1 00:00:00,000 --> 00:00:04,320 The coefficients n-choose-k that we calculated in the 2 00:00:04,320 --> 00:00:08,220 previous segment are known as the binomial coefficients. 3 00:00:08,220 --> 00:00:11,060 They are intimately related to certain probabilities 4 00:00:11,060 --> 00:00:15,590 associated with coin tossing models, the so-called binomial 5 00:00:15,590 --> 00:00:16,930 probabilities. 6 00:00:16,930 --> 00:00:19,670 This is going to be our subject. 7 00:00:19,670 --> 00:00:25,540 We consider a coin which we toss n times in a row, 8 00:00:25,540 --> 00:00:28,190 independently. 9 00:00:28,190 --> 00:00:31,840 For each one of the tosses of this coin, we assume that 10 00:00:31,840 --> 00:00:34,360 there is a certain probability, p, that the 11 00:00:34,360 --> 00:00:38,290 result is heads, which of course, implies that the 12 00:00:38,290 --> 00:00:42,340 probability of obtaining tails in any particular toss is 13 00:00:42,340 --> 00:00:45,180 going to be 1 minus p. 14 00:00:45,180 --> 00:00:49,740 The question we want to address is the following. 15 00:00:49,740 --> 00:00:53,400 We want to calculate the probability that in those n 16 00:00:53,400 --> 00:00:56,010 independent coin tosses, we're going to 17 00:00:56,010 --> 00:00:58,835 observe exactly k heads. 18 00:01:02,060 --> 00:01:05,980 Let us start working our way towards the solution to this 19 00:01:05,980 --> 00:01:10,600 problem by looking first at a simple setting 20 00:01:10,600 --> 00:01:13,300 and then move on. 21 00:01:13,300 --> 00:01:16,600 So let us answer this simple question. 22 00:01:16,600 --> 00:01:19,530 What is the probability that we observe 23 00:01:19,530 --> 00:01:21,940 this particular sequence? 24 00:01:21,940 --> 00:01:26,780 Of course here we take n equal to six, and we wish to 25 00:01:26,780 --> 00:01:29,020 calculate this probability. 26 00:01:29,020 --> 00:01:32,200 Now, because we have assumed that the coin tosses are 27 00:01:32,200 --> 00:01:35,250 independent, we can multiply probabilities. 28 00:01:35,250 --> 00:01:38,000 So the probability of this sequence is equal to the 29 00:01:38,000 --> 00:01:42,690 probability that the first toss is heads times the 30 00:01:42,690 --> 00:01:45,850 probability that the second toss is tails, which is 1 31 00:01:45,850 --> 00:01:49,860 minus p, times the probability that the third toss is tails, 32 00:01:49,860 --> 00:01:53,460 which is 1 minus p, times the probability of heads, times 33 00:01:53,460 --> 00:01:54,820 the probability of heads, times the 34 00:01:54,820 --> 00:01:56,289 probability of heads. 35 00:01:56,289 --> 00:02:01,290 And by collecting terms, this is p to the 4th times 1 minus 36 00:02:01,290 --> 00:02:02,620 p to the second power. 37 00:02:05,880 --> 00:02:11,840 More generally, if I give you a particular sequence of heads 38 00:02:11,840 --> 00:02:15,910 and tails, as in this example, and I ask you, what is the 39 00:02:15,910 --> 00:02:18,710 probability that this particular sequence is 40 00:02:18,710 --> 00:02:25,160 observed, then by generalizing from this answer or from the 41 00:02:25,160 --> 00:02:31,170 derivation of this answer, you see that you're going to get p 42 00:02:31,170 --> 00:02:33,615 to the power number of heads. 43 00:02:36,120 --> 00:02:39,940 And the reason is that each time that there's a head 44 00:02:39,940 --> 00:02:43,220 showing up in this sequence, there's a corresponding factor 45 00:02:43,220 --> 00:02:46,570 of p in this numerical answer. 46 00:02:46,570 --> 00:02:50,450 And then there are factors associated with tails. 47 00:02:50,450 --> 00:02:54,050 Each tail contributes a factor of 1 minus p. 48 00:02:54,050 --> 00:02:57,980 And so we're going to have here 1 minus p to a power 49 00:02:57,980 --> 00:02:59,910 equal to the number of tails. 50 00:03:04,050 --> 00:03:08,270 Now, if I ask you about the probability of a particular 51 00:03:08,270 --> 00:03:11,720 sequence and that particular sequence has happened to have 52 00:03:11,720 --> 00:03:15,400 exactly k heads, what is the probability of that sequence? 