that, you get say the cards you draw are 1, 5, 3, jack, and 8.*2733

*I’m not even going to worry about the difference suits, the clubs, spades, or hearts, or diamonds.*2747

*But suppose you get a 1, 5, 3, jack, or 8.*2753

*You could also had drawn the jack and then the 3 and then the 1 and then the 5 and 8.*2758

*Once those cards are in your hand, that looks like the same poker hand because you can shuffle them around in your hand.*2764

*What we did when we are counting all the different ways to draw the 5 cards*2772

*is we counted these poker hand separately, even though they are really the same poker hand.*2776

*The problem is they were over counting.*2785

*We are counting the same poker hand multiple times.*2787

*We have to take this answer and we have to divide by something and we have to divide by *2791

*the number of ways that each poker hand got counted.*2797

*Now, let us just think about a particular poker hand.*2803

*Let us think about this one in particular.*2806

*1, 5, 3, jack, 8, let us think about the number of different ways that that poker hand was counted in our original list.*2807

*Just think about how many different ways could you order those 5 cards.*2820

*There are 5 choices for which card could have been first, assuming that we are talking about that particular hand.*2826

*There are 5 ways, the 5 choices for the first card.*2834

*Once you have picked that first card, there are 4 choices left for the second, *2839

*and then 3 choices left for the third, and then 2 choices left for the 4th, and then the 5th card would be fixed.*2846

*Remember, we are talking about a particular poker hand like 1, 5, 3, jack, 8.*2854

*We are talking those 5 cards, how many different ways could those 5 cards be rearranged*2860

*and how many different ways that they get counted in this original list.*2866

*The answer is 5 × 4 × 3 × 2 × 1.*2871

*We have to divide that out of our list because each poker hand, each group of 5 cards got counted.*2878

*1 × 2 × 3 × 4 × 5 is 5! ways in our original list.*2901

*We have to divide out by 5! in order to get an accurate number of the different possible poker hands.*2909

*That is the total number of outcomes.*2919

*There is a clever way to simplify this using factorials.*2921

*Notice that the denominator here is 5!.*2926

*The numerator is 52 × 51 × 50 × 49 × 48.*2933

*That looks like we are starting to build up 52!, but what is missing there is 47 × 46, on that down to 1.*2942

*If I just multiply those on, I have to divide them out again.*2953

*47 × 46 on down to 1.*2957

*Another way to write that is as 52! ÷ 47!.*2963

*That is another way to list the total number of hands.*2974

*This is actually something we are going to learn about in the next lesson.*2978

*The total number of ways to choose 5 cards from 52 is, the notation we are going to use is 52 choose 5.*2981

*There are some other notations along with this notation.*2994

*It is a little confusing because different textbooks use different notations.*2998

*C of 52 5, what it means is this factorial formula.*3002

*This is read as 52 choose 5.*3008

*What it means is this factorial formula, you expanded it out by doing 52! ÷ 5!, and this 47! *3015

*really came from the fact that 47 is 52 -5.*3026

*That is where that came from because that 47 came from this 47 here, which came from this 47,*3033

*which came from the fact that we had already used up 5 of the factors of 52! there.*3039

*The number of outcomes here is 52 choose 5.*3047

*This is our new notation, 52 choose 5.*3051

*It is the way you calculate it is as 52! ÷ 5! ÷ 47!.*3055

*Let me recap this because we have to keep going on the next side here.*3067

*I have not even talked about the event that we are interested in yet.*3070

*Our experiment is drawing the cards.*3073

*The number of ways to draw the cards, there are 52 ways to get your first card.*3075

*Once you draw on that card, there is only 51 card left in the deck, so there are 51 ways to get the second card.*3079

*There are 50 cards left in the deck, 50 ways, 49 and 48.*3086

*But the problem is that that over counts because you can get the same poker hand in different possible orders.*3090

*In order to fix that, we looked in a single poker hand, a typical poker hand, *3099

*and we figured out that the number of ways that that could have been ordered was 5 × 4 × 3 × 2 × 1.*3106

*That is why we had to divide out by 5 × 4 × 3 × 2 × 1, it is because each poker hand*3113

*that counted that many ways in our initial list up here.*3120

*Each poker hand, we over counted by a factor of 5!.*3126

*That is our number of outcomes but then as a clever way to write the numerator, we wrote that as 52! ÷ 47!.*3132

*That is where this part of our answer comes from.*3142

*The 47 comes from 52 -5 and then this 5! Is where we got that 5! in the denominator there.*3145

