1 00:00:00,000 --> 00:00:00,250 2 00:00:00,250 --> 00:00:02,310 In this problem, we will be helping Romeo and Juliet meet 3 00:00:02,310 --> 00:00:03,420 up for a date. 4 00:00:03,420 --> 00:00:06,450 And in the process, also we'll review some concepts in basic 5 00:00:06,450 --> 00:00:09,950 probability theory, including sample spaces 6 00:00:09,950 --> 00:00:12,840 and probability laws. 7 00:00:12,840 --> 00:00:16,550 This problem, the basic setup is that Romeo and Juliet are 8 00:00:16,550 --> 00:00:18,040 trying to meet up for a date. 9 00:00:18,040 --> 00:00:19,510 And let's say they're trying to meet up for 10 00:00:19,510 --> 00:00:20,910 lunch tomorrow at noon. 11 00:00:20,910 --> 00:00:22,610 But they're not necessarily punctual. 12 00:00:22,610 --> 00:00:27,840 So they may arrive on time with a delay of 0, or they may 13 00:00:27,840 --> 00:00:31,860 actually be up to 1 hour late and arrive at 1:00 PM. 14 00:00:31,860 --> 00:00:34,800 So the other thing that we assume in this problem is that 15 00:00:34,800 --> 00:00:38,380 all pairs of arrival times-- so the time that Romeo arrives 16 00:00:38,380 --> 00:00:40,290 paired with the time they Juliet arrives-- 17 00:00:40,290 --> 00:00:43,090 all of these pairs are equally likely. 18 00:00:43,090 --> 00:00:45,090 And I've put this in quotes, because we haven't really 19 00:00:45,090 --> 00:00:46,710 specify exactly what this means. 20 00:00:46,710 --> 00:00:49,450 And we'll come back to that in a little bit. 21 00:00:49,450 --> 00:00:52,320 The last important thing is that each person will wait for 22 00:00:52,320 --> 00:00:54,730 15 minutes for the other person to arrive. 23 00:00:54,730 --> 00:00:56,780 If within that 15-minute window the other person 24 00:00:56,780 --> 00:01:00,660 doesn't arrive, then they'll give up and they'll end up not 25 00:01:00,660 --> 00:01:02,480 meeting up for lunch. 26 00:01:02,480 --> 00:01:05,630 So to solve this problem, let's first try to set up a 27 00:01:05,630 --> 00:01:09,550 sample space and come up with a probability law to describe 28 00:01:09,550 --> 00:01:12,370 this scenario. 29 00:01:12,370 --> 00:01:15,350 And let's actually start with a simpler 30 00:01:15,350 --> 00:01:16,520 version of this problem. 31 00:01:16,520 --> 00:01:21,330 And instead of assuming that they can arrive at any delay 32 00:01:21,330 --> 00:01:25,510 between 0 and 1 hour, let's pretend instead that Romeo and 33 00:01:25,510 --> 00:01:29,000 Juliet can only arrive in 15-minute increments. 34 00:01:29,000 --> 00:01:33,420 So Romeo can arrive on time with a delay 0, or be 15 35 00:01:33,420 --> 00:01:35,630 minutes late, 30 minutes late, 45 minutes 36 00:01:35,630 --> 00:01:36,870 late, or one hour late. 37 00:01:36,870 --> 00:01:39,630 But none of the other times are possible. 38 00:01:39,630 --> 00:01:40,880 And the same thing for Juliet. 39 00:01:40,880 --> 00:01:43,720 40 00:01:43,720 --> 00:01:46,020 Let's start out with just the simple case first, because it 41 00:01:46,020 --> 00:01:49,220 helps us get the intuition for the problem, and it's an 42 00:01:49,220 --> 00:01:50,470 easier case to analyze. 43 00:01:50,470 --> 00:01:53,250 44 00:01:53,250 --> 00:01:55,530 So it's actually easy to visualize this. 45 00:01:55,530 --> 00:01:58,760 It's a nice visual tool to group this sample 46 00:01:58,760 --> 00:02:00,420 space into a grid. 47 00:02:00,420 --> 00:02:03,400 So the horizontal axis here represents the arrival time of 48 00:02:03,400 --> 00:02:05,670 Romeo, and the vertical axis represents the 49 00:02:05,670 --> 00:02:07,870 arrival time of Juliet. 