WEBVTT 00:00:00.040 --> 00:00:02.460 The following content is provided under a Creative 00:00:02.460 --> 00:00:03.870 Commons license. 00:00:03.870 --> 00:00:06.910 Your support will help MIT OpenCourseWare continue to 00:00:06.910 --> 00:00:10.560 offer high quality educational resources for free. 00:00:10.560 --> 00:00:13.460 To make a donation, or view additional materials from 00:00:13.460 --> 00:00:19.290 hundreds of MIT courses, visit MIT OpenCourseWare at 00:00:19.290 --> 00:00:21.996 ocw.mit.edu. 00:00:21.996 --> 00:00:22.490 PROFESSOR: OK. 00:00:22.490 --> 00:00:26.380 So today's lecture will be on the subject of counting. 00:00:26.380 --> 00:00:29.620 So counting, I guess, is a pretty simple affair 00:00:29.620 --> 00:00:33.060 conceptually, but it's a topic that can also get 00:00:33.060 --> 00:00:34.410 to be pretty tricky. 00:00:34.410 --> 00:00:37.840 The reason we're going to talk about counting is that there's 00:00:37.840 --> 00:00:41.000 a lot of probability problems whose solution actually 00:00:41.000 --> 00:00:45.050 reduces to successfully counting the cardinalities of 00:00:45.050 --> 00:00:46.110 various sets. 00:00:46.110 --> 00:00:49.270 So we're going to see the basic, simplest methods that 00:00:49.270 --> 00:00:52.350 one can use to count systematically in various 00:00:52.350 --> 00:00:53.660 situations. 00:00:53.660 --> 00:00:56.600 So in contrast to previous lectures, we're not going to 00:00:56.600 --> 00:00:59.700 introduce any significant new concepts of a 00:00:59.700 --> 00:01:01.460 probabilistic nature. 00:01:01.460 --> 00:01:04.340 We're just going to use the probability tools that we 00:01:04.340 --> 00:01:05.680 already know. 00:01:05.680 --> 00:01:08.460 And we're going to apply them in situations where there's 00:01:08.460 --> 00:01:10.840 also some counting involved. 00:01:10.840 --> 00:01:13.000 Now, today we're going to just touch the 00:01:13.000 --> 00:01:14.490 surface of this subject. 00:01:14.490 --> 00:01:16.650 There's a whole field of mathematics called 00:01:16.650 --> 00:01:19.920 combinatorics who are people who actually spend their whole 00:01:19.920 --> 00:01:24.220 lives counting more and more complicated sets. 00:01:24.220 --> 00:01:27.580 We were not going to get anywhere close to the full 00:01:27.580 --> 00:01:31.360 complexity of the field, but we'll get just enough tools 00:01:31.360 --> 00:01:36.260 that allow us to address problems of the type that one 00:01:36.260 --> 00:01:39.730 encounters in most common situations. 00:01:39.730 --> 00:01:43.250 So the basic idea, the basic principle is something that 00:01:43.250 --> 00:01:45.400 we've already discussed. 00:01:45.400 --> 00:01:49.820 So counting methods apply in situations where we have 00:01:49.820 --> 00:01:53.660 probabilistic experiments with a finite number of outcomes 00:01:53.660 --> 00:01:56.070 and where every outcome-- 00:01:56.070 --> 00:01:57.770 every possible outcome-- 00:01:57.770 --> 00:02:00.260 has the same probability of occurring. 00:02:00.260 --> 00:02:04.440 So we have our sample space, omega, and it's got a bunch of 00:02:04.440 --> 00:02:06.520 discrete points in there. 00:02:06.520 --> 00:02:10.539 And the cardinality of the set omega is some capital N. So, 00:02:10.539 --> 00:02:14.960 in particular, we assume that the sample points are equally 00:02:14.960 --> 00:02:18.490 likely, which means that every element of the sample space 00:02:18.490 --> 00:02:22.420 has the same probability equal to 1 over N. 00:02:22.420 --> 00:02:26.650 And then we are interested in a subset of the sample space, 00:02:26.650 --> 00:02:29.860 call it A. And that subset consists 00:02:29.860 --> 00:02:31.350 of a number of elements. 00:02:31.350 --> 00:02:36.080 Let the cardinality of that subset be equal to little n. 00:02:36.080 --> 00:02:39.370 And then to find the probability of that set, all 00:02:39.370 --> 00:02:42.060 we need to do is to add the probabilities of the 00:02:42.060 --> 00:02:43.610 individual elements. 00:02:43.610 --> 00:02:47.030 There's little n elements, and each one has probability one 00:02:47.030 --> 00:02:50.340 over capital N. And that's the answer. 00:02:50.340 --> 00:02:53.250 So this means that to solve problems in this context, all 00:02:53.250 --> 00:02:56.580 that we need to be able to do is to figure out the number 00:02:56.580 --> 00:03:00.800 capital N and to figure out the number little n. 00:03:00.800 --> 00:03:04.330 Now, if somebody gives you a set by just giving you a list 00:03:04.330 --> 00:03:07.660 and gives you another set, again, giving you a list, it's 00:03:07.660 --> 00:03:09.120 easy to count there element. 00:03:09.120 --> 00:03:11.510 You just count how much there is on the list. 00:03:11.510 --> 00:03:15.030 But sometimes the sets are described in some more 00:03:15.030 --> 00:03:20.590 implicit way, and we may have to do a little bit more work. 00:03:20.590 --> 00:03:22.360 There's various tricks that are 00:03:22.360 --> 00:03:24.410 involved in counting properly. 00:03:24.410 --> 00:03:27.080 And the most common one is to-- 00:03:27.080 --> 00:03:31.350 when you consider a set of possible outcomes, to describe 00:03:31.350 --> 00:03:33.800 the construction of those possible outcomes through a 00:03:33.800 --> 00:03:35.440 sequential process. 00:03:35.440 --> 00:03:38.130 So think of a probabilistic experiment that involves a 00:03:38.130 --> 00:03:42.560 number of stages, and in each one of the stages there's a 00:03:42.560 --> 00:03:45.890 number of possible choices that there may be. 00:03:45.890 --> 00:03:48.630 The overall experiment consists of carrying out all 00:03:48.630 --> 00:03:49.930 the stages to the end. 00:03:52.855 --> 00:03:55.830 And the number of points in the sample space is how many 00:03:55.830 --> 00:03:59.920 final outcomes there can be in this multi-stage experiment. 00:03:59.920 --> 00:04:02.910 So in this picture we have an experiment in which of the 00:04:02.910 --> 00:04:07.230 first stage we have four choices. 00:04:07.230 --> 00:04:11.640 In the second stage, no matter what happened in the first 00:04:11.640 --> 00:04:16.010 stage, the way this is drawn we have three choices. 00:04:16.010 --> 00:04:19.720 No matter whether we ended up here, there, or there, we have 00:04:19.720 --> 00:04:22.860 three choices in the second stage. 00:04:22.860 --> 00:04:27.830 And then there's a third stage and at least in this picture, 00:04:27.830 --> 00:04:31.520 no matter what happened in the first two stages, in the third 00:04:31.520 --> 00:04:35.930 stage we're going to have two possible choices. 00:04:35.930 --> 00:04:41.070 So how many leaves are there at the end of this tree? 00:04:41.070 --> 00:04:42.170 That's simple. 00:04:42.170 --> 00:04:45.460 It's just the product of these three numbers. 00:04:45.460 --> 00:04:48.520 The number of possible leaves that we have out there is 4 00:04:48.520 --> 00:04:50.430 times 3 times 2. 00:04:50.430 --> 00:04:54.030 Number of choices at each stage gets multiplied, and 00:04:54.030 --> 00:04:57.680 that gives us the number of overall choices. 00:04:57.680 --> 00:05:01.230 So this is the general rule, the general trick that we are 00:05:01.230 --> 00:05:03.520 going to use over and over. 00:05:03.520 --> 00:05:07.660 So let's apply it to some very simple problems as a warm up. 00:05:07.660 --> 00:05:10.530 How many license plates can you make if you're allowed to 00:05:10.530 --> 00:05:17.140 use three letters and then followed by four digits? 00:05:17.140 --> 00:05:20.020 At least if you're dealing with the English alphabet, you 00:05:20.020 --> 00:05:23.460 have 26 choices for the first letter. 00:05:23.460 --> 00:05:27.100 Then you have 26 choices for the second letter. 00:05:27.100 --> 00:05:30.300 And then 26 choices for the third letter. 