For more information, please see full course syllabus of Probability

For more information, please see full course syllabus of Probability

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### Inclusion & Exclusion

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
- Inclusion/Exclusion: Two Events 0:09
- Inclusion/Exclusion: Two Events
- Inclusion/Exclusion: Three Events 2:30
- Inclusion/Exclusion: Three Events
- Example I: Inclusion & Exclusion 6:24
- Example II: Inclusion & Exclusion 11:01
- Example III: Inclusion & Exclusion 18:41
- Example IV: Inclusion & Exclusion 28:24
- Example V: Inclusion & Exclusion 39:33

### Introduction to Probability Online Course

### Transcription: Inclusion & Exclusion

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

*Today, we are going to talk about the rule of inclusion and exclusion.*0005

*The rule of inclusion, exclusion is a way of counting the union of two or more events.*0010

*Let me show you what is going on here.*0018

*I'm just going to show you the version of inclusion, exclusion of two events first.*0022

*Then I’m going to show you the merging with three events in a moment.*0027

*The idea here is that you got two events here and I will call them in red and blue circles here.*0030

*We are trying to count the union in there.*0038

*I’m going to call these events A and B.*0041

*I’m trying to account the combined area between A and B.*0044

*The rule of inclusion/exclusion says that if you want to count the union of A and B,*0049

*what you do is you count A separately by itself and then you count B separately by itself.*0056

*If you do that, what you have done is you have counted everything in A.*0062

*You will count all this area right here and then you count everything in B.*0067

*You have counted all this red area right here.*0071

*The problem is that you have over counted the stuff that is common between A and B.*0075

*You have over counted all this stuff in the middle here, the intersection of A and B.*0081

*To fix that, you subtract that off.*0086

*That is where we get this last term in the formula.*0089

*You subtract off the intersection of A and B.*0093

*That is the rule of inclusion/exclusion for two events.*0097

*This is also useful if you solve it the other way around, if you solve for the intersection instead of the union.*0100

*If you just take this same rule and you move the intersections to the other side and*0107

*move the union over to the right hand side, then you get a parallel rule*0112

*which tells us that you can count the intersection of A and B by first counting A + B, and then by subtracting off the union.*0117

*It is basically the same rule but you swap the rules of intersections and unions.*0130

*That is the inclusion/exclusion rule for two events.*0136

*We will see some examples in the problems later on where you get some practice counting those things.*0139

*First, we want to go ahead and look at the inclusion/exclusion rule for three events.*0145

*That one is a little more complicated.*0152

*I will draw more complicated picture here because we are going to have three different events going on.*0154

*We will try to count all three.*0161

*There is my A, there is my B, they have to be the same size which is good because mine are not.*0164

*There is my C.*0174

*We are trying to count the union of the three events.*0176

*We are trying to count the area that is covered by all three circles here.*0179

*First, we count all the area inside A.*0183

*That is all that area right there, all the blue areas.*0187

*That is what is going on here with the A right there.*0191

*Then, we count all the area inside B.*0195

*Let me shade that in red.*0197

*We count all that area right there, that is the B.*0201

*I will color the area for C in green.*0205

*We count all that area in green here.*0211

*That is where that term comes from.*0216

*If you look, we have over counted a lot of the area in here.*0218

*We have counted a lot in this area more than once.*0222

*First of all, we look at the common area between A and B, the intersection of A and B.*0225

*That is A intersect B right there.*0232

*It looks like we counted it twice.*0235

*We counted it once in red and once in green.*0238

*We have to subtract it off.*0241

*I'm going to show you that subtraction term right here.*0243

*That is where we subtract off the intersection of A and B.*0247

*And then A intersect C over here is colored both blue and green, that also got counted twice.*0250

*I will subtract that area off right here A intersect C.*0261

*And then a similar thing happens where B intersects C, is this area right here.*0265

*The red and green areas, that is B intersect C.*0271

*That counted twice so I will subtract that off.*0273

*It gets a little more complicated because there is one area in the middle which I color in yellow.*0279