*This is our new notation, 52 choose 5.*3157

*It is sometimes written with these elongated parentheses.*3161

*It is sometimes written with this C.*3165

*C stands for combinations there.*3169

*By the way, this 52 choose 5, it looks like a fraction but it is not a fraction.*3172

*It is not the same as 52/5.*3176

*Do not put the bar there, do not turn it into a fraction because those are totally different things.*3180

*That is just the number of outcomes, we still need to count all the outcomes in the event that we are interested in.*3185

*The event was that we draw all spades.*3192

*Let us try and figure that out.*3196

*I will try and figure that out in the next slide.*3197

*But in fact, it would d be fairly easy if you remember the strategy that we used here.*3200

*Remember that 52 choose 5 is all the ways you can draw 5 cards from 52 cards total.*3206

*Keeping going with this example, our event is A is ways to get a poker hand, a 5 card poker hand of all spades.*3217

*Here is the clever way to think about it.*3243

*Remember how we drew our hand in the first place, the 52 choose 5.*3245

*Think about if they are all going to be spades.*3249

*If they are all going to be spades then we are really saying it is the number of ways*3254

*to choose 5 spade out of how many spades are there total?*3263

*There are 13, the ace through the king of spades, out of 13.*3275

*According to our logic before that is exactly 13 choose 5.*3283

*If we wanted to express that in factorials, that is 13! ÷ 5! ÷ 13 -5! which is 13! ÷ 5! × 8!.*3291

*That is our number of ways to choose 5 spades out of, there are only 13 in the deck.*3310

*That is the number of ways that we can get 5 spades.*3322

*Our probability of A is the number of outcomes in A ÷ the number of outcomes in S.*3325

*We already calculated both of those, it is 13 choose 5 ÷ the total number of poker hands we can get is 52 choose 5.*3351

*We could simplify that down if we wanted.*3365

*13! ÷ 5! × 8!.*3367

*We already saw above that 52 choose 5 is 52! ÷ 5! × 47!.*3373

*We could simplify that a bit further.*3383

*We can cancel the 5! and flip up the denominator.*3385

*13! ÷ 52!.*3389

*13! ÷ 8! × 52!.*3399

*The 47 would flip up to the numerator and the 5! would cancel.*3405

*I'm going to make no attempt to try to actually calculate that as a number.*3413

*It would be a very small number.*3418

*But that is our probability, if you draw 5 cards out of the deck that you will get all 5 cards will be spades.*3419

*That is of the end of our 6th example there.*3434

*Let me recap just quickly what we did on the second slide here.*3437

*Our event was the number ways to get a poker hand of all spades.*3441

*Using our strategy from before that is 13 choose 5, remember we have learned that notation on the previous slide.*3445

*If you do not believe that, another way to think about that is to say how many ways can we get all 5 spades.*3452

*If you are choosing 5 spades, there are 13 ways to get the first one.*3464

*Once you have picked one spade, there are 12 ways after that to get the second one.*3468

*11 ways once you picked the first two, × 10 × 9.*3473

*There is your 5 spades but that over counts because the 5 spades could have come in any order.*3478

*The number of possible orders they could arrive in is 5 × 4 × 3 × 2 × 1.*3485

*That simplifies, we have 5! in the denominator and the numerator we can cleverly write it as 13! ÷ 8!.*3493

*Because we are multiplying all the numbers down from 13 but we are cutting off all the ones down from 8.*3506

*We can get that as 13! ÷ 8!.*3515

*That is another way to get to that same answer there for the event that we are interested in.*3519

*To find the probability, we just take the number of ways to get to our event ÷ *3525

*the total number of outcomes 13 choose 5 ÷ 52 choose 5.*3530

*We get a kind of cumbersome expression involving factorials in our answer.*3535

*That wraps up our last example and that wraps up our introductory lecture here on www.educator.com, for probability.*3543

*We went over some terminology, we practiced some basic probability using counting techniques.*3550

*On the next set of lectures, we will develop these counting techniques further *3556

*and we will have complicated probability problems.*3560

*I hope you will stick around for that.*3564

*These are the probability lectures here on www.educator.com and my name is Will Murray.*3565

*Thanks, bye. *3570

1 answer

Last reply by: Dr. William Murray

Wed Jul 13, 2016 6:08 PM

Post by Ravi Bala on July 12 at 10:24:38 AM

Hey Dr. Murray, I was just wondering what kinds of probability questions will be found in the ACT and SAT

1 answer

Last reply by: Dr. William Murray

Tue Feb 23, 2016 5:36 PM

Post by Ali AlSaedi on February 23 at 10:53:57 AM

Thank you very much, you made it so clear and easy to understand.