00:05:30.300 --> 00:05:31.750 And then we start the digits. 00:05:31.750 --> 00:05:34.970 We have 10 choices for the first digit, 10 choices for 00:05:34.970 --> 00:05:37.780 the second digit, 10 choices for the third, 10 choices for 00:05:37.780 --> 00:05:40.030 the last one. 00:05:40.030 --> 00:05:43.010 Let's make it a little more complicated, suppose that 00:05:43.010 --> 00:05:47.100 we're interested in license plates where no letter can be 00:05:47.100 --> 00:05:50.270 repeated and no digit can be repeated. 00:05:50.270 --> 00:05:53.040 So you have to use different letters, different digits. 00:05:53.040 --> 00:05:55.110 How many license plates can you make? 00:05:55.110 --> 00:05:56.780 OK, let's choose the first letter, 00:05:56.780 --> 00:05:59.130 and we have 26 choices. 00:05:59.130 --> 00:06:02.350 Now, I'm ready to choose my second letter, how many 00:06:02.350 --> 00:06:04.610 choices do I have? 00:06:04.610 --> 00:06:08.090 I have 25, because I already used one letter. 00:06:08.090 --> 00:06:11.900 I have the 25 remaining letters to choose from. 00:06:11.900 --> 00:06:14.330 For the next letter, how many choices? 00:06:14.330 --> 00:06:17.030 Well, I used up two of my letters, so I 00:06:17.030 --> 00:06:19.840 only have 24 available. 00:06:19.840 --> 00:06:22.950 And then we start with the digits, 10 choices for the 00:06:22.950 --> 00:06:26.910 first digit, 9 choices for the second, 8 for the third, 7 for 00:06:26.910 --> 00:06:28.160 the last one. 00:06:30.480 --> 00:06:31.730 All right. 00:06:31.730 --> 00:06:38.710 So, now, let's bring some symbols in a related problem. 00:06:38.710 --> 00:06:44.190 You are given a set that consists of n elements and 00:06:44.190 --> 00:06:47.040 you're supposed to take those n elements and 00:06:47.040 --> 00:06:50.050 put them in a sequence. 00:06:50.050 --> 00:06:52.290 That is to order them. 00:06:52.290 --> 00:06:56.240 Any possible ordering of those elements is called a 00:06:56.240 --> 00:06:57.560 permutation. 00:06:57.560 --> 00:07:02.630 So for example, if we have the set 1, 2, 3, 4, a possible 00:07:02.630 --> 00:07:05.105 permutation is the list 2, 3, 4, 1. 00:07:08.560 --> 00:07:10.680 That's one possible permutation. 00:07:10.680 --> 00:07:13.180 And there's lots of possible permutations, of course, the 00:07:13.180 --> 00:07:15.560 question is how many are there. 00:07:15.560 --> 00:07:20.390 OK, let's think about building this permutation by choosing 00:07:20.390 --> 00:07:21.460 one at a time. 00:07:21.460 --> 00:07:26.330 Which of these elements goes into each one of these slots? 00:07:26.330 --> 00:07:28.740 How many choices for the number that goes into the 00:07:28.740 --> 00:07:31.160 first slot or the elements? 00:07:31.160 --> 00:07:34.430 Well, we can choose any one of the available elements, so we 00:07:34.430 --> 00:07:35.680 have n choices. 00:07:38.650 --> 00:07:42.220 Let's say this element goes here, having used up that 00:07:42.220 --> 00:07:45.730 element, we're left with n minus 1 elements and we can 00:07:45.730 --> 00:07:49.530 pick any one of these and bring it into the second slot. 00:07:49.530 --> 00:07:52.340 So here we have n choices, here we're going to have n 00:07:52.340 --> 00:07:55.940 minus 1 choices, then how many we put there will 00:07:55.940 --> 00:07:58.060 have n minus 2 choices. 00:07:58.060 --> 00:08:00.220 And you go down until the end. 00:08:00.220 --> 00:08:02.160 What happens at this point when you are to 00:08:02.160 --> 00:08:03.880 pick the last element? 00:08:03.880 --> 00:08:06.650 Well, you've used n minus of them, there's only one 00:08:06.650 --> 00:08:08.070 left in your bag. 00:08:08.070 --> 00:08:09.520 You're forced to use that one. 00:08:09.520 --> 00:08:13.920 So the last stage, you're going to have only one choice. 00:08:13.920 --> 00:08:18.050 So, basically, the number of possible permutations is the 00:08:18.050 --> 00:08:21.860 product of all integers from n down to one, or 00:08:21.860 --> 00:08:24.020 from one up to n. 00:08:24.020 --> 00:08:26.550 And there's a symbol that we use for this number, it's 00:08:26.550 --> 00:08:29.210 called n factorial. 00:08:29.210 --> 00:08:32.990 So n factorial is the number of permutations of n objects. 00:08:32.990 --> 00:08:37.320 The number of ways that you can order n objects that are 00:08:37.320 --> 00:08:39.100 given to you. 00:08:39.100 --> 00:08:42.100 Now, a different equation. 00:08:42.100 --> 00:08:44.310 We have n elements. 00:08:44.310 --> 00:08:48.680 Let's say the elements are 1, 1,2, up to n. 00:08:48.680 --> 00:08:51.310 And it's a set. 00:08:51.310 --> 00:08:54.490 And we want to create a subset. 00:08:54.490 --> 00:08:58.460 How many possible subsets are there? 00:08:58.460 --> 00:09:02.950 So speaking of subsets means looking at each one of the 00:09:02.950 --> 00:09:06.880 elements and deciding whether you're going to put it in to 00:09:06.880 --> 00:09:08.440 subsets or not. 00:09:08.440 --> 00:09:13.240 For example, I could choose to put 1 in, but 2 I'm not 00:09:13.240 --> 00:09:17.100 putting it in, 3 I'm not putting it in, 4 I'm putting 00:09:17.100 --> 00:09:18.630 it, and so on. 00:09:18.630 --> 00:09:21.200 So that's how you create a subset. 00:09:21.200 --> 00:09:23.660 You look at each one of the elements and you say, OK, I'm 00:09:23.660 --> 00:09:27.310 going to put it in the subset, or I'm not going to put it. 00:09:27.310 --> 00:09:30.900 So think of these as consisting of stages. 00:09:30.900 --> 00:09:33.240 At each stage you look at one element, and you 00:09:33.240 --> 00:09:35.090 make a binary decision. 00:09:35.090 --> 00:09:38.410 Do I put it in the subset, or not? 00:09:38.410 --> 00:09:41.940 So therefore, how many subsets are there? 00:09:41.940 --> 00:09:45.060 Well, I have two choices for the first element. 00:09:45.060 --> 00:09:47.740 Am I going to put in the subset, or not? 00:09:47.740 --> 00:09:50.630 I have two choices for the next element, and so on. 00:09:53.450 --> 00:09:57.390 For each one of the elements, we have two choices. 00:09:57.390 --> 00:10:02.090 So the overall number of choices is 2 to the power n. 00:10:02.090 --> 00:10:03.710 So, conclusion-- 00:10:03.710 --> 00:10:10.150 the number of subsets, often n element set, is 2 to the n. 00:10:15.050 --> 00:10:20.430 So in particular, if we take n equal to 1, let's check that 00:10:20.430 --> 00:10:22.190 our answer makes sense. 00:10:22.190 --> 00:10:26.420 If we have n equal to one, how many subsets does it have? 00:10:26.420 --> 00:10:29.675 So we're dealing with a set of just one. 00:10:29.675 --> 00:10:30.925 What are the subsets? 00:10:33.830 --> 00:10:37.290 One subset is this one. 00:10:37.290 --> 00:10:41.530 Do we have other subsets of the one element set? 00:10:41.530 --> 00:10:43.920 Yes, we have the empty set. 00:10:43.920 --> 00:10:44.870 That's the second one. 00:10:44.870 --> 00:10:48.860 These are the two possible subsets of 00:10:48.860 --> 00:10:50.850 this particular set. 00:10:50.850 --> 00:10:56.790 So 2 subsets when n is equal to 1, that checks the answer. 00:10:56.790 --> 00:10:58.040 All right. 00:11:00.290 --> 00:11:07.590 OK, so having gone so far, we can do our first example now. 00:11:07.590 --> 00:11:12.990 So we are given a die and we're going 00:11:12.990 --> 00:11:16.620 to roll it 6 times. 00:11:16.620 --> 00:11:20.030 OK, let's make some assumptions about the rolls. 00:11:20.030 --> 00:11:29.560 Let's assume that the rolls are independent, and that the 00:11:29.560 --> 00:11:30.985 die is also fair. 00:11:34.180 --> 00:11:38.110 So this means that the probability of any particular 00:11:38.110 --> 00:11:40.350 outcome of the die rolls-- 00:11:40.350 --> 00:11:43.960 for example, so we have 6 rolls, one particular outcome 00:11:43.960 --> 00:11:48.760 could be 3,3,1,6,5. 00:11:48.760 --> 00:11:51.220 So that's one possible outcome. 