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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 and0107
move the union over to the right hand side, then you get a parallel rule0112
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 separately0360
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 English0385
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 then0564
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 down0586
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 Math0608
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 up0623
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 and0645
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 because0667
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 history1058
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 out1074
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 is1129
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 like1291
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 formula1691
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 numbers2139
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 numbers2243
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
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(A∩B) = P(A)*P(B)
is it since we are doing probability? Little explanation would help a lot.
Thank you