50 00:02:07,870 --> 00:02:17,680 And so, for example, this point here would represent 51 00:02:17,680 --> 00:02:20,690 Romeo arriving 15 minutes late and Juliet 52 00:02:20,690 --> 00:02:23,320 arriving 30 minutes late. 53 00:02:23,320 --> 00:02:24,950 So this is our sample space now. 54 00:02:24,950 --> 00:02:28,270 This is our omega. 55 00:02:28,270 --> 00:02:31,180 And now let's try to assign a probability law. 56 00:02:31,180 --> 00:02:34,170 And we'll continue to assume that all pairs of arrival 57 00:02:34,170 --> 00:02:35,620 times are equally likely. 58 00:02:35,620 --> 00:02:40,180 And now we can actually specifically specify what this 59 00:02:40,180 --> 00:02:41,380 term means. 60 00:02:41,380 --> 00:02:44,540 And in particular, we'll be invoking the discrete uniform 61 00:02:44,540 --> 00:02:48,390 law, which basically says that all of these points, which are 62 00:02:48,390 --> 00:02:53,790 just outcomes in our probabilistic experiment-- 63 00:02:53,790 --> 00:02:56,670 all of these outcomes are equally likely. 64 00:02:56,670 --> 00:03:00,060 And so since there are 25 of them, each one of these 65 00:03:00,060 --> 00:03:03,860 outcomes has a probability of 1 over 25. 66 00:03:03,860 --> 00:03:06,250 So now we've specified our sample space and our 67 00:03:06,250 --> 00:03:07,440 probability law. 68 00:03:07,440 --> 00:03:09,840 So now let's try to answer the question, what is the 69 00:03:09,840 --> 00:03:11,650 probability that Romeo and Juliet will meet 70 00:03:11,650 --> 00:03:13,730 up for their date? 71 00:03:13,730 --> 00:03:16,520 So all that amounts to now is just identifying which of 72 00:03:16,520 --> 00:03:21,820 these 25 outcomes results in Romeo and Juliet arriving 73 00:03:21,820 --> 00:03:23,860 within 15 minutes of each other. 74 00:03:23,860 --> 00:03:26,080 So let's start with this one that I've picked out. 75 00:03:26,080 --> 00:03:29,470 If Romeo arrives 15 minutes late and Juliet arrives 30 76 00:03:29,470 --> 00:03:32,640 minutes late, then they will arrive within 15 minutes of 77 00:03:32,640 --> 00:03:33,090 each other. 78 00:03:33,090 --> 00:03:38,370 So this outcome does result in the two of them meeting. 79 00:03:38,370 --> 00:03:42,840 And so we can actually highlight all of these. 80 00:03:42,840 --> 00:03:55,170 And it turns out that these outcomes that I'm highlighting 81 00:03:55,170 --> 00:03:58,630 result in the two them arriving within 15 minutes of 82 00:03:58,630 --> 00:04:00,560 each other. 83 00:04:00,560 --> 00:04:05,230 So because each one has a probability of 1 over 25, all 84 00:04:05,230 --> 00:04:07,210 we really need to do now is just count how many 85 00:04:07,210 --> 00:04:08,340 outcomes there are. 86 00:04:08,340 --> 00:04:16,230 So there's 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13. 87 00:04:16,230 --> 00:04:21,640 So the probability in the end is for the discrete case. 88 00:04:21,640 --> 00:04:22,580 The discrete case-- 89 00:04:22,580 --> 00:04:25,600 I'm referring to the case where we simplified it and 90 00:04:25,600 --> 00:04:28,880 considered only arrival times with increments of 15 minutes. 91 00:04:28,880 --> 00:04:36,300 In this case, the probability is 13 over 25. 92 00:04:36,300 --> 00:04:39,200 So now we have an idea of how to solve this problem. 93 00:04:39,200 --> 00:04:42,340 It amounts to basically coming up with a sample space, a 94 00:04:42,340 --> 00:04:45,010 probability law, and then identifying the events of 95 00:04:45,010 --> 00:04:47,820 interest in calculating the probability of that event. 