00:11:51.220 --> 00:11:54.060 What's the probability of this outcome? 00:11:54.060 --> 00:11:57.470 There's probability 1/6 that this happens, 1/6 that this 00:11:57.470 --> 00:12:00.530 happens, 1/6 that this happens, and so on. 00:12:00.530 --> 00:12:04.250 So the probability that the outcome is this 00:12:04.250 --> 00:12:07.540 is 1/6 to the sixth. 00:12:10.970 --> 00:12:13.690 What did I use to come up with this answer? 00:12:13.690 --> 00:12:17.550 I used independence, so I multiplied the probability of 00:12:17.550 --> 00:12:20.360 the first roll gives me a 2, times the probability that the 00:12:20.360 --> 00:12:22.990 second roll gives me a 3, and so on. 00:12:22.990 --> 00:12:26.790 And then I used the assumption that the die is fair, so that 00:12:26.790 --> 00:12:30.300 the probability of 2 is 1/6, the probably of 3 00:12:30.300 --> 00:12:32.240 is 1/6, and so on. 00:12:32.240 --> 00:12:34.810 So if I were to spell it out, it's the probability that we 00:12:34.810 --> 00:12:37.740 get the 2 in the first roll, times the probability of 3 in 00:12:37.740 --> 00:12:40.850 the second roll, times the probability of the 00:12:40.850 --> 00:12:42.800 5 in the last roll. 00:12:42.800 --> 00:12:46.455 So by independence, I can multiply probabilities. 00:12:46.455 --> 00:12:49.530 And because the die is fair, each one of these numbers is 00:12:49.530 --> 00:12:53.910 1/6 to the sixth. 00:12:53.910 --> 00:12:58.170 And so the same calculation would apply no matter what 00:12:58.170 --> 00:13:00.610 numbers I would put in here. 00:13:00.610 --> 00:13:03.920 So all possible outcomes are equally likely. 00:13:06.630 --> 00:13:08.400 Let's start with this. 00:13:08.400 --> 00:13:12.650 So since all possible outcomes are equally likely to find an 00:13:12.650 --> 00:13:15.870 answer to a probability question, if we're dealing 00:13:15.870 --> 00:13:21.430 with some particular event, so the event is that all rolls 00:13:21.430 --> 00:13:22.900 give different numbers. 00:13:22.900 --> 00:13:31.160 That's our event A. And our sample space is some set 00:13:31.160 --> 00:13:32.790 capital omega. 00:13:32.790 --> 00:13:35.890 We know that the answer is going to be the cardinality of 00:13:35.890 --> 00:13:40.200 the set A, divided by the cardinality of the set omega. 00:13:40.200 --> 00:13:42.830 So let's deal with the easy one first. 00:13:42.830 --> 00:13:45.960 How many elements are there in the sample space? 00:13:45.960 --> 00:13:48.880 How many possible outcomes are there when you 00:13:48.880 --> 00:13:51.570 roll a dice 6 times? 00:13:51.570 --> 00:13:54.600 You have 6 choices for the first roll. 00:13:54.600 --> 00:13:57.840 You have 6 choices for the second roll and so on. 00:13:57.840 --> 00:14:00.330 So the overall number of outcomes is going 00:14:00.330 --> 00:14:03.950 to be 6 to the sixth. 00:14:03.950 --> 00:14:08.200 So number of elements in the sample space is 6 00:14:08.200 --> 00:14:10.470 to the sixth power. 00:14:10.470 --> 00:14:14.820 And I guess this checks with this. 00:14:14.820 --> 00:14:18.480 We have 6 to the sixth outcomes, each one has this 00:14:18.480 --> 00:14:20.570 much probability, so the overall 00:14:20.570 --> 00:14:23.230 probability is equal to one. 00:14:23.230 --> 00:14:24.460 Right? 00:14:24.460 --> 00:14:28.690 So the probability of an individual outcome is one over 00:14:28.690 --> 00:14:32.400 how many possible outcomes we have, which is this. 00:14:32.400 --> 00:14:33.810 All right. 00:14:33.810 --> 00:14:36.620 So how about the numerator? 00:14:36.620 --> 00:14:42.430 We are interested in outcomes in which the numbers that we 00:14:42.430 --> 00:14:44.585 get are all different. 00:14:48.770 --> 00:14:54.080 So what is an outcome in which the numbers are all different? 00:14:54.080 --> 00:14:56.310 So the die has 6 faces. 00:14:56.310 --> 00:14:58.200 We roll it 6 times. 00:14:58.200 --> 00:15:00.340 We're going to get 6 different numbers. 00:15:00.340 --> 00:15:03.780 This means that we're going to exhaust all the possible 00:15:03.780 --> 00:15:07.640 numbers, but they can appear in any possible sequence. 00:15:07.640 --> 00:15:13.190 So an outcome that makes this event happen is a list of the 00:15:13.190 --> 00:15:16.250 numbers from 1 to 6, but arranged in 00:15:16.250 --> 00:15:18.160 some arbitrary order. 00:15:18.160 --> 00:15:23.200 So the possible outcomes that make event A happen are just 00:15:23.200 --> 00:15:25.990 the permutations of the numbers from 1 to 6. 00:15:31.070 --> 00:15:33.900 One possible outcome that makes our events to happen-- 00:15:33.900 --> 00:15:35.440 it would be this. 00:15:39.000 --> 00:15:42.050 Here we have 6 possible numbers, but any other list of 00:15:42.050 --> 00:15:44.070 this kind in which none of the numbers is 00:15:44.070 --> 00:15:46.650 repeated would also do. 00:15:46.650 --> 00:15:51.660 So number of outcomes that make the event happen is the 00:15:51.660 --> 00:15:53.980 number of permutations of 6 elements. 00:15:53.980 --> 00:15:56.060 So it's 6 factorial. 00:15:56.060 --> 00:15:59.340 And so the final answer is going to be 6 factorial 00:15:59.340 --> 00:16:02.830 divided by 6 to the sixth. 00:16:02.830 --> 00:16:06.580 All right, so that's a typical way that's one solves problems 00:16:06.580 --> 00:16:07.800 of this kind. 00:16:07.800 --> 00:16:10.660 We know how to count certain things. 00:16:10.660 --> 00:16:14.260 For example, here we knew how to count permutations, and we 00:16:14.260 --> 00:16:16.830 used our knowledge to count the elements of the set that 00:16:16.830 --> 00:16:18.080 we need to deal with. 00:16:24.380 --> 00:16:30.970 So now let's get to a slightly more difficult problem. 00:16:30.970 --> 00:16:37.390 We're given once more a set with n elements. 00:16:40.620 --> 00:16:46.000 We already know how many subsets that set has, but now 00:16:46.000 --> 00:16:50.940 we would be interested in subsets that have exactly k 00:16:50.940 --> 00:16:54.480 elements in them. 00:16:54.480 --> 00:17:05.819 So we start with our big set that has n elements, and we 00:17:05.819 --> 00:17:11.890 want to construct a subset that has k elements. 00:17:11.890 --> 00:17:14.200 Out of those n I'm going to choose k 00:17:14.200 --> 00:17:16.079 and put them in there. 00:17:16.079 --> 00:17:18.180 In how many ways can I do this? 00:17:18.180 --> 00:17:20.710 More concrete way of thinking about this problem-- 00:17:20.710 --> 00:17:24.960 you have n people in some group and you want to form a 00:17:24.960 --> 00:17:28.270 committee by picking people from that group, and you want 00:17:28.270 --> 00:17:31.020 to form a committee with k people. 00:17:31.020 --> 00:17:32.460 Where k is a given number. 00:17:32.460 --> 00:17:34.670 For example, a 5 person committee. 00:17:34.670 --> 00:17:37.510 How many 5 person committees are possible if you're 00:17:37.510 --> 00:17:39.450 starting with 100 people? 00:17:39.450 --> 00:17:40.960 So that's what we want to count. 00:17:40.960 --> 00:17:44.210 How many k element subsets are there? 00:17:44.210 --> 00:17:48.030 We don't yet know the answer, but let's give a name to it. 00:17:48.030 --> 00:17:52.210 And the name is going to be this particular symbol, which 00:17:52.210 --> 00:17:55.220 we read as n choose k. 00:17:55.220 --> 00:18:00.130 Out of n elements, we want to choose k of them. 00:18:00.130 --> 00:18:02.000 OK. 00:18:02.000 --> 00:18:04.810 That may be a little tricky. 00:18:04.810 --> 00:18:10.170 So what we're going to do is to instead figure out a 00:18:10.170 --> 00:18:15.590 somewhat easier problem, which is going to be-- 00:18:15.590 --> 00:18:20.840 in how many ways can I pick k out of these people and puts 00:18:20.840 --> 00:18:25.450 them in a particular order? 00:18:25.450 --> 00:18:30.430 So how many possible ordered lists can I make that consist 00:18:30.430 --> 00:18:31.840 of k people? 