WEBVTT 00:00:02.120 --> 00:00:04.630 We will now use counting to solve a simple 00:00:04.630 --> 00:00:06.510 probabilistic problem. 00:00:06.510 --> 00:00:10.280 We have in our hands an ordinary six-sided die which 00:00:10.280 --> 00:00:12.570 we are going to roll six times. 00:00:12.570 --> 00:00:15.700 So this is our probabilistic experiment. 00:00:15.700 --> 00:00:17.950 And we're interested in the probability of a certain 00:00:17.950 --> 00:00:21.810 event, the event that the six rolls result 00:00:21.810 --> 00:00:24.250 in different numbers. 00:00:24.250 --> 00:00:28.390 So let us give a name to that event and call it event A. So 00:00:28.390 --> 00:00:33.440 we wish to calculate the probability of this event. 00:00:33.440 --> 00:00:36.600 But before we can even get started answering this 00:00:36.600 --> 00:00:38.920 question, we need a probabilistic model. 00:00:38.920 --> 00:00:41.500 We need to make some assumptions, and the 00:00:41.500 --> 00:00:45.200 assumption that we're going to make is that all outcomes of 00:00:45.200 --> 00:00:48.130 this experiment are equally likely. 00:00:48.130 --> 00:00:51.750 This is going to place us within a discrete uniform 00:00:51.750 --> 00:00:54.310 probabilistic model so that we can calculate 00:00:54.310 --> 00:00:55.830 probabilities by counting. 00:00:55.830 --> 00:00:59.950 In particular, as we discussed earlier in this lecture, the 00:00:59.950 --> 00:01:04.530 probability of an event A is going to be the number of 00:01:04.530 --> 00:01:09.520 elements of the set A, the number of outcomes that make 00:01:09.520 --> 00:01:15.930 event A occur, divided by the total number of possible 00:01:15.930 --> 00:01:19.150 outcomes, which is the number of elements 00:01:19.150 --> 00:01:20.400 in our sample space. 00:01:24.700 --> 00:01:28.150 So let us start with the denominator, and let us look 00:01:28.150 --> 00:01:31.080 at the typical outcomes of this experiment. 00:01:31.080 --> 00:01:43.080 A typical outcome is something like this sequence, 00:01:43.080 --> 00:01:49.130 2, 3, 4, 3, 6, 2. 00:01:49.130 --> 00:01:51.460 That's one possible outcome. 00:01:51.460 --> 00:01:54.539 How many outcomes of this kind are there? 00:01:54.539 --> 00:01:57.860 Well, we have 6 choices for the result of the first roll, 00:01:57.860 --> 00:02:01.070 6 choices for the result of the second roll, and so on. 00:02:01.070 --> 00:02:05.100 And since we have a total of 6 rolls, this means that there 00:02:05.100 --> 00:02:10.500 is a total of 6 to the 6th power possible outcomes, 00:02:10.500 --> 00:02:14.010 according to the Counting Principle. 00:02:14.010 --> 00:02:18.579 And by the way, since we have so many possible outcomes and 00:02:18.579 --> 00:02:20.890 we assume that they are equally likely, the 00:02:20.890 --> 00:02:24.400 probability of each one of them would be 1 00:02:24.400 --> 00:02:27.090 over 6 to the 6th. 00:02:27.090 --> 00:02:31.340 Incidentally, that's the same number, the same answer, you 00:02:31.340 --> 00:02:35.079 would get if you were to assume, instead of assuming 00:02:35.079 --> 00:02:38.310 directly that all outcomes are equally likely, to just assume 00:02:38.310 --> 00:02:44.380 that the different rolls are rolls of a fair six-sided die, 00:02:44.380 --> 00:02:49.480 so the probability of getting a 2 is 1/6, and also that the 00:02:49.480 --> 00:02:53.170 different rolls are independent of each other. 00:02:53.170 --> 00:02:55.730 So in that case, the probability, let's say, of 00:02:55.730 --> 00:02:58.050 this particular sequence would be the probability of 00:02:58.050 --> 00:03:01.630 obtaining a 2, which is 1/6, times the probability that we 00:03:01.630 --> 00:03:05.840 get a 3 at the next roll, which is 1/6, times 1/6 times 00:03:05.840 --> 00:03:10.080 1/6 and so on, and we get the same answer, 1 over 6 to 6th. 00:03:10.080 --> 00:03:14.070 So we see that this assumption of all outcomes being equally 00:03:14.070 --> 00:03:17.310 likely has an alternative interpretation in terms of 00:03:17.310 --> 00:03:22.380 having a fair die which is rolled independently 6 times. 00:03:22.380 --> 00:03:27.470 Now, let us look at the event of interest, A. What is a 00:03:27.470 --> 00:03:32.450 typical element of A? 00:03:32.450 --> 00:03:40.010 A typical element of A is a sequence of 6 rolls in which 00:03:40.010 --> 00:03:42.640 no number gets repeated. 00:03:42.640 --> 00:03:47.880 So, for example, it could be a sequence of results of this 00:03:47.880 --> 00:03:51.130 kind, where each number appears just once. 00:03:51.130 --> 00:03:54.770 So all the numbers appear exactly once in this sequence. 00:03:54.770 --> 00:03:58.450 So what we need here is basically to have a 00:03:58.450 --> 00:04:02.680 permutation of the numbers 1 up to 6. 00:04:02.680 --> 00:04:06.160 So these 6 numbers have to appear in an arbitrary order. 00:04:06.160 --> 00:04:10.040 In how many ways can we order 6 elements? 00:04:10.040 --> 00:04:13.050 This is the number of permutations of a set of 6 00:04:13.050 --> 00:04:15.460 elements and, as we discussed earlier, this 00:04:15.460 --> 00:04:17.310 is equal to 6 factorial. 00:04:17.310 --> 00:04:20.940 So we have now counted the number of outcomes that make 00:04:20.940 --> 00:04:23.990 event A happen, which is 6 factorial. 00:04:23.990 --> 00:04:27.830 And by calculating this ratio, we have obtained the 00:04:27.830 --> 00:04:31.820 probability of the desired event. 00:04:31.820 --> 00:04:35.065 You can now pause and try to solve a problem 00:04:35.065 --> 00:04:36.450 of a similar kind.