*This area right here, this area in the middle.*0286

*I will see if I can describe that in yellow.*0290

*That is A intersect B intersect C.*0303

*What happened was that originally got counted 3 separate times and it got counted in A,*0313

*it got counted in B, it got counted in C.*0319

*But then, it got subtracted off three separate times.*0322

*When we subtracted off A intersect B and A intersect C, B intersect C.*0326

*We kind of counted it three times and then we subtracted it off three times.*0329

*The result that it got counted 0 times.*0336

*That area did not get counted at all in the final analysis.*0339

*What we have to do is add that area back in and that is where this final term comes for the inclusion/exclusion formula.*0343

*A intersect B intersect C, we have to add it back in one more time to make sure it is counted exactly once in the final formula.*0351

*Our formula, ultimately, if you want to count A union B union C, what you do is you count A, B, and C separately*0360

*and you subtract off each of their intersections and then you have to add back in the intersection of all three.*0369

*It is a bit complicated, that formula, we will see some practice in the exercises.*0375

*Let us go ahead and try some examples out.*0382

*First example here is, we are at a small college and apparently, there are 150 freshmen here taking English*0385

*and 120 freshmen at this college are taking Math.*0392

*There might be some overlap, in fact it tells us that there are 90 freshmen taking both classes.*0396

*The question is how many are taking at least one of English and Math?*0402

*How many freshmen will there be taking at least one of English and Math?*0406

*To solve this, I'm going to set up some events here.*0411

*I’m going to say A is the freshmen taking English.*0414

*I will just write English for short here.*0418

*B is the freshmen taking Math.*0422

*We want to count how many are taking at least one of the two which is the union of the two sets,*0428

*because it is all the people taking English or Math or both.*0435

*What we want to count here is A union B.*0438

*That is straightforward application of our inclusion/exclusion formula.*0445

*Let me remind you what that was.*0450

*The way you count A union B is you count A and then you count B.*0452

*And now you have over counted the intersection so you want to subtract off the intersection - A intersect B.*0456

*In this case, all the numbers are just given to us right here in the problem.*0463

*The number of people taking English was 150.*0466

*The number of people taking math is 120.*0471

*The number of people taking both of them, that is the intersection is 90.*0475

*We have 270 - 90 and that simplifies down to 180 freshman here are taking either English or Math or both of them.*0481

*The other way to look at this is to draw one of our Venn diagrams and then fill in the numbers on each of these.*0493

*We have a certain number of people taking English.*0503

*A certain number of people taking Math.*0506

*There is English right there and there is Math.*0509

*We have a certain number people taking both of them.*0513

*Let me fill in the people taking both of them first.*0515

*There are 90 people taking both of them.*0518

*There is a 150 taking English.*0522

*We already accounted for many in the overlap.*0525

*There must be 60 more of them just taking English but not Math.*0527

*In math, we got 90 that we already accounted for in the overlap but 120 totals.*0532

*There must be 30 out here just taking Math.*0539

*If you try to figure out how many there are taking either one or both, it would be 60 + 90 + 30.*0542

*And of course that gives you the same number we got as 180 students taking either one.*0551

*Let me recap quickly how we got those answers.*0560

*We set up events for the students taking English and math and then*0564

*we just use our straightforward inclusion/exclusion for two events.*0567

*The number of people in A union B is number in A + number in B - the number in both A and B which is the intersection.*0572

*Those numbers come straight out of the stem of the problem and we add them up and we get 180.*0580

*This was a little pictorial way to illustrate it and figure out directly if we break down*0586

*how many students are taking both, that is 90 taking both and then the 60 really came from doing 150 – 90,*0592

*because there are 150 taking English and we know that 90 of them are already accounted for taking Math.*0601

*This 30, in a similar fashion came from 120 - 90 because there are 120 taking Math*0608