00:18:31.840 --> 00:18:35.240 By ordered, I mean that we take those k people and we say 00:18:35.240 --> 00:18:38.010 this is the first person in the community. 00:18:38.010 --> 00:18:39.600 That's the second person in the committee. 00:18:39.600 --> 00:18:42.070 That's the third person in the committee and so on. 00:18:42.070 --> 00:18:46.100 So in how many ways can we do this? 00:18:46.100 --> 00:18:50.840 Out of these n, we want to choose just k of them and put 00:18:50.840 --> 00:18:52.010 them in slots. 00:18:52.010 --> 00:18:53.970 One after the other. 00:18:53.970 --> 00:18:57.480 So this is pretty much like the license plate problem we 00:18:57.480 --> 00:19:00.680 solved just a little earlier. 00:19:00.680 --> 00:19:06.390 So we have n choices for who we put as the top person in 00:19:06.390 --> 00:19:07.350 the community. 00:19:07.350 --> 00:19:11.490 We can pick anyone and have them be the first person. 00:19:11.490 --> 00:19:13.000 Then I'm going to choose the second 00:19:13.000 --> 00:19:14.640 person in the committee. 00:19:14.640 --> 00:19:16.700 I've used up 1 person. 00:19:16.700 --> 00:19:21.530 So I'm going to have n minus 1 choices here. 00:19:21.530 --> 00:19:25.840 And now, at this stage I've used up 2 people, so I have n 00:19:25.840 --> 00:19:28.640 minus 2 choices here. 00:19:28.640 --> 00:19:31.240 And this keeps going on. 00:19:31.240 --> 00:19:34.110 Well, what is going to be the last number? 00:19:34.110 --> 00:19:36.310 Is it's n minus k? 00:19:36.310 --> 00:19:39.980 Well, not really. 00:19:39.980 --> 00:19:44.090 I'm starting subtracting numbers after the second one, 00:19:44.090 --> 00:19:48.250 so by the end I will have subtracted k minus 1. 00:19:48.250 --> 00:19:54.800 So that's how many choices I will have for the last person. 00:19:54.800 --> 00:19:58.270 So this is the number of ways-- 00:19:58.270 --> 00:20:02.390 the product of these numbers there gives me the number of 00:20:02.390 --> 00:20:08.420 ways that I can create ordered lists consisting of k people 00:20:08.420 --> 00:20:11.700 out of the n that we started with. 00:20:11.700 --> 00:20:15.120 Now, you can do a little bit of algebra and check that this 00:20:15.120 --> 00:20:17.910 expression here is the same as that expression. 00:20:17.910 --> 00:20:19.200 Why is this? 00:20:19.200 --> 00:20:22.520 This factorial has all the products from 1 up to n. 00:20:22.520 --> 00:20:25.140 This factorial has all the products from 1 00:20:25.140 --> 00:20:26.710 up to n minus k. 00:20:26.710 --> 00:20:28.240 So you get cancellations. 00:20:28.240 --> 00:20:31.860 And what's left is all the products starting from the 00:20:31.860 --> 00:20:37.610 next number after here, which is this particular number. 00:20:37.610 --> 00:20:42.350 So the number of possible ways of creating such ordered lists 00:20:42.350 --> 00:20:46.330 is n factorial divided by n minus k factorial. 00:20:49.480 --> 00:20:53.180 Now, a different way that I could make an ordered list-- 00:20:53.180 --> 00:20:57.950 instead of picking the people one at a time, I could first 00:20:57.950 --> 00:21:01.700 choose my k people who are going to be in the committee, 00:21:01.700 --> 00:21:04.080 and then put them in order. 00:21:04.080 --> 00:21:07.680 And tell them out of these k, you are the first, you are the 00:21:07.680 --> 00:21:10.010 second, you are the third. 00:21:10.010 --> 00:21:12.590 Starting with this k people, in how many 00:21:12.590 --> 00:21:15.580 ways can I order them? 00:21:15.580 --> 00:21:18.150 That's the number of permutations. 00:21:20.820 --> 00:21:25.180 Starting with a set with k objects, in how many ways can 00:21:25.180 --> 00:21:28.340 I put them in a specific order? 00:21:28.340 --> 00:21:31.140 How many specific orders are there? 00:21:31.140 --> 00:21:32.390 That's basically the question. 00:21:32.390 --> 00:21:34.600 In how many ways can I permute these k 00:21:34.600 --> 00:21:36.290 people and arrange them. 00:21:36.290 --> 00:21:38.450 So the number of ways that you can do 00:21:38.450 --> 00:21:42.660 this step is k factorial. 00:21:42.660 --> 00:21:48.330 So in how many ways can I start with a set with n 00:21:48.330 --> 00:21:52.020 elements, go through this process, and end up with a 00:21:52.020 --> 00:21:55.560 sorted list with k elements? 00:21:55.560 --> 00:21:57.620 By the rule that-- 00:21:57.620 --> 00:22:02.160 when we have stages, the total number of stages is how many 00:22:02.160 --> 00:22:05.370 choices we had in the first stage, times how many choices 00:22:05.370 --> 00:22:08.510 we had in the second stage. 00:22:08.510 --> 00:22:12.670 The number of ways that this process can happen is this 00:22:12.670 --> 00:22:14.890 times that. 00:22:14.890 --> 00:22:18.340 This is a different way that that process could happen. 00:22:18.340 --> 00:22:22.640 And the number of possible of ways is this number. 00:22:22.640 --> 00:22:27.600 No matter which way we carry out that process, in the end 00:22:27.600 --> 00:22:34.610 we have the possible ways of arranging k people out of the 00:22:34.610 --> 00:22:36.770 n that we started with. 00:22:36.770 --> 00:22:40.730 So the final answer that we get when we count should be 00:22:40.730 --> 00:22:44.220 either this, or this times that. 00:22:44.220 --> 00:22:47.620 Both are equally valid ways of counting, so both should give 00:22:47.620 --> 00:22:49.050 us the same answer. 00:22:49.050 --> 00:22:52.950 So we get this equality here. 00:22:52.950 --> 00:22:56.370 So these two expressions corresponds to two different 00:22:56.370 --> 00:23:01.660 ways of constructing ordered lists of k people starting 00:23:01.660 --> 00:23:05.580 with n people initially. 00:23:05.580 --> 00:23:09.120 And now that we have this relation, we can send the k 00:23:09.120 --> 00:23:11.540 factorial to the denominator. 00:23:11.540 --> 00:23:13.940 And that tells us what that number, n choose 00:23:13.940 --> 00:23:16.250 k, is going to be. 00:23:16.250 --> 00:23:20.060 So this formula-- it's written here in red, because you're 00:23:20.060 --> 00:23:22.150 going to see it a zillion times until 00:23:22.150 --> 00:23:23.740 the end of the semester-- 00:23:23.740 --> 00:23:25.950 they are called the binomial coefficients. 00:23:31.170 --> 00:23:34.600 And they tell us the number of possible ways that we can 00:23:34.600 --> 00:23:38.380 create a k element subset, starting with a 00:23:38.380 --> 00:23:41.270 set that has n elements. 00:23:41.270 --> 00:23:44.430 It's always good to do a sanity check to formulas by 00:23:44.430 --> 00:23:46.710 considering extreme cases. 00:23:46.710 --> 00:23:52.810 So let's take the case where k is equal to n. 00:23:56.820 --> 00:23:59.420 What's the right answer in this case? 00:23:59.420 --> 00:24:02.905 How many n elements subsets are there out 00:24:02.905 --> 00:24:04.950 of an element set? 00:24:04.950 --> 00:24:07.580 Well, your subset needs to include every one. 00:24:07.580 --> 00:24:09.400 You don't have any choices. 00:24:09.400 --> 00:24:10.750 There's only one choice. 00:24:10.750 --> 00:24:12.600 It's the set itself. 00:24:12.600 --> 00:24:15.700 So the answer should be equal to 1. 00:24:15.700 --> 00:24:19.980 That's the number of n element subsets, starting with a set 00:24:19.980 --> 00:24:21.340 with n elements. 00:24:21.340 --> 00:24:25.250 Let's see if the formula gives us the right answer. 00:24:25.250 --> 00:24:31.750 We have n factorial divided by k, which is n in our case-- 00:24:31.750 --> 00:24:32.630 n factorial. 00:24:32.630 --> 00:24:36.700 And then n minus k is 0 factorial. 00:24:36.700 --> 00:24:42.070 So if our formula is correct, we should have this equality. 00:24:42.070 --> 00:24:45.620 And what's the way to make that correct? 