2 answers

Last reply by: Dr. William Murray

Tue Feb 2, 2016 4:28 PM

Post by Krishna Kumar on January 31 at 03:43:21 PM

Hello Dr.Murray,

When I play this lecture on an iPad, in lesson 1 the lecture stops at example 4 outcomes section. I cannot play beyond this point. I have tried it on both chrome and safari and encounter the same issue.

Is educator.com expected to work on an iPad? If so, can you help me figure out the issue?

Thank you

1 answer

Last reply by: Dr. William Murray

Fri Jul 10, 2015 12:42 PM

Post by R Abdullah on July 10, 2015

Hi Dr. Murray,

I'm really excited about this Probability course, especially after the wonderful Calculus 2 videos you made. I was just wondering if you can recommend a textbook that follows this course or at least covers the main topics.

Thank you,

Rasheed Abdullah

1 answer

Last reply by: Dr. William Murray

Mon May 11, 2015 2:42 PM

Post by Muhammad Asad Ullah MOAVIA on May 11, 2015

Professor Dr.William Murry, you are the best Professor I have ever seen!

3 answers

Last reply by: Dr. William Murray

Mon May 11, 2015 2:41 PM

Post by Micheal Bingham on May 6, 2015

Hi, your lectures are wonderful and I enjoy them a lot, may I ask, How long did your Ph.D take? I'm interested in obtaining a Ph.D and would like to wonder how many years does it usually take. Do you think the Ph.D is worth it given most colleges will already have well established jobs from their Bachelor's Degree?

3 answers

Last reply by: Dr. William Murray

Wed Feb 25, 2015 4:01 PM

Post by jason varner on February 22, 2015

Dr. Murray,

My initial instinct about example number 6 was to multiply: 13/52 * 12/52 * 11/52 * 10/52 * 9/52. I thought this would account for how many spades are available each time you pull a card. The exact answer I get is very close the exact answer (decimal) that you got, but not identical. Is this a coincidence?

Thank You

1 answer

Last reply by: Dr. William Murray

Tue Aug 5, 2014 3:16 PM

Post by Saria Abbas on July 21, 2014

I have a couple of past papers but no answers to them, some are visual mechanisms, is there a way I could send you questions and/workings out for feedback?

1 answer

Last reply by: Dr. William Murray

Sat Jul 5, 2014 5:58 PM

Post by Philippe Tremblay on June 30, 2014

For Example V:

There is a 1/2 probability that we stop at any step. Therefore the probability of flipping at least 3 times is 1/2 * 1/2.

Anything wrong with that approach?

1 answer

Last reply by: Dr. William Murray

Fri Jun 27, 2014 5:06 PM

Post by Thuy Nguyen on June 26, 2014

1 is not a prime number though, so it would instead be 14/36 for the question: The probability of prime numbers for two dice roll.

1 answer

Last reply by: Dr. William Murray

Fri May 30, 2014 3:56 PM

Post by Narin Gopaul on May 30, 2014

Good morning DR.

I have a quick question

Would this course be similar to the Probability and statistics for engineers?

3 answers

Last reply by: Dr. William Murray

Thu Apr 10, 2014 7:46 PM

Post by Ali Momeni on March 30, 2014

Dr Murray,

In example 3: why doesn't the total number of outcomes of the sample space decrease? Since you don't have any replacement for the first 20 cards that are drawn shouldn't the equation go 11/(52-already drawn cards). Thank you for your great work. I got an A in calc 2 because of your fantastic explanations.

All the best,

Ali

1 answer

Last reply by: Dr. William Murray

Thu Mar 27, 2014 4:43 PM

Post by shawn page on March 23, 2014

I have a question that states on avg customers enter a bank at a rate of 5 in 3 mins. I am then asked - what s the probability that 10 customers arrive in the next 3 mins. What lecture can help me with this?

1 answer

Last reply by: Dr. William Murray

Thu Mar 27, 2014 4:40 PM

Post by julius mogyorossy on March 19, 2014

Does your probability class have to do with quantum physics, please email me your answer to, juliusmogyorossy@yahoo.com, so I know you answered it, if you choose to answer it, thanks.