1 00:00:02,680 --> 00:00:07,432 Our next rule for probability concerns the probabilities of disjunctions. 2 00:00:07,432 --> 00:00:11,925 That is, the probability that either one thing or another will happen. 3 00:00:11,925 --> 00:00:16,743 And as with the case of the probability of conjunctions, we have two cases. 4 00:00:16,743 --> 00:00:22,016 You need a different rule when these two events are mutually exclusive than when 5 00:00:22,016 --> 00:00:26,444 they're not mutually exclusive. We're going to look first at cases that 6 00:00:26,444 --> 00:00:30,545 are mutually exclusive. And what that means is simply that they 7 00:00:30,545 --> 00:00:34,712 cannot occur together. The probability of them both occurring is 8 00:00:34,712 --> 00:00:36,900 zero. It can't happen. 9 00:00:36,900 --> 00:00:42,352 Now, the formula for that simpler case says that if two events are mutually 10 00:00:42,352 --> 00:00:48,385 exclusive, then the probability of either one event or the other event occurring is 11 00:00:48,385 --> 00:00:54,419 simply the sum of the probability of the first event occurring and the probability 12 00:00:54,419 --> 00:01:00,339 of the second event occurring. And we can symbolize it using the same 13 00:01:00,339 --> 00:01:05,380 symbols we've used before as the probability of hypothesis one or 14 00:01:05,380 --> 00:01:10,728 hypothesis two is equal to the probability of hypothesis one plus the 15 00:01:10,728 --> 00:01:14,291 probability of hypothesis two. Simple enough. 16 00:01:14,291 --> 00:01:18,987 Let's look at an example. What's the probability that I will pick 17 00:01:18,987 --> 00:01:23,539 either an ace of diamonds or an ace of hearts out of this deck? 18 00:01:23,539 --> 00:01:28,820 Well, the probability that I pick an ace of diamonds is one in four. 19 00:01:28,820 --> 00:01:33,600 The probability that I'll pick an ace of hearts is one in four. 20 00:01:33,600 --> 00:01:39,821 So, the probability that I'll pick either one or the other is one in four plus one 21 00:01:39,821 --> 00:01:42,856 in four, which is two in four, which is 0.5. 22 00:01:42,856 --> 00:01:47,864 Two times out of four, 0.5. So, this formula makes perfect sense when 23 00:01:47,864 --> 00:01:51,733 it's exclusive, when you cannot pick both an ace of 24 00:01:51,733 --> 00:01:55,300 diamonds and an ace of hearts on a single pick. 25 00:01:56,720 --> 00:02:01,080 Here's another example. It's a special case. 26 00:02:01,080 --> 00:02:07,659 We've been talking so far about permutations which are ordered sets so 27 00:02:07,659 --> 00:02:14,795 the two events in a conjunction would flipping heads first and then, tails 28 00:02:14,795 --> 00:02:18,367 second. We can also look at combinations where 29 00:02:18,367 --> 00:02:22,692 order doesn't matter. Flipping one heads and one tails out of 30 00:02:22,692 --> 00:02:26,238 two flips but it doesn't matter what the order is. 31 00:02:26,238 --> 00:02:31,414 Then, we have heads and tails in two flips, which means either a heads and 32 00:02:31,414 --> 00:02:37,016 then a tails or a tails and then a heads. So, we've got a disjunction and we can 33 00:02:37,016 --> 00:02:42,263 figure out that the probability of this combination, which is unordered, by 34 00:02:42,263 --> 00:02:46,092 looking at the disjunction of those two possibilities. 35 00:02:46,092 --> 00:02:51,977 Here's a chart showing the possibilities. If you take heads on flip one, you are in 36 00:02:51,977 --> 00:02:56,190 the left-hand column. Tails on the flip one, you are in the 37 00:02:56,190 --> 00:03:00,403 right-hand column. Head on flip two, you are in the top row. 38 00:03:00,403 --> 00:03:04,460 Tails on flip two, you are in the bottom row. 39 00:03:04,460 --> 00:03:12,525 Which of these, for the two flips, includes a heads and a tails where the 40 00:03:12,525 --> 00:03:17,846 order doesn't matter? And the answer is, the upper right has 41 00:03:17,846 --> 00:03:25,150 tails on flip one and heads on flip two. And the lower left has tails on flip two 42 00:03:25,150 --> 00:03:30,471 and heads on flip one. So, it's those two possibilities, this 43 00:03:30,471 --> 00:03:37,325 one and this one, out of the four that include the combination, one heads and 44 00:03:37,325 --> 00:03:43,158 one tails in any other order. And the disjunction is calculated by 45 00:03:43,158 --> 00:03:47,418 saying, what's the probability of this, 0.25, one in four. 46 00:03:47,418 --> 00:03:52,941 And the probability of this, 0.25, one in four and you add them together. 