00:24:45.620 --> 00:24:47.880 Well, it depends what kind of meaning do we 00:24:47.880 --> 00:24:49.420 give to this symbol? 00:24:49.420 --> 00:24:53.510 How do we define zero factorial? 00:24:53.510 --> 00:24:55.750 I guess in some ways it's arbitrary. 00:24:55.750 --> 00:24:58.110 We're going to define it in a way that makes 00:24:58.110 --> 00:24:59.640 this formula right. 00:24:59.640 --> 00:25:03.870 So the definition that we will be using is that whenever you 00:25:03.870 --> 00:25:08.700 have 0 factorial, it's going to stand for the number 1. 00:25:08.700 --> 00:25:12.030 So let's check that this is also correct, at the other 00:25:12.030 --> 00:25:13.380 extreme case. 00:25:13.380 --> 00:25:17.670 If we let k equal to 0, what does the formula give us? 00:25:17.670 --> 00:25:20.710 It gives us, again, n factorial divided by 0 00:25:20.710 --> 00:25:23.090 factorial times n factorial. 00:25:23.090 --> 00:25:27.560 According to our convention, this again is equal to 1. 00:25:27.560 --> 00:25:33.680 So there is one subset of our set that we started with that 00:25:33.680 --> 00:25:35.260 has zero elements. 00:25:35.260 --> 00:25:37.450 Which subset is it? 00:25:37.450 --> 00:25:39.980 It's the empty set. 00:25:39.980 --> 00:25:45.190 So the empty set is the single subset of the set that we 00:25:45.190 --> 00:25:49.360 started with that happens to have exactly zero elements. 00:25:49.360 --> 00:25:52.510 So the formula checks in this extreme case as well. 00:25:52.510 --> 00:25:55.820 So we're comfortable using it. 00:25:55.820 --> 00:26:01.180 Now these factorials and these coefficients are really messy 00:26:01.180 --> 00:26:03.050 algebraic objects. 00:26:03.050 --> 00:26:07.740 There's lots of beautiful identities that they satisfy, 00:26:07.740 --> 00:26:10.350 which you can prove algebraically sometimes by 00:26:10.350 --> 00:26:13.930 using induction and having cancellations happen 00:26:13.930 --> 00:26:15.310 all over the place. 00:26:15.310 --> 00:26:17.780 But it's really messy. 00:26:17.780 --> 00:26:22.540 Sometimes you can bypass those calculations by being clever 00:26:22.540 --> 00:26:24.630 and using your understanding of what these 00:26:24.630 --> 00:26:26.620 coefficients stand for. 00:26:26.620 --> 00:26:31.490 So here's a typical example. 00:26:31.490 --> 00:26:35.450 What is the sum of those binomial coefficients? 00:26:35.450 --> 00:26:40.130 I fix n, and sum over all possible cases. 00:26:40.130 --> 00:26:44.110 So if you're an algebra genius, you're going to take 00:26:44.110 --> 00:26:49.830 this expression here, plug it in here, and then start doing 00:26:49.830 --> 00:26:51.460 algebra furiously. 00:26:51.460 --> 00:26:54.970 And half an hour later, you may get the right answer. 00:26:54.970 --> 00:26:56.425 But now let's try to be clever. 00:26:59.470 --> 00:27:01.380 What does this really do? 00:27:01.380 --> 00:27:04.200 What does that formula count? 00:27:04.200 --> 00:27:07.280 We're considering k element subsets. 00:27:07.280 --> 00:27:09.040 That's this number. 00:27:09.040 --> 00:27:12.360 And we're considering the number of k element subsets 00:27:12.360 --> 00:27:14.840 for different choices of k. 00:27:14.840 --> 00:27:18.890 The first term in this sum counts how many 0-element 00:27:18.890 --> 00:27:20.450 subsets we have. 00:27:20.450 --> 00:27:23.680 The next term in this sum counts how many 1-element 00:27:23.680 --> 00:27:24.660 subsets we have. 00:27:24.660 --> 00:27:30.010 The next term counts how many 2-element subsets we have. 00:27:30.010 --> 00:27:33.130 So in the end, what have we counted? 00:27:33.130 --> 00:27:36.660 We've counted the total number of subsets. 00:27:36.660 --> 00:27:38.430 We've considered all possible cardinalities. 00:27:43.420 --> 00:27:46.850 We've counted the number of subsets of size k. 00:27:46.850 --> 00:27:49.740 We've considered all possible sizes k. 00:27:49.740 --> 00:27:52.230 The overall count is going to be the 00:27:52.230 --> 00:27:54.356 total number of subsets. 00:27:57.880 --> 00:28:00.800 And we know what this is. 00:28:00.800 --> 00:28:03.740 A couple of slides ago, we discussed that this number is 00:28:03.740 --> 00:28:05.480 equal to 2 to the n. 00:28:05.480 --> 00:28:11.550 So, nice, clean and simple answer, which is easy to guess 00:28:11.550 --> 00:28:15.110 once you give an interpretation to the 00:28:15.110 --> 00:28:17.580 algebraic expression that you have in front of you. 00:28:21.610 --> 00:28:22.280 All right. 00:28:22.280 --> 00:28:27.410 So let's move again to sort of an example in which those 00:28:27.410 --> 00:28:31.960 binomial coefficients are going to show up. 00:28:31.960 --> 00:28:34.700 So here's the setting-- 00:28:34.700 --> 00:28:40.900 n independent coin tosses, and each coin toss has a 00:28:40.900 --> 00:28:45.770 probability, P, of resulting in heads. 00:28:45.770 --> 00:28:48.320 So this is our probabilistic experiment. 00:28:48.320 --> 00:28:51.200 Suppose we do 6 tosses. 00:28:51.200 --> 00:28:53.980 What's the probability that we get this particular sequence 00:28:53.980 --> 00:28:56.320 of outcomes? 00:28:56.320 --> 00:28:59.800 Because of independence, we can multiply probability. 00:28:59.800 --> 00:29:02.400 So it's going to be the probability that the first 00:29:02.400 --> 00:29:05.570 toss results in heads, times the probability that the 00:29:05.570 --> 00:29:08.770 second toss results in tails, times the probability that the 00:29:08.770 --> 00:29:12.050 third one results in tails, times probability of heads, 00:29:12.050 --> 00:29:14.610 times probability of heads, times probability of heads, 00:29:14.610 --> 00:29:20.930 which is just P to the fourth times (1 minus P) squared. 00:29:20.930 --> 00:29:24.360 So that's the probability of this particular sequence. 00:29:24.360 --> 00:29:26.980 How about a different sequence? 00:29:26.980 --> 00:29:32.830 If I had 4 tails and 2 heads, but in a different order-- 00:29:39.130 --> 00:29:42.870 let's say if we considered this particular outcome-- 00:29:42.870 --> 00:29:45.480 would the answer be different? 00:29:45.480 --> 00:29:49.020 We would still have P, times P, times P, times P, times (1 00:29:49.020 --> 00:29:51.070 minus P), times (1 minus P). 00:29:51.070 --> 00:29:54.670 We would get again, the same answer. 00:29:54.670 --> 00:29:59.510 So what you observe from just this example is that, more 00:29:59.510 --> 00:30:03.240 generally, the probability of obtaining a particular 00:30:03.240 --> 00:30:08.930 sequence of heads and tails is P to a power, equal to the 00:30:08.930 --> 00:30:10.300 number of heads. 00:30:10.300 --> 00:30:12.240 So here we had 4 heads. 00:30:12.240 --> 00:30:15.100 So there's P to the fourth showing up. 00:30:15.100 --> 00:30:21.116 And then (1 minus P) to the power number of tails. 00:30:21.116 --> 00:30:26.970 So every k head sequence-- 00:30:26.970 --> 00:30:32.310 every outcome in which we have exactly k heads, has the same 00:30:32.310 --> 00:30:37.180 probability, which is going to be P to the k, (1 minus p), to 00:30:37.180 --> 00:30:38.930 the (n minus k). 00:30:38.930 --> 00:30:43.310 This is the probability of any particular sequence that has 00:30:43.310 --> 00:30:44.980 exactly k heads. 00:30:44.980 --> 00:30:46.980 So that's the probability of a particular 00:30:46.980 --> 00:30:48.920 sequence with k heads. 00:30:48.920 --> 00:30:53.160 So now let's ask the question, what is the probability that 00:30:53.160 --> 00:30:57.980 my experiment results in exactly k heads, but in some 00:30:57.980 --> 00:30:59.930 arbitrary order? 00:30:59.930 --> 00:31:02.500 So the heads could show up anywhere. 00:31:02.500 --> 00:31:04.080 So there's a number of different ways 00:31:04.080 --> 00:31:05.370 that this can happen. 00:31:05.370 --> 00:31:11.220 What's the overall probability that this event takes place? 