1 00:00:00,000 --> 00:00:01,040 2 00:00:01,040 --> 00:00:04,710 In this problem, we're given an urn with n balls in it, out 3 00:00:04,710 --> 00:00:08,510 of which m balls are red balls. 4 00:00:08,510 --> 00:00:13,370 To visualize it, we can draw a box that represents the set of 5 00:00:13,370 --> 00:00:14,990 all n balls. 6 00:00:14,990 --> 00:00:18,310 Somewhere in the middle or somewhere else we have a cut, 7 00:00:18,310 --> 00:00:21,750 such that to the left we have all the red balls (there are 8 00:00:21,750 --> 00:00:24,670 m), and non-red balls. 9 00:00:24,670 --> 00:00:27,100 Let's for now call it black balls. 10 00:00:27,100 --> 00:00:30,250 That is n minus m. 11 00:00:30,250 --> 00:00:34,450 Now, from this box, we are to draw k balls, and we'd like to 12 00:00:34,450 --> 00:00:37,900 know the probability that i out of those k 13 00:00:37,900 --> 00:00:39,455 balls are red balls. 14 00:00:39,455 --> 00:00:42,690 15 00:00:42,690 --> 00:00:44,670 For the rest of the problem, we'll refer to this 16 00:00:44,670 --> 00:00:50,120 probability as p-r, where r stands for the red balls. 17 00:00:50,120 --> 00:00:54,180 So from this picture, we know that we're going to draw a 18 00:00:54,180 --> 00:00:59,330 subset of the balls, such that i of them are red, and the 19 00:00:59,330 --> 00:01:03,920 remaining k minus i are black. 20 00:01:03,920 --> 00:01:06,830 And we'll like to know what is the probability that this 21 00:01:06,830 --> 00:01:09,020 event would occur. 22 00:01:09,020 --> 00:01:13,360 To start, we define our sample space, omega, as the set of 23 00:01:13,360 --> 00:01:17,150 all ways to draw k balls out of n balls. 24 00:01:17,150 --> 00:01:19,540 We found a simple counting argument -- we know that size 25 00:01:19,540 --> 00:01:23,750 of our sample space has n-choose-k, which is the total 26 00:01:23,750 --> 00:01:25,710 number of ways to draw k balls out of n balls. 27 00:01:25,710 --> 00:01:28,300 28 00:01:28,300 --> 00:01:30,450 Next, we'd like to know how many of those samples 29 00:01:30,450 --> 00:01:32,630 correspond to the event that we're interested in. 30 00:01:32,630 --> 00:01:35,700 In particular, we would like to know c, which is equal to 31 00:01:35,700 --> 00:01:38,670 the number of ways to get i red balls after 32 00:01:38,670 --> 00:01:40,800 we draw the k balls. 33 00:01:40,800 --> 00:01:44,500 To do so, we'll break c into a product of two numbers -- 34 00:01:44,500 --> 00:01:46,660 let's call it a times b -- 35 00:01:46,660 --> 00:01:52,680 where a is the total number of ways to select i red balls out 36 00:01:52,680 --> 00:01:54,520 of m red balls. 37 00:01:54,520 --> 00:02:02,530 So the number of ways to get i out of m red balls. 38 00:02:02,530 --> 00:02:05,820 Going back to the picture, this corresponds to the total 39 00:02:05,820 --> 00:02:09,340 number of ways to get these balls. 40 00:02:09,340 --> 00:02:15,000 And similarly, we define b as the total number of ways to 41 00:02:15,000 --> 00:02:23,150 get the remaining k minus i balls out of the set n minus m 42 00:02:23,150 --> 00:02:25,340 black balls. 43 00:02:25,340 --> 00:02:29,890 This corresponds to the total number of ways to select the 44 00:02:29,890 --> 00:02:34,820 subset right here in the right side of the box. 45 00:02:34,820 --> 00:02:38,080 Now as you can see, once we have a and b, we multiply them 46 00:02:38,080 --> 00:02:41,600 together, and this yields the total number of ways 47 00:02:41,600 --> 00:02:43,470 to get i red balls. 