47 00:03:52,941 --> 00:03:59,252 Because the probability of doing either one or the other is going to be the sum, 48 00:03:59,252 --> 00:04:04,222 just like the formula says. Next, we have to look at the general 49 00:04:04,222 --> 00:04:10,760 formula that applies whether or not the two events are mutually exclusive. 50 00:04:10,760 --> 00:04:15,379 And that's pretty simple. We're going to call it 3G for the 51 00:04:15,379 --> 00:04:20,983 generalized version of rule three. What it says is that the probability of 52 00:04:20,983 --> 00:04:26,586 either one event occurring or the other event occurring is the sum of the 53 00:04:26,586 --> 00:04:32,796 probability of the first event occurring plus the probability of the second event 54 00:04:32,796 --> 00:04:36,810 occurring minus the probability that they both occur. 55 00:04:36,810 --> 00:04:43,584 Or in symbols, the probability of h1 or h2 is equal to the probability of h1 plus 56 00:04:43,584 --> 00:04:49,690 the probability of h2 minus the probability of the conjunction of h1 and 57 00:04:49,690 --> 00:04:55,963 h2. So, we're going to have to use that previous rule for conjunction to figure 58 00:04:55,963 --> 00:05:01,233 out what gets subtracted. The more basic question is why are we 59 00:05:01,233 --> 00:05:06,920 subtracting at the first place? And you can think about that in this 60 00:05:06,920 --> 00:05:10,695 example. What's the probability of getting heads 61 00:05:10,695 --> 00:05:16,517 on either the first flip or the second flip in two flips of a fare coin? Well, 62 00:05:16,517 --> 00:05:22,261 one way to figure this out might be to say, well, what's the probability of 63 00:05:22,261 --> 00:05:26,506 getting on the first flip? Well, that's 0.5. 64 00:05:26,506 --> 00:05:30,238 What's the probability of getting on the second flip? 65 00:05:30,238 --> 00:05:33,689 Well, that's 0.5. So, let's add them together and the 66 00:05:33,689 --> 00:05:38,619 probability of getting it on either the first flip or the second flip is.5=1. 67 00:05:38,619 --> 00:05:41,577 0.5 + 0.5 = 1. So, it's absolutely certain that were 68 00:05:41,577 --> 00:05:47,423 going to get a heads on either the first flip or the second flip of two flips of a 69 00:05:47,423 --> 00:05:50,804 fair coin. But you know that's not right because 70 00:05:50,804 --> 00:05:55,574 sometimes you get two tails in a row. So, that can't be the right way to 71 00:05:55,574 --> 00:05:59,520 calculate the probability. That's why you need to subtract the 72 00:05:59,520 --> 00:06:04,020 conjunction because otherwise, you're going to get the wrong answer. 73 00:06:04,020 --> 00:06:10,490 The right way then is 0.5 heads on the first flip, 0.5 heads on the second flip 74 00:06:10,490 --> 00:06:16,365 minus the probability of both. But the question is, why do you have to 75 00:06:16,365 --> 00:06:20,707 subtract? And this little chart should show you why 76 00:06:20,707 --> 00:06:25,986 you have to subtract. Because the probability of getting heads 77 00:06:25,986 --> 00:06:29,562 on the first flip is the left-hand column, 78 00:06:29,562 --> 00:06:33,562 that's the upper one. You get heads on the first flip. 79 00:06:33,562 --> 00:06:36,744 The lower one, you get heads on the first flip. 80 00:06:36,744 --> 00:06:42,340 The probability of getting heads on the second flip is given by the top row. 81 00:06:42,340 --> 00:06:46,700 The first one, you get heads on the second flip. 82 00:06:46,700 --> 00:06:50,800 The second one, we get the heads on the second flip. 83 00:06:50,800 --> 00:06:59,942 So, if you want to add them together, you are adding together this one and you're 84 00:06:59,942 --> 00:07:06,711 adding it to this one, that gives you 0.5 of getting heads on the first flip. Then, 85 00:07:06,711 