*but we know that we already accounted for 90 of them also taking English.*0617

*That tells us exactly how many students are in each group and we can add those up*0623

*to find the total number of students taking English or Math.*0628

*Now, I want you to hang onto these numbers for the next example because in the next example,*0632

*we are going to stay at the same small college and we still have the same number of people taking English and Math.*0636

*We are going to add in a third subject which is History and*0645

*we are going to have to use our inclusion/exclusion formula for three events.*0647

*We will use the same number so make sure you understand these numbers before you move on to example 2.*0653

*In example 2, we are at the same college that we were in for example 1.*0663

*If you have not just watched the video for example 1, go back and watch that one first because*0667

*I'm going to keep using the same numbers that we figure out for example 1 in example 2, at the same college.*0672

*We are going to use the same events as well.*0679

*A is the event of a student taking English and B is the event of a student taking Math.*0682

*And now we are introducing a third event, C is the event of a student taking history.*0694

*And let us see, we are trying to find out how many are taking at least one of the three?*0701

*That would be the union of our three events.*0708

*We are going to count that.*0712

*We are going to use some of the numbers from example 1.*0714

*Remember to get the numbers from example 1.*0716

*I'm going to write out the formula for inclusion/exclusion for three events.*0718

*That is A union B union C.*0724

*It was this complicated formula.*0728

*It is the number things in A + the number of things in B + the number of things in C.*0730

*Then, you have to subtract off all the intersection.*0737

*A intersect B - A intersect C - B intersect C.*0739

*As we saw on one of the beginning slides in this lecture, just a couple of slides ago.*0747

*They have to add back in the intersection of all three of them, A intersect B intersect C.*0752

*I want to go through and fill in all those numbers.*0761

*Some of these were given to us in the first example, example 1.*0763

*Go back and check those numbers, if you do not know where these come from.*0768

*The number of people taking English was 158, we are told that in example 1.*0771

*Remember, the people taking math was 120 and now we are told that the same college right here,*0776

*100 people are taking history.*0782

*Remember, people taking both English and Math, we found that out in the previous problem was 90.*0785

*A intersect C is English and History.*0792

*We are told that that is 80 people taking both English and History.*0796

*Let me subtract off 80 there.*0804

*And then, math and history is 75.*0806

*And A intersect B intersect C is the people taking all three Math, English and History.*0810

*We are told that there are 60 people taking all three classes.*0819

*I’m going to add that 60 back in.*0822

*It is just a quick matter of adding up the arithmetic and if you do,*0825

*I think I already checked these numbers 270, 370, 280, 200, 125 + 60 is 185.*0831

*That is how many students are taking all three classes.*0848

*We are also asked how many are taking only history?*0852

*I think the easiest way to figure that out is by drawing a diagram.*0855

*Let me draw a diagram of all three classes represented there, math, English, and history.*0861

*We will see if we can figure out just exactly how many students go in each different category there.*0868

*I have got a group of students taking English, a group of students taking math, and a group of students taking history.*0877

*There is my English group, there is my math group, and there is my history group.*0890

*They are perfect circles but that is alright.*0894

*I want to find whatever numbers I can now and I know that 60 are taking all three classes.*0896

*I will write it from the inside out here.*0901

*I know that 60 people are taking all three classes.*0904

*I know from the previous example, from example 1, that there were 90 students taking both English and math.*0908

*And we already accounted for 60 of them.*0917

*That means that 30 students left over here in the English and math.*0919

*In English and history, there are 80 students total.*0923

*We already accounted for 60 of them so there must be 20 here that are taking English and history but not math.*0927

*For math and history, we got 75 total.*0934

*60 of them are also taking English that means 15 of them are taking math and history but not English.*0938

*Let me go ahead and figure and fill in the others.*0944

*For English, I see that I already got 110 accounted for but there is a 150 English students total.*0947

*It must be 40 more outside here.*0957

*From math, I see I got 90 + 15, a 105 total but there were 120 people taking Math.*0960