53 00:03:15,400 --> 00:03:18,340 Well, we already calculated what it is. 54 00:03:18,340 --> 00:03:22,920 It is the previous answer, except we use the symbol k 55 00:03:22,920 --> 00:03:27,560 instead of just writing out explicitly "number of heads." 56 00:03:27,560 --> 00:03:31,960 And the number of tails is the number of tosses minus how 57 00:03:31,960 --> 00:03:36,400 many tosses resulted in heads. 58 00:03:36,400 --> 00:03:40,810 Now, we're ready to consider the actual problem that we 59 00:03:40,810 --> 00:03:43,100 want to solve, which is calculate the 60 00:03:43,100 --> 00:03:46,010 probability of k heads. 61 00:03:46,010 --> 00:03:50,270 The event of obtaining k heads can happen in 62 00:03:50,270 --> 00:03:52,490 many different ways. 63 00:03:52,490 --> 00:03:58,280 Any particular k-head sequence makes that event to occur. 64 00:03:58,280 --> 00:04:04,590 Any particular k-head sequence has a probability equal to 65 00:04:04,590 --> 00:04:06,840 this expression. 66 00:04:06,840 --> 00:04:11,170 The overall probability of k heads is going to be the 67 00:04:11,170 --> 00:04:18,230 probability of any particular k-head sequence, times the 68 00:04:18,230 --> 00:04:22,170 number of k-head sequences that we have. 69 00:04:27,490 --> 00:04:31,800 Now, the reason why we can carry out this argument is the 70 00:04:31,800 --> 00:04:36,090 fact that any k-head sequence has the same probability. 71 00:04:36,090 --> 00:04:39,520 Otherwise, we wouldn't be able to write down an answer which 72 00:04:39,520 --> 00:04:41,980 is just the product of two terms. 73 00:04:41,980 --> 00:04:44,600 But because every k-head sequence has the same 74 00:04:44,600 --> 00:04:47,860 probability, to find the overall probability, we take 75 00:04:47,860 --> 00:04:52,050 the probability of each one of them and multiply it with the 76 00:04:52,050 --> 00:04:55,360 number of how many of these we have. 77 00:04:55,360 --> 00:05:00,110 So to make further progress, now we need to calculate the 78 00:05:00,110 --> 00:05:03,550 number of possible k-head sequences. 79 00:05:03,550 --> 00:05:05,920 How many are there? 80 00:05:05,920 --> 00:05:09,340 Well, specifying a k-head sequence is 81 00:05:09,340 --> 00:05:11,160 the same as the following. 82 00:05:11,160 --> 00:05:14,580 You think of having n time slots. 83 00:05:14,580 --> 00:05:17,590 These time slots corresponds to the different 84 00:05:17,590 --> 00:05:20,240 tosses of your coin. 85 00:05:20,240 --> 00:05:24,410 And to specify a k-head sequence, you need to tell me 86 00:05:24,410 --> 00:05:31,610 which ones of these slots happen to contain a head. 87 00:05:31,610 --> 00:05:35,420 You need to tell me k of those slots. 88 00:05:35,420 --> 00:05:40,800 So in other words, what you're doing is you're specifying a 89 00:05:40,800 --> 00:05:45,460 subset of the set of these n slots, a 90 00:05:45,460 --> 00:05:48,290 subset that has k elements. 91 00:05:48,290 --> 00:05:53,070 You need to choose k of the slots out of the n and tell me 92 00:05:53,070 --> 00:05:55,630 that those k slots have heads. 93 00:05:55,630 --> 00:05:59,560 That's the way of specifying a particular k-head sequence. 94 00:05:59,560 --> 00:06:02,630 So what's the number of k-head sequences? 95 00:06:02,630 --> 00:06:05,760 Well, it's the same as the number of ways that you can 96 00:06:05,760 --> 00:06:11,110 choose k slots out of the n slots, which is our binomial 97 00:06:11,110 --> 00:06:14,410 coefficient, n-choose-k. 98 00:06:14,410 --> 00:06:18,420 Therefore, the answer to our problem is this expression 99 00:06:18,420 --> 00:06:25,590 here, times n-choose-k, which is shown up here. 100 00:06:25,590 --> 00:06:30,080 At this point, we can pause and consider a simple question 101 00:06:30,080 --> 00:06:34,040 to check your understanding of the binomial probabilities.