96 00:04:47,820 --> 00:04:51,590 So now let's actually solve the problem that we really are 97 00:04:51,590 --> 00:04:57,420 interested in, which is that instead of confining Romeo and 98 00:04:57,420 --> 00:04:59,830 Juliet to arrive in only 15-minute minute increments, 99 00:04:59,830 --> 00:05:01,860 really, time is continuous, and Romeo and Juliet can 100 00:05:01,860 --> 00:05:02,940 arrive at any time. 101 00:05:02,940 --> 00:05:05,900 So they don't necessarily have to arrive 15 minutes late. 102 00:05:05,900 --> 00:05:09,470 Romeo could arrive 15 minutes and 37 seconds late 103 00:05:09,470 --> 00:05:11,830 if he wanted to. 104 00:05:11,830 --> 00:05:15,350 So now our new sample space is actually just, instead of only 105 00:05:15,350 --> 00:05:19,610 these 25 points in the grid, it's this entire square. 106 00:05:19,610 --> 00:05:22,700 So any point within the square could be a possible pair of 107 00:05:22,700 --> 00:05:26,710 meeting times between Romeo and Juliet. 108 00:05:26,710 --> 00:05:29,740 So that is our new sample space, our new omega. 109 00:05:29,740 --> 00:05:32,280 And now let's assign a new probability law. 110 00:05:32,280 --> 00:05:34,840 And now, instead of being in the discrete world, we're in 111 00:05:34,840 --> 00:05:36,050 the continuous world. 112 00:05:36,050 --> 00:05:38,740 And the analogy here is to consider 113 00:05:38,740 --> 00:05:40,910 probabilities as areas. 114 00:05:40,910 --> 00:05:43,970 So the area of this entire square is one. 115 00:05:43,970 --> 00:05:46,950 And that also corresponds to the probability of omega, the 116 00:05:46,950 --> 00:05:48,170 sample space. 117 00:05:48,170 --> 00:05:52,370 And imagine just spreading probability evenly across this 118 00:05:52,370 --> 00:05:55,230 square so that the probability of any event-- 119 00:05:55,230 --> 00:05:56,440 which in this case would just be any 120 00:05:56,440 --> 00:05:58,190 shape within this square-- 121 00:05:58,190 --> 00:06:02,760 is exactly equal to the area of that shape. 122 00:06:02,760 --> 00:06:05,090 So now that is our new sample space and our 123 00:06:05,090 --> 00:06:06,460 new probability law. 124 00:06:06,460 --> 00:06:08,850 So what we have to do now is just to identify the event of 125 00:06:08,850 --> 00:06:12,160 interest, which is still the event that Romeo and Juliet 126 00:06:12,160 --> 00:06:14,810 arrive within 15 minutes of each other. 127 00:06:14,810 --> 00:06:17,130 So let's do that. 128 00:06:17,130 --> 00:06:19,840 If Romeo and Juliet arrive both on time, then obviously 129 00:06:19,840 --> 00:06:20,790 they'll meet. 130 00:06:20,790 --> 00:06:24,130 And if Romeo's on time and Juliet is 15 minutes late, 131 00:06:24,130 --> 00:06:25,520 then they will still meet. 132 00:06:25,520 --> 00:06:31,110 And in fact, any pairs of meeting times between these 133 00:06:31,110 --> 00:06:34,190 would still work, because now Romeo can be on time, and 134 00:06:34,190 --> 00:06:37,790 Juliet can arrive at any time between 0 and 15 minutes late. 135 00:06:37,790 --> 00:06:43,870 But you notice that if Juliet is even a tiny bit later than 136 00:06:43,870 --> 00:06:46,460 15 minutes, then they won't end up meeting. 137 00:06:46,460 --> 00:06:50,970 So this segment here is part of the event of interest. 138 00:06:50,970 --> 00:06:54,720 And similarly, this segment here is 139 00:06:54,720 --> 00:06:56,560 also part of the event. 