*There must be 15 left over.*0966

*And for history, I see that I got 20 + 60 + 15 that is 80 + 15.*0969

*95 students total.*0975

*It said that 100 freshmen are taking history.*0978

*I have accounted for 95 of them.*0984

*That must mean that there is 5 extra students left over here.*0985

*I think the question said, how many are taking only history?*0990

*The answer to that is that 5 students are taking history but not math or English.*0994

*That is coming from that 5, right there.*1007

*Let me remind you how we get everything here.*1019

*We got three events here, we got English, math, and history.*1021

*We are using our inclusion/exclusion rule for three events.*1024

*And that is the formula that we had on one of the earlier slides for the inclusion/exclusion rule for three events, A union B union C.*1028

*You add the individual events, you subtract off the intersections, and then you add back in the three way intersection.*1036

*And I just fill then all the numbers.*1044

*Some of these numbers I got from example 1 because it was the same example.*1046

*Then, I filled in the new numbers that we are given here in example 2.*1051

*And I add them up and I got 185.*1056

*Another way to do that is to set up this diagram here and setup circles for English, math and history*1058

*and figure out what the numbers of students are in each of these categories.*1066

*To do that, you really want to work from the inside out.*1072

*We start the 60 students in all three, and then we work our way out into figure out*1074

*how many students are in each intersection, and we get those by subtracting.*1080

*For example, this 20 came from the fact that there were 80 students taking English and history.*1083

*We already accounted for 60 of them taking all three classes.*1091

*That 20 was 80 -60.*1097

*We have figured out this 30 right here and this 15 right here.*1100

*By doing some more subtraction, we figure out that there are 40 students just taking English,*1104

*15 students just taking Math, and 5 students just taking history which was the answer to the second part of the problem there.*1109

*For example 3, we are going to keep going in a college setting but this time we are going to look at student ID numbers.*1123

*In this particular small college, they range from 000 to 999 which really means there is*1129

*a thousand numbers available because 1 through 999 and then one more for 000.*1134

*The question is how many of these numbers have at least 1-3 and at least 1-4 in them?*1141

*I want to set up some events to solve this.*1148

*Let me go ahead and describe my events here.*1150

*A is going to be the set of all numbers that have at least 1-3 in them.*1154

*B will be the set of numbers that have at least 1-4 in them.*1167

*By the way, some notation that I'm using here that you might not seem is this colon equals notation.*1177

*Colon equals just means, when I'm defining a set.*1184

*That means A is defined to be, whatever is appearing on the right.*1188

*Defined to be whatever is appearing on the right.*1198

*That is what that colon equals notation means.*1203

*It is a notation that I borrowed from the Computer Sciences.*1206

*It is very useful when you are programming to say this variable is defined to be some value.*1209

*That is why I mean by that colon equals.*1214

*If you do not like it, you can just use and equal sign and it essentially means the same thing.*1217

*Let me keep going here.*1225

*We are going to use inclusion/exclusion here and we are going to try to count A intersect B.*1226

*We have been asked to find the number of ids that have at least 1-3 and at least 1-4.*1235

*That really means we are going to count A intersect B.*1243

*But inclusion/exclusion say we can count that if we can find A by itself and B by itself, and A union B.*1247

*Each one of those is a little problem here.*1255

*Let us try to count each one of those.*1257

*A by itself, how many numbers have at least 1-3 in them?*1259

*The easiest way to count that is to count the complement.*1266

*It is 1000 numbers total - the numbers with no 3’s in them.*1269

*Let us think about how many numbers have no 3’s in them.*1286

*That means you are trying to build a three digit number and you are allowed to use any digit you like*1291

*except you cannot use the digit 3.*1297

*You got three digits here, if you cannot use the digit 3 and you have really only got 9 choices left,*1300

*0-9 except the 3 for each of these possibilities.*1307

*There are 9 possibilities here, 9 possibilities for the second digit and 9 possibilities for the third digit.*1312