00:31:11.220 --> 00:31:15.560 So the probability of an event taking place is the sum of the 00:31:15.560 --> 00:31:19.390 probabilities of all the individual ways that 00:31:19.390 --> 00:31:22.020 the event can occur. 00:31:22.020 --> 00:31:24.410 So it's the sum of the probabilities of all the 00:31:24.410 --> 00:31:27.650 outcomes that make the event happen. 00:31:27.650 --> 00:31:31.420 The different ways that we can obtain k heads are the number 00:31:31.420 --> 00:31:37.940 of different sequences that contain exactly k heads. 00:31:37.940 --> 00:31:44.430 We just figured out that any sequence with exactly k heads 00:31:44.430 --> 00:31:47.110 has this probability. 00:31:47.110 --> 00:31:51.110 So to do this summation, we just need to take the common 00:31:51.110 --> 00:31:56.030 probability of each individual k head sequence, times how 00:31:56.030 --> 00:31:58.140 many terms we have in this sum. 00:32:01.320 --> 00:32:07.020 So what we're left to do now is to figure out how many k 00:32:07.020 --> 00:32:09.990 head sequences are there. 00:32:09.990 --> 00:32:15.448 How many outcomes are there in which we have exactly k heads. 00:32:18.940 --> 00:32:21.270 OK. 00:32:21.270 --> 00:32:27.590 So what are the ways that I can describe to you a sequence 00:32:27.590 --> 00:32:30.600 with k heads? 00:32:30.600 --> 00:32:34.970 I can take my n slots that corresponds to 00:32:34.970 --> 00:32:36.220 the different tosses. 00:32:42.920 --> 00:32:45.420 I'm interested in particular sequences that 00:32:45.420 --> 00:32:47.750 have exactly k heads. 00:32:47.750 --> 00:32:53.590 So what I need to do is to choose k slots and assign 00:32:53.590 --> 00:32:54.850 heads to them. 00:33:05.530 --> 00:33:11.580 So to specify a sequence that has exactly k heads is the 00:33:11.580 --> 00:33:17.380 same thing as drawing this picture and telling you which 00:33:17.380 --> 00:33:23.640 are the k slots that happened to have heads. 00:33:23.640 --> 00:33:30.110 So I need to choose out of those n slots, k of them, and 00:33:30.110 --> 00:33:31.635 assign them heads. 00:33:31.635 --> 00:33:35.290 In how many ways can I choose this k slots? 00:33:35.290 --> 00:33:41.640 Well, it's the question of starting with a set of n slots 00:33:41.640 --> 00:33:47.080 and choosing k slots out of the n available. 00:33:47.080 --> 00:33:55.540 So the number of k head sequences is the same as the 00:33:55.540 --> 00:34:04.300 number of k element subsets of the set of slots that we 00:34:04.300 --> 00:34:10.520 started with, which are the n slots 1 up to n. 00:34:10.520 --> 00:34:12.800 We know what that number is. 00:34:12.800 --> 00:34:18.770 We counted, before, the number of k element subsets, starting 00:34:18.770 --> 00:34:20.290 with a set with n elements. 00:34:20.290 --> 00:34:23.030 And we gave a symbol to that number, which is that 00:34:23.030 --> 00:34:24.850 thing, n choose k. 00:34:24.850 --> 00:34:28.110 So this is the final answer that we obtain. 00:34:28.110 --> 00:34:32.449 So these are the so-called binomial probabilities. 00:34:32.449 --> 00:34:35.190 And they gave us the probabilities for different 00:34:35.190 --> 00:34:39.580 numbers of heads starting with a fair coin that's being 00:34:39.580 --> 00:34:42.050 tossed a number of times. 00:34:42.050 --> 00:34:46.170 This formula is correct, of course, for reasonable values 00:34:46.170 --> 00:34:52.370 of k, meaning its correct for k equals 0, 1, up to n. 00:34:52.370 --> 00:34:57.650 If k is bigger than n, what's the probability of k heads? 00:34:57.650 --> 00:35:01.340 If k is bigger than n, there's no way to obtain k heads, so 00:35:01.340 --> 00:35:03.480 that probability is, of course, zero. 00:35:03.480 --> 00:35:07.610 So these probabilities only makes sense for the numbers k 00:35:07.610 --> 00:35:10.405 that are possible, given that we have n tosses. 00:35:13.200 --> 00:35:16.850 And now a question similar to the one we had in 00:35:16.850 --> 00:35:18.480 the previous slide. 00:35:18.480 --> 00:35:22.910 If I write down this summation-- 00:35:22.910 --> 00:35:28.240 even worse algebra than the one in the previous slide-- 00:35:28.240 --> 00:35:35.840 what do you think this number will turn out to be? 00:35:35.840 --> 00:35:39.930 It should be 1 because this is the probability 00:35:39.930 --> 00:35:42.930 of obtaining k heads. 00:35:42.930 --> 00:35:45.470 When we do the summation, what we're doing is we're 00:35:45.470 --> 00:35:48.550 considering the probability of 0 heads, plus the probability 00:35:48.550 --> 00:35:50.780 of 1 head, plus the probability of 2 heads, plus 00:35:50.780 --> 00:35:52.420 the probability of n heads. 00:35:52.420 --> 00:35:54.780 We've exhausted all the possibilities in our 00:35:54.780 --> 00:35:55.720 experiment. 00:35:55.720 --> 00:35:58.730 So the overall probability, when you exhaust all 00:35:58.730 --> 00:36:01.160 possibilities, must be equal to 1. 00:36:01.160 --> 00:36:04.180 So that's yet another beautiful formula that 00:36:04.180 --> 00:36:06.960 evaluates into something really simple. 00:36:06.960 --> 00:36:11.460 And if you tried to prove this identity algebraically, of 00:36:11.460 --> 00:36:16.030 course, you would have to suffer quite a bit. 00:36:16.030 --> 00:36:20.130 So now armed with the binomial probabilities, we can do the 00:36:20.130 --> 00:36:21.380 harder problems. 00:36:23.480 --> 00:36:27.340 So let's take the same experiment again. 00:36:27.340 --> 00:36:32.610 We flip a coin independently 10 times. 00:36:32.610 --> 00:36:37.450 So these 10 tosses are independent. 00:36:37.450 --> 00:36:40.000 We flip it 10 times. 00:36:40.000 --> 00:36:43.985 We don't see the result, but somebody comes and tells us, 00:36:43.985 --> 00:36:47.890 you know, there were exactly 3 heads in the 10 00:36:47.890 --> 00:36:49.688 tosses that you had. 00:36:49.688 --> 00:36:50.930 OK? 00:36:50.930 --> 00:36:53.280 So a certain event happened. 00:36:53.280 --> 00:36:57.450 And now you're asked to find the probability of another 00:36:57.450 --> 00:37:01.400 event, which is that the first 2 tosses were heads. 00:37:01.400 --> 00:37:08.990 Let's call that event A. OK. 00:37:08.990 --> 00:37:14.320 So are we in the setting of discrete 00:37:14.320 --> 00:37:16.850 uniform probability laws? 00:37:16.850 --> 00:37:21.760 When we toss a coin multiple times, is it the case that all 00:37:21.760 --> 00:37:24.130 outcomes are equally likely? 00:37:24.130 --> 00:37:27.850 All sequences are equally likely? 00:37:27.850 --> 00:37:30.515 That's the case if you have a fair coin-- 00:37:30.515 --> 00:37:32.630 that all sequences are equally likely. 00:37:32.630 --> 00:37:37.170 But if your coin is not fair, of course, heads/heads is 00:37:37.170 --> 00:37:39.630 going to have a different probability than tails/tails. 00:37:39.630 --> 00:37:43.720 If your coin is biased towards heads, then heads/heads is 00:37:43.720 --> 00:37:45.330 going to be more likely. 00:37:45.330 --> 00:37:49.440 So we're not quite in the uniform setting. 00:37:49.440 --> 00:37:53.450 Our overall sample space, omega, does not have equally 00:37:53.450 --> 00:37:55.680 likely elements. 00:37:55.680 --> 00:37:57.880 Do we care about that? 00:37:57.880 --> 00:37:59.700 Not necessarily. 00:37:59.700 --> 00:38:04.570 All the action now happens inside the event B that we are 00:38:04.570 --> 00:38:06.510 told has occurred. 00:38:06.510 --> 00:38:10.000 So we have our big sample space, omega. 00:38:10.000 --> 00:38:13.860 Elements of that sample space are not equally likely. 00:38:13.860 --> 00:38:17.390 We are told that a certain event B occurred. 00:38:17.390 --> 00:38:21.830 And inside that event B, we're asked to find the conditional 00:38:21.830 --> 00:38:26.100 probability that A has also occurred. 00:38:26.100 --> 00:38:30.850 Now here's the lucky thing, inside the event B, all 00:38:30.850 --> 00:38:33.270 outcomes are equally likely. 