140 00:06:56,560 --> 00:06:59,230 And if you take this exercise and extend it, you can 141 00:06:59,230 --> 00:07:07,040 actually verify that the event of interest is this strip 142 00:07:07,040 --> 00:07:11,640 shape in the middle of the square. 143 00:07:11,640 --> 00:07:14,180 Which, if you think about it, makes sense, because you want 144 00:07:14,180 --> 00:07:17,470 the arrival times between Romeo and Juliet to be close 145 00:07:17,470 --> 00:07:21,050 to each other, so you would be expect it to be somewhere 146 00:07:21,050 --> 00:07:25,080 close to a diagonal in this square. 147 00:07:25,080 --> 00:07:27,190 So now we have our event of interest. 148 00:07:27,190 --> 00:07:28,980 We have our sample space and our probability law. 149 00:07:28,980 --> 00:07:31,550 So all we have to do now is just calculate what this 150 00:07:31,550 --> 00:07:33,010 probability is. 151 00:07:33,010 --> 00:07:36,110 And we've already said that the probability in this 152 00:07:36,110 --> 00:07:38,930 probability law is just areas. 153 00:07:38,930 --> 00:07:41,980 So now it actually just boils down to not a probability 154 00:07:41,980 --> 00:07:44,390 problem, but a problem in geometry. 155 00:07:44,390 --> 00:07:47,740 So to calculate this area, you can do it in lots of ways. 156 00:07:47,740 --> 00:07:51,110 One way is to calculate the area of the square, which is 157 00:07:51,110 --> 00:07:55,520 1, and subtract the areas of these two triangles. 158 00:07:55,520 --> 00:07:58,600 So let's do that. 159 00:07:58,600 --> 00:08:00,980 So in the continuous case, the probability of meeting is 160 00:08:00,980 --> 00:08:04,620 going to be 1 minus the area of this triangle. 161 00:08:04,620 --> 00:08:10,320 The base here is 3/4 and 3/4, so it's 1/2 162 00:08:10,320 --> 00:08:13,810 times 3/4 times 3/4. 163 00:08:13,810 --> 00:08:15,340 That's the area of one of these triangles. 164 00:08:15,340 --> 00:08:17,830 There's two of them, so we'll multiply by two. 165 00:08:17,830 --> 00:08:26,200 And we end up with 1 minus 9/16, or 7/16 166 00:08:26,200 --> 00:08:29,250 as our final answer. 167 00:08:29,250 --> 00:08:33,490 So in this problem, we've reviewed some basic concepts 168 00:08:33,490 --> 00:08:36,360 of probability, and that's also helped us solve this 169 00:08:36,360 --> 00:08:39,100 problem of helping Romeo and Juliet meet up for a date. 170 00:08:39,100 --> 00:08:41,320 And if you wanted to, you could even extend this problem 171 00:08:41,320 --> 00:08:44,130 even further and turn it on its head. 172 00:08:44,130 --> 00:08:47,090 And instead of calculating given that they arrive within 173 00:08:47,090 --> 00:08:49,070 15 minutes of each other, what is the probability that 174 00:08:49,070 --> 00:08:51,730 they'll meet, let's say that Romeo really wants to meet up 175 00:08:51,730 --> 00:08:56,895 with Juliet, and he wants to assure himself a least, say, a 176 00:08:56,895 --> 00:08:58,800 90% chance of meeting Juliet. 177 00:08:58,800 --> 00:09:02,550 Then you can ask, if he wants to have at least a 90% chance 178 00:09:02,550 --> 00:09:05,630 of meeting her, how long should he be willing to wait? 179 00:09:05,630 --> 00:09:08,990 And so that's the flip side of the problem. 180 00:09:08,990 --> 00:09:11,860 And you can see that with just some basic concepts of 181 00:09:11,860 --> 00:09:13,800 probability, you can answer some already pretty 182 00:09:13,800 --> 00:09:16,060 interesting problems. 183 00:09:16,060 --> 00:09:18,310 So I hope this problem was interesting, and we'll 184 00:09:18,310 --> 00:09:19,560 see you next time. 185 00:09:19,560 --> 00:09:20,747