*This is, in total, 1000 – 9³.*1320

*We are throwing out all the numbers that do not have any 3's in them,*1327

*leaving us exactly the numbers that have at least 1-3 in them.*1332

*9³ is 729, this is 1000 – 729.*1336

*That is, 1000 – 729 is 271.*1344

*For B, what is all the numbers that have at least 1 -4?*1352

*Exactly the same reasoning applies, just instead of kicking out all the ones with the 3 in them,*1357

*we are going to kick out the ones with 4 in them.*1364

*To count those, we have 9 digits available because we kicked out the 4, instead of kicking out the 3.*1368

*That is going work out exactly the same way.*1373

*1000 – 9³ and again that is going to work out to 271.*1376

*Since, I would like to know calculate A intersect B, the preliminary step to doing that is to calculate A union B,*1384

*which means the number of ids that have at least 1-3 or at least 1-4 or both.*1394

*That is what we have to try to count.*1418

*That is quite difficult to count directly.*1420

*It is hard to count the number of ids that have at least 1-3 or at least 1-4.*1423

*The easy way to count it is to work backwards and start with 1000 numbers total and subtract off the complement of that set,*1428

*which is all the numbers that have no 3’s and no 4’s.*1437

*Let us try to count all the numbers with no 3’s and no 4’s.*1454

*That is kind of similar to what we get above.*1458

*We have a three digit number, we have all the digits available to us except there are no 3’s and no 4’s.*1460

*There is 8 possibilities for each digit, 8 times 8 times 8.*1468

*This is 1000 -8³.*1473

*8³, if you know your powers of 2 very well turn out to be 512.*1478

*1000 -512 and that is 488.*1483

*Finally, we are in a position to use our formula for inclusion/exclusion.*1494

*We are trying to count A intersect B.*1497

*We want the numbers that have a 3 and have a 4 in them.*1499

*Inclusion/exclusion says you add up all the a's, all the b’s, and then you subtract off the union.*1503

*That was our second formula for inclusion/exclusion back on the very first slide of this lecture.*1513

*You can go back and check that out.*1519

*In this case, we have 271.*1520

*Let me make that a little more obvious, what I’m writing there.*1525

*271 + 271 – 488.*1528

*271 + 271 is 5420 – 488, that is 42 + 12 which is 54.*1536

*That is the number of student ID numbers that will have both a 3 and a 4 in them.*1551

*They have at least 1-3 and 1-4 in them.*1558

*That is our answer.*1562

*Let me just highlight the key steps there.*1565

*First thing here was to set up some events.*1567

*We set up, we define an event A to be all the numbers with at least 1-3.*1570

*Event B is all the numbers with at least 1-4.*1576

*We are planning to use this inclusion/exclusion formula.*1579

*We want to count the and of something which means we want to count an intersection.*1584

*Our inclusion/exclusion formula for an intersection says that you have to add up the individual sets and then subtract off the union.*1590

*I have got to count all the things in the individual sets and in the union.*1597

*To find the individual sets, A is all the numbers with at least 1-3.*1602

*It is quite tricky to count directly but it is easy to count the complement of that.*1607

*That is what we are doing here.*1611

*We are counting all the things with no 3’s in them.*1613

*That is really a complement there.*1616

*To get the number with no 3’s, you are building a number out of the 9 remaining digits, 0 through 9 but you cannot use a 3.*1620

*There are 9 choices for each decimal place and that is why we got 9³ there.*1628

*1000 -9³ simplifies down to 271.*1635

*These are all the numbers that have at least 1-4.*1639

*That is exactly the same reasoning, you are building a number out of 3 digits but you are not allowed to use 4.*1641

*You end up with 271 again.*1647

*A union B is all the numbers that have at least 1-3 or at least 1-4.*1651

*Again, that is quite a difficult thing to count directly but you can count the compliment.*1658