00:38:35.920 --> 00:38:40.710 The outcomes inside B are the sequences of 10 tosses that 00:38:40.710 --> 00:38:42.970 have exactly 3 heads. 00:38:42.970 --> 00:38:47.370 Every 3-head sequence has this probability. 00:38:47.370 --> 00:38:50.790 So the elements of B are equally 00:38:50.790 --> 00:38:52.760 likely with each other. 00:38:55.800 --> 00:39:01.030 Once we condition on the event B having occurred, what 00:39:01.030 --> 00:39:03.740 happens to the probabilities of the different outcomes 00:39:03.740 --> 00:39:05.430 inside here? 00:39:05.430 --> 00:39:09.790 Well, conditional probability laws keep the same proportions 00:39:09.790 --> 00:39:11.710 as the unconditional ones. 00:39:11.710 --> 00:39:15.930 The elements of B were equally likely when we started, so 00:39:15.930 --> 00:39:21.590 they're equally likely once we are told that B has occurred. 00:39:21.590 --> 00:39:26.440 So to do with this problem, we need to just transport us to 00:39:26.440 --> 00:39:30.680 this smaller universe and think about what's happening 00:39:30.680 --> 00:39:32.920 in that little universe. 00:39:32.920 --> 00:39:36.150 In that little universe, all elements of 00:39:36.150 --> 00:39:39.930 B are equally likely. 00:39:39.930 --> 00:39:43.860 So to find the probability of some subset of that set, we 00:39:43.860 --> 00:39:47.250 only need to count the cardinality of B, and count 00:39:47.250 --> 00:39:51.090 the cardinality of A. So let's do that. 00:39:51.090 --> 00:39:53.780 Number of outcomes in B-- 00:39:53.780 --> 00:40:00.290 in how many ways can we get 3 heads out of 10 tosses? 00:40:00.290 --> 00:40:03.190 That's the number we considered before, and 00:40:03.190 --> 00:40:06.250 it's 10 choose 3. 00:40:06.250 --> 00:40:11.020 This is the number of 3-head sequences 00:40:11.020 --> 00:40:13.840 when you have 10 tosses. 00:40:13.840 --> 00:40:20.580 Now let's look at the event A. The event A is that the first 00:40:20.580 --> 00:40:26.150 2 tosses where heads, but we're living now inside this 00:40:26.150 --> 00:40:31.220 universe B. Given that B occurred, how many elements 00:40:31.220 --> 00:40:34.760 does A have in there? 00:40:34.760 --> 00:40:41.536 In how many ways can A happen inside the B universe. 00:40:41.536 --> 00:40:46.860 If you're told that the first 2 were heads-- 00:40:46.860 --> 00:40:49.470 sorry. 00:40:49.470 --> 00:40:54.540 So out of the outcomes in B that have 3 heads, how many 00:40:54.540 --> 00:40:56.630 start with heads/heads? 00:40:56.630 --> 00:41:00.370 Well, if it starts with heads/heads, then the only 00:41:00.370 --> 00:41:04.830 uncertainty is the location of the third head. 00:41:04.830 --> 00:41:07.940 So we started with heads/heads, we're going to 00:41:07.940 --> 00:41:13.020 have three heads, the question is, where is that third head 00:41:13.020 --> 00:41:14.090 going to be. 00:41:14.090 --> 00:41:16.540 It has eight possibilities. 00:41:16.540 --> 00:41:20.940 So slot 1 is heads, slot 2 is heads, the third heads can be 00:41:20.940 --> 00:41:22.140 anywhere else. 00:41:22.140 --> 00:41:25.020 So there's 8 possibilities for where the third 00:41:25.020 --> 00:41:26.270 head is going to be. 00:41:29.630 --> 00:41:31.660 OK. 00:41:31.660 --> 00:41:36.720 So what we have counted here is really the cardinality of A 00:41:36.720 --> 00:41:43.450 intersection B, which is out of the elements in B, how many 00:41:43.450 --> 00:41:49.410 of them make A happen, divided by the cardinality of B. And 00:41:49.410 --> 00:41:53.860 that gives us the answer, which is going to be 10 choose 00:41:53.860 --> 00:41:57.530 3, divided by 8. 00:41:57.530 --> 00:42:01.330 And I should probably redraw a little bit of the picture that 00:42:01.330 --> 00:42:02.510 they have here. 00:42:02.510 --> 00:42:06.690 The set A is not necessarily contained in B. It could also 00:42:06.690 --> 00:42:14.750 have stuff outside B. So the event that the first 2 tosses 00:42:14.750 --> 00:42:18.690 are heads can happen with a total of 3 heads, but it can 00:42:18.690 --> 00:42:22.650 also happen with a different total number of heads. 00:42:22.650 --> 00:42:27.340 But once we are transported inside the set B, what we need 00:42:27.340 --> 00:42:32.460 to count is just this part of A. It's A intersection B and 00:42:32.460 --> 00:42:35.180 compare it with the total number of elements in the set 00:42:35.180 --> 00:42:40.310 B. Did I write it the opposite way? 00:42:40.310 --> 00:42:41.700 Yes. 00:42:41.700 --> 00:42:46.330 So this is 8 over 10 choose 3. 00:42:49.260 --> 00:42:49.640 OK. 00:42:49.640 --> 00:42:52.965 So we're going to close with a more difficult problem now. 00:42:57.920 --> 00:43:00.580 OK. 00:43:00.580 --> 00:43:05.650 This business of n choose k has to do with starting with a 00:43:05.650 --> 00:43:11.350 set and picking a subset of k elements. 00:43:11.350 --> 00:43:15.080 Another way of thinking of that is that we start with a 00:43:15.080 --> 00:43:20.770 set with n elements and you choose a subset that has k, 00:43:20.770 --> 00:43:24.350 which means that there's n minus k that are left. 00:43:24.350 --> 00:43:29.980 Picking a subset is the same as partitioning our set into 00:43:29.980 --> 00:43:32.510 two pieces. 00:43:32.510 --> 00:43:36.010 Now let's generalize this question and start counting 00:43:36.010 --> 00:43:38.150 partitions in general. 00:43:38.150 --> 00:43:42.570 Somebody gives you a set that has n elements. 00:43:42.570 --> 00:43:44.670 Somebody gives you also certain numbers-- 00:43:44.670 --> 00:43:49.400 n1, n2, n3, let's say, n4, where these 00:43:49.400 --> 00:43:53.740 numbers add up to n. 00:43:53.740 --> 00:43:58.740 And you're asked to partition this set into four subsets 00:43:58.740 --> 00:44:01.450 where each one of the subsets has this particular 00:44:01.450 --> 00:44:02.580 cardinality. 00:44:02.580 --> 00:44:08.250 So you're asking to cut it into four pieces, each one 00:44:08.250 --> 00:44:11.100 having the prescribed cardinality. 00:44:11.100 --> 00:44:15.090 In how many ways can we do this partitioning? 00:44:15.090 --> 00:44:19.370 n choose k was the answer when we partitioned in two pieces, 00:44:19.370 --> 00:44:21.910 what's the answer more generally? 00:44:21.910 --> 00:44:26.230 For a concrete example of a partition, you have your 52 00:44:26.230 --> 00:44:32.120 card deck and you deal, as in bridge, by giving 13 cards to 00:44:32.120 --> 00:44:34.000 each one of the players. 00:44:34.000 --> 00:44:38.080 Assuming that the dealing is done fairly and with a well 00:44:38.080 --> 00:44:43.790 shuffled deck of cards, every particular partition of the 52 00:44:43.790 --> 00:44:50.590 cards into four hands, that is four subsets of 13 each, 00:44:50.590 --> 00:44:52.380 should be equally likely. 00:44:52.380 --> 00:44:56.140 So we take the 52 cards and we partition them into subsets of 00:44:56.140 --> 00:44:58.550 13, 13, 13, and 13. 00:44:58.550 --> 00:45:01.020 And we assume that all possible partitions, all 00:45:01.020 --> 00:45:04.240 possible ways of dealing the cards are equally likely. 00:45:04.240 --> 00:45:07.560 So we are again in a setting where we can use counting, 00:45:07.560 --> 00:45:10.410 because all the possible outcomes are equally likely. 00:45:10.410 --> 00:45:14.050 So an outcome of the experiment is the hands that 00:45:14.050 --> 00:45:17.070 each player ends up getting. 00:45:17.070 --> 00:45:20.170 And when you get the cards in your hands, it doesn't matter 00:45:20.170 --> 00:45:22.000 in which order that you got them. 00:45:22.000 --> 00:45:25.460 It only matters what cards you have on you. 00:45:25.460 --> 00:45:31.160 So it only matters which subset of the cards you got. 00:45:31.160 --> 00:45:31.590 All right. 