48 00:02:43,470 --> 00:02:48,335 To compute what these numbers are, we see that a is equal to 49 00:02:48,335 --> 00:02:54,280 m-choose-i number of ways to get i red balls, and b is n 50 00:02:54,280 --> 00:03:01,570 minus m, the total number of black balls, choose k minus i, 51 00:03:01,570 --> 00:03:06,160 the balls that are not red within those k balls. 52 00:03:06,160 --> 00:03:09,670 Now putting everything back, we have p-r, the probability 53 00:03:09,670 --> 00:03:14,260 we set out to compute, is equal to c, the size of the 54 00:03:14,260 --> 00:03:20,250 event, divided by the size of the entire sample space. 55 00:03:20,250 --> 00:03:22,700 From the previous calculations, we know that c 56 00:03:22,700 --> 00:03:28,410 is equal to a times b, which is then equal to m-choose-i 57 00:03:28,410 --> 00:03:33,000 times (n minus m)-choose-(k minus i). 58 00:03:33,000 --> 00:03:38,460 And on the denominator, we have the entire sample space 59 00:03:38,460 --> 00:03:41,110 is a size n-choose-k. 60 00:03:41,110 --> 00:03:44,810 And that completes our derivation. 61 00:03:44,810 --> 00:03:48,080 Now let's look at a numerical example of this problem. 62 00:03:48,080 --> 00:03:51,630 Here, let's say we have a deck of 52 cards. 63 00:03:51,630 --> 00:03:57,290 And we draw a box with n equals 52, out of which we 64 00:03:57,290 --> 00:04:00,400 know that there are 4 aces. 65 00:04:00,400 --> 00:04:04,660 So we'll call these the left side of the box, which is we 66 00:04:04,660 --> 00:04:09,130 have m equals 4 aces. 67 00:04:09,130 --> 00:04:12,640 Now if we were to draw seven cards-- 68 00:04:12,640 --> 00:04:15,550 call it k equal to 7-- 69 00:04:15,550 --> 00:04:18,620 and we'd like to know what is the probability that out of 70 00:04:18,620 --> 00:04:21,340 the 7 cards, we have 3 aces. 71 00:04:21,340 --> 00:04:24,610 72 00:04:24,610 --> 00:04:29,420 Using the notation we did earlier, if we were to draw a 73 00:04:29,420 --> 00:04:33,000 circle representing the seven cards, we want to know what is 74 00:04:33,000 --> 00:04:36,550 the probability that we have 3 aces in the left side of the 75 00:04:36,550 --> 00:04:41,650 box and 4 non-aces for the remainder of the deck. 76 00:04:41,650 --> 00:04:46,570 In particular, we'll call i equal to 3. 77 00:04:46,570 --> 00:04:49,990 So by this point, we've cast the problem of drawing cards 78 00:04:49,990 --> 00:04:53,810 from the deck in the same way as we did earlier of drawing 79 00:04:53,810 --> 00:04:55,210 balls from an urn. 80 00:04:55,210 --> 00:04:59,970 And from the expression right here, which we computed 81 00:04:59,970 --> 00:05:03,630 earlier, we can readily compute the probability of 82 00:05:03,630 --> 00:05:06,280 having 3 aces. 83 00:05:06,280 --> 00:05:10,360 In particular, we just have to substitute into the expression 84 00:05:10,360 --> 00:05:16,370 right here the value of m equal to 4, n equal to 52, k 85 00:05:16,370 --> 00:05:19,980 equal to 7, finally, i equal to 3. 86 00:05:19,980 --> 00:05:29,300 So we have 4-choose-3 times n minus m, in this case would be 87 00:05:29,300 --> 00:05:36,260 48, choose k minus i, will be 4, and on the denominator, we 88 00:05:36,260 --> 00:05:41,750 have 52 total number of cards, choosing 7 cards. 89 00:05:41,750 --> 00:05:42,810 That gives us [the] 90 00:05:42,810 --> 00:05:44,440 numerical answer [for] 91 00:05:44,440 --> 00:05:47,890 the probability of getting 3 aces when we draw 7 cards. 92 00:05:47,890 --> 00:05:49,140