*The compliment means that you would have no 3’s and no 4’s.*1663

*That is A union B complement right there.*1668

*That means you are trying to build a number using all the digits except no 3’s and no 4’s.*1672

*You are allowed to use 8 digits here, 8 digits here, 8 digits here, and you end up with 8³ numbers.*1678

*I will subtract that from 1000 because that was the complement of what we want.*1685

*You end up with 488.*1689

*And that is just a matter of dropping those numbers into our inclusion/exclusion formula*1691

*and simplifying down to 54 student ID numbers is our final answer there.*1696

*Example 4, we have to figure out how many whole numbers between 1 and 1000 are divisible by 2, 3, or 5?*1707

*Again, this is going to be inclusion/exclusion.*1715

*We are going to have three events here.*1717

*Let me go ahead and define what the events are.*1719

*A is going to be the set of all numbers that are divisible by 2.*1721

*Remember that notation with a colon equals, that means define to be.*1734

*B is defined to be the set of all numbers divisible by 3.*1738

*C is the set of all numbers divisible by 5.*1748

*We are going to use inclusion/exclusion.*1752

*We are trying to find the union of three events here because we want all the numbers that are divisible by at least 1 of 2, 3, or 5.*1754

*You see or, you know you are counting union, A union B union C.*1773

*Let me go ahead and write out the formula for the inclusion/exclusion formula for the union of three events.*1781

*We discover this in the second slide of this topic.*1787

*You can go back and check that out, if you do not remember it.*1791

*It is everything in A + everything in B + everything in C.*1793

*And I have to subtract off the intersections.*1803

*A intersect B - A intersects C - B intersect C.*1813

*And now you have to add in the intersection of all three, A intersect B intersect C.*1816

*Now, we have to think how big each of these sets are.*1825

*A is the set of numbers divisible by 2.*1830

*How many numbers between 1 and 1000 are divisible by 2?*1833

*Since, every other number is divisible by 2, the size of A is 1000 ÷ 2 which is 500.*1837

*There is 500 numbers that are divisible by 2, 500 even numbers.*1852

*How many are divisible by 3?*1856

*It is essential 1000 ÷ 3 but that is not a whole number.*1858

*What happens is it does not quite work because we have multiples of 3 every third number.*1866

*But then at the end, we just get some extra numbers that do not give us anything.*1871

*I’m going to round that down.*1876

*This is the floor function notation.*1879

*It just means I'm running it down to 999/3 and the reason I pick 999 is because it is a multiple of 3.*1881

*That gives me 333 numbers divisible by 3.*1892

*C is divisible by 5, how many numbers are divisible by 5?*1900

*1000/5 and that is a whole number, that is just 200, + 200 here.*1905

*A intersect B is where it starts to get interesting because A intersect B means, it is divisible by 2 and it is divisible by 3.*1913

*Since 2 and 3 are relatively prime, if it is divisible by both of them, A intersect B really means that is divisible by 6.*1922

*If it is divisible by both 2 and 3 then it is divisible by 6.*1937

*It is a multiple of 6.*1941

*How many numbers between 1 and 1000 are divisible by 6?*1942

*Every 6th number is divisible by 6.*1946

*You really have to look at 1000 ÷ 6.*1950

*That is not a whole number.*1954

*I’m going to use this 4 notation to round down.*1955

*The largest number below 1000 that is divisible by 6 is 996.*1960

*We can kind of throw out everything after 996 and just see how many multiples of 6 there are between 1 and 996.*1966

*If we divide that by 6, 996/6 turns out to be 166.*1974

*That is how many numbers there are between 1 and 1000 that are divisible by 6.*1983

*That is what it means to be divisible by 2 and by 3.*1988

*We are going to subtract of 166 here.*1994

*For A intersect C, that is divisible by 2 and divisible by 5.*1999

*If it is divisible by two and by 5 then you are divisible by 10.*2005

*We are going to ask how many multiples of 10 are there between 1 and 1000?*2014

*And of course, there are 1000 ÷ 10.*2019

*I do not have to round that down since it is a whole number.*2024

*There is 100 of those, -100.*2026

*Finally, for B intersect C, that is not finally.*2031

*I’m running out of space here.*2034

*Let me carve out some space down here for myself.*2039

*For B intersect C, it would be the set of numbers that are divisible by both 3 and 5.*2043