00:45:31.590 --> 00:45:35.820 So what's the cardinality of the sample space in this 00:45:35.820 --> 00:45:37.160 experiment? 00:45:37.160 --> 00:45:42.010 So let's do it for the concrete numbers that we have 00:45:42.010 --> 00:45:49.390 for the problem of partitioning 52 cards. 00:45:49.390 --> 00:45:52.540 So think of dealing as follows-- you shuffle the deck 00:45:52.540 --> 00:45:56.250 perfectly, and then you take the top 13 cards and give them 00:45:56.250 --> 00:45:57.740 to one person. 00:45:57.740 --> 00:46:03.230 In how many possible hands are there for that person? 00:46:03.230 --> 00:46:08.970 Out of the 52 cards, I choose 13 at random and give them to 00:46:08.970 --> 00:46:10.680 the first person. 00:46:10.680 --> 00:46:13.260 Having done that, what happens next? 00:46:13.260 --> 00:46:16.260 I'm left with 39 cards. 00:46:16.260 --> 00:46:20.600 And out of those 39 cards, I pick 13 of them and give them 00:46:20.600 --> 00:46:22.250 to the second person. 00:46:22.250 --> 00:46:25.790 Now I'm left with 26 cards. 00:46:25.790 --> 00:46:30.920 Out of those 26, I choose 13, give them to the third person. 00:46:30.920 --> 00:46:34.040 And for the last person there isn't really any choice. 00:46:34.040 --> 00:46:37.890 Out of the 13, I have to give that person all 13. 00:46:37.890 --> 00:46:40.230 And that number is just equal to 1. 00:46:40.230 --> 00:46:43.530 So we don't care about it. 00:46:43.530 --> 00:46:43.910 All right. 00:46:43.910 --> 00:46:48.270 So next thing you do is to write down the formulas for 00:46:48.270 --> 00:46:49.450 these numbers. 00:46:49.450 --> 00:46:52.450 So, for example, here you would have 52 factorial, 00:46:52.450 --> 00:46:55.880 divided by 13 factorial, times 39 00:46:55.880 --> 00:46:59.040 factorial, and you continue. 00:46:59.040 --> 00:47:01.310 And then there are nice cancellations that happen. 00:47:01.310 --> 00:47:05.120 This 39 factorial is going to cancel the 39 factorial that 00:47:05.120 --> 00:47:07.020 comes from there, and so on. 00:47:07.020 --> 00:47:10.200 After you do the cancellations and all the algebra, you're 00:47:10.200 --> 00:47:13.380 left with this particular answer, which is the number of 00:47:13.380 --> 00:47:18.120 possible partitions of 52 cards into four players where 00:47:18.120 --> 00:47:21.710 each player gets exactly 13 hands. 00:47:21.710 --> 00:47:25.140 If you were to generalize this formula to the setting that we 00:47:25.140 --> 00:47:29.200 have here, the more general formula is-- 00:47:29.200 --> 00:47:33.840 you have n factorial, where n is the number of objects that 00:47:33.840 --> 00:47:39.770 you are distributing, divided by the product of the 00:47:39.770 --> 00:47:41.840 factorials of the-- 00:47:41.840 --> 00:47:46.000 OK, here I'm doing it for the case where we split 00:47:46.000 --> 00:47:49.310 it into four sets. 00:47:49.310 --> 00:47:53.740 So that would be the answer when we partition a set into 00:47:53.740 --> 00:47:57.780 four subsets of prescribed cardinalities. 00:47:57.780 --> 00:48:00.120 And you can guess how that formula would generalize if 00:48:00.120 --> 00:48:03.590 you want to split it into five sets or six sets. 00:48:03.590 --> 00:48:03.950 OK. 00:48:03.950 --> 00:48:10.190 So far we just figured out the size of the sample space. 00:48:10.190 --> 00:48:14.660 Now we need to look at our event, which is the event that 00:48:14.660 --> 00:48:20.640 each player gets an ace, let's call that event A. In how many 00:48:20.640 --> 00:48:22.800 ways can that event happens? 00:48:22.800 --> 00:48:26.970 How many possible hands are there in which every player 00:48:26.970 --> 00:48:29.350 has exactly one ace? 00:48:29.350 --> 00:48:33.100 So I need to think about the sequential process by which I 00:48:33.100 --> 00:48:36.860 distribute the cards so that everybody gets exactly one 00:48:36.860 --> 00:48:40.440 ace, and then try to think in how many ways can that 00:48:40.440 --> 00:48:42.210 sequential process happen. 00:48:42.210 --> 00:48:45.660 So one way of making sure that everybody gets exactly one ace 00:48:45.660 --> 00:48:46.730 is the following-- 00:48:46.730 --> 00:48:51.210 I take the four aces and I distribute them randomly to 00:48:51.210 --> 00:48:53.970 the four players, but making sure that each one gets 00:48:53.970 --> 00:48:55.580 exactly one ace. 00:48:55.580 --> 00:48:57.510 In how many ways can that happen? 00:48:57.510 --> 00:49:02.210 I take the ace of spades and I send it to a random person out 00:49:02.210 --> 00:49:03.210 of the four. 00:49:03.210 --> 00:49:07.430 So there's 4 choices for this. 00:49:07.430 --> 00:49:10.280 Then I'm left with 3 aces to distribute. 00:49:10.280 --> 00:49:14.050 That person already gotten an ace. 00:49:14.050 --> 00:49:17.270 I take the next ace, and I give it to one of 00:49:17.270 --> 00:49:19.590 the 3 people remaining. 00:49:19.590 --> 00:49:22.480 So there's 3 choices for how to do that. 00:49:22.480 --> 00:49:26.970 And then for the next ace, there's 2 people who have not 00:49:26.970 --> 00:49:28.640 yet gotten an ace, and they give it 00:49:28.640 --> 00:49:30.770 randomly to one of them. 00:49:30.770 --> 00:49:36.760 So these are the possible ways of distributing for the 4 00:49:36.760 --> 00:49:42.040 aces, so that each person gets exactly one. 00:49:42.040 --> 00:49:44.230 It's actually the same as this problem. 00:49:44.230 --> 00:49:48.930 Starting with a set of four things, in how many ways can I 00:49:48.930 --> 00:49:53.430 partition them into four subsets where the first set 00:49:53.430 --> 00:49:56.220 has one element, the second has one element, the third one 00:49:56.220 --> 00:49:58.270 has another element, and so on. 00:49:58.270 --> 00:50:05.710 So it agrees with that formula by giving us 4 factorial. 00:50:05.710 --> 00:50:06.040 OK. 00:50:06.040 --> 00:50:09.400 So there are different ways of distributing the aces. 00:50:09.400 --> 00:50:11.760 And then there's different ways of distributing the 00:50:11.760 --> 00:50:13.460 remaining 48 cards. 00:50:13.460 --> 00:50:15.110 How many ways are there? 00:50:15.110 --> 00:50:18.920 Well, I have 48 cards that I'm going to distribute to four 00:50:18.920 --> 00:50:22.760 players by giving 12 cards to each one. 00:50:22.760 --> 00:50:26.430 It's exactly the same question as the one we had here, except 00:50:26.430 --> 00:50:30.230 that now it's 48 cards, 12 to each person. 00:50:30.230 --> 00:50:33.400 And that gives us this particular count. 00:50:33.400 --> 00:50:39.216 So putting all that together gives us the different ways 00:50:39.216 --> 00:50:43.350 that we can distribute the cards to the four players so 00:50:43.350 --> 00:50:45.910 that each one gets exactly one ace. 00:50:45.910 --> 00:50:48.600 The number of possible ways is going to be this four 00:50:48.600 --> 00:50:54.610 factorial, coming from here, times this number-- 00:50:54.610 --> 00:50:56.890 this gives us the number of ways that the event of 00:50:56.890 --> 00:50:58.760 interest can happen-- 00:50:58.760 --> 00:51:02.760 and then the denominator is the cardinality of our sample 00:51:02.760 --> 00:51:04.930 space, which is this number. 00:51:04.930 --> 00:51:07.560 So this looks like a horrible mess. 00:51:07.560 --> 00:51:10.590 It turns out that this expression does simplify to 00:51:10.590 --> 00:51:13.180 something really, really simple. 00:51:13.180 --> 00:51:16.420 And if you look at the textbook for this problem, you 00:51:16.420 --> 00:51:18.750 will see an alternative derivation that gives you a 00:51:18.750 --> 00:51:22.720 short cut to the same numerical answer. 00:51:22.720 --> 00:51:23.160 All right. 00:51:23.160 --> 00:51:25.240 So that basically concludes chapter one. 00:51:25.240 --> 00:51:29.940 From next time we're going to consider introducing random 00:51:29.940 --> 00:51:32.950 variables and make the subject even more interesting.