*Those are being divisible by 15.*2056

*I have to figure out how many numbers between 1 and 1000 are divisible by 15?*2059

*Again, it is just all the multiples of 15.*2064

*It is every 15th number.*2066

*It is 1000/15 except that is not a whole number.*2068

*We are going to throw out the last few numbers and I’m going to round down.*2075

*When I cut it off at the last multiple of 15 before 1000 which is 990.*2078

*990 ÷ 15 and that turns out be, 90 ÷ 15 is 6, 900 ÷ 15 is 60.*2085

*That is 66 there for B intersect C.*2096

*Finally, A intersect B intersect C, that means you are divisible by 2 and 3 and 5,*2102

*which means you are divisible by the least common multiple of 2 and 3 and 5 which is 30.*2110

*I want to find out how many numbers there are between about 1 and 1000 that are divisible by 30.*2126

*Essentially, I just divide 1000 by 30.*2131

*But again, it is not a whole number.*2133

*I’m going to round down this one sided bracket notation.*2135

*It is the floor function, it means you round down because you are cutting off any numbers*2139

*at the end that would not be divisible by 30.*2144

*The last multiple of 30 before 14000 is also 990.*2146

*Let me throw away all the numbers between 990 and 1000.*2153

*I will just keep the ones up through 990.*2156

*990 ÷ 30 is 33.*2161

*That is the set of numbers divisible by 2, 3, and 5.*2165

*I want to add at the end here, 33.*2170

*Now, it is just a matter of doing the arithmetic.*2175

*500 + 333 + 200 is 1033 - 166 -100 -66 + 33.*2178

*1033 + 33 -66 give us an even 1000 - 100 is 900 -166 is 734.*2198

*That is our answer there.*2212

*That is the number of whole numbers between 1 and 1000 that would be divisible by 2 or 3 or 5,*2214

*or some combination of those prime numbers.*2223

*That is our answer, let me show you again the steps we followed there.*2229

*We first set up three events there.*2233

*A is the stuff divisible by 2.*2235

*B is the stuff divisible by 3.*2238

*C is the stuff divisible by 5.*2241

*I’m going to use the formula for inclusion/exclusion for 3 events to find all the numbers*2243

*that are divisible by at least one of those things.*2250

*I’m counting a union there.*2254

*I’m counting all the stuff in A + all the stuff in B + all the stuff in C.*2256

*Let me remind you how we counted that.*2261

*A is all the multiples of 2.*2264

*To see how many multiples of 2 there are, you just divide 1000 by 2 and you get 500.*2266

*That is where that 500 came from.*2272

*B is the multiples of 3.*2274

*1000/3 does not go quite evenly, you have to round down to 333.*2276

*C is 1000/5, the multiples of 5, there are 200 of them.*2283

*A intersect B, you are looking at multiples of 2 and 3 because it is an intersection, it is N.*2288

*If it is divisible by 2 and 3, then it is divisible by 6.*2295

*We have to find the multiples of 6.*2300

*We look at 1000 ÷ 6 which is not a whole number.*2303

*Throw out all the numbers at the very end which are not visible by 6 anyway.*2307

*And we just look at the last multiple of 6 is 996, divide that by 6 and get 166 multiples of 6 there.*2311

*Similarly, A and C means divisible by 2 and 5, you are divisible by 10.*2319

*Every 10th number is divisible by 10, there is 100 of them.*2325

*B and C divisible by 3 and 5, these are divisible by 15.*2330

*1000/15, once you round down is 66.*2335

*That is where that 66 come from.*2338

*A intersect B intersect C means you are divisible by all three numbers, 2, 3, and 5, which makes you divisible by 30.*2341

*1000/30, we are going to throw out all the numbers at the end that are not multiples of 30.*2350

*We will just stop at 990, the last multiple of 30.*2356

*And then counting up to there, we get 990/30 is 33.*2359

*That is just a matter of simplifying the numbers down and doing the arithmetic and coming up with our answer of 734 numbers.*2364

*In our final example here, we are going out to a busy restaurant.*2375

*They are serving 200 customers that night.*2379

*I'm looking at it from restaurants point of view.*2382

*125 of their customers ordered appetizers and 110 ordered desserts.*2385

*170 of those customers ordered at least one of an appetizer and or a dessert.*2391

*The question is how many ordered both appetizers and desserts?*2399

*Quickly, I need to set up some events here.*2403

*My colon equals, remember, means to find the B.*2408

*I’m just cutting it for short and say A is the set of all people who ordered an appetizer.*2410

*I can remember how to spell appetizer, it would help.*2419

*B is the set of all the people that ordered dessert.*2424

*And we are asked how many ordered both?*2430

*Both means we are looking for people who ordered an appetizer and a dessert.*2434

*That is the intersection.*2438

*We are trying to calculate how many people ordered both?*2440

*Our original rule for inclusion/exclusion on two events, if you go back and look at the very first slide in this lecture,*2445

*we had the formula A intersect B.*2452

*The number of things in A intersect B is the number of things in A + the number of things in B - the number of things in the union.*2455

*We can calculate all of these directly from the problem stem.*2465

*A is the number of people who had appetizers and it tells us that is 125.*2469

*It tells us that up here.*2474

*B is the number people who ordered desserts, there are 110 of them.*2476

*But 170 people ordered at least one, that is the union right there.*2480

*At least one means a union.*2485

*We are going to subtract off 170 in here.*2487

*Let us just do the arithmetic.*2491

*We get 235 -170 is 65 people.*2493

*We must have had 65 people ordering both an appetizer and desert at this particular restaurant.*2501

*Ordered both, that means they are in the intersection of A and B.*2510

*They are in both A and B.*2515

*That was probably easier than some of the other problems here.*2519

*Let me make sure that all the steps are really clear.*2523

*First thing do is to set up events A and B, people who ordered appetizer, people who ordered dessert.*2525

*And then we are asked how many people ordered both which means we are counting an intersection.*2531

*We are going to count an intersection.*2536

*We are going to use the formula for inclusion/exclusion that we had back in the very first slide.*2539

*You count the individual events and then you subtract off the union.*2544

*And we know the size of those because it is given to us in the stem of the problem.*2549

*125 ordered appetizers, 110 ordered desserts, and 170 ordered at least one.*2553

*That is the union right there, it is the 170.*2560

*We just run the arithmetic here and we end up with 65 people ordering both appetizers and desserts.*2563

*65 people in the intersection there.*2572

*Interesting point about this problem is that this 200 customers total in the restaurant appears to be a red herring.*2575

*It does not appear to be relevant at all to solving the problem.*2583

*You do not always have to use every number in the problem to get your answer.*2587

*Often the way, problems in homework exercises are set up.*2592

*You use every number but it is not always true.*2596

*Sometimes there is some red herring information there.*2598

*That wraps up our lecture on inclusion/exclusion.*2602

*I hope you will stick around for some more lectures.*2606

*We got some good stuff coming up on independence and on Bayes' rule in the next couple of lectures.*2608

*These are the probability lecture series here on www.educator.com.*2614

*My name is Will Murray, thank you for watching, bye.*2618

1 answer

Last reply by: Dr. William Murray

Mon Jun 23, 2014 7:25 PM

Post by Sitora Muhamedova on June 21, 2014

I am a bit confuse because in the beginning of the course you had similar equation but a different way to get it.

you had:

P(Aâˆ©B) = P(A)*P(B)

is it since we are doing probability? Little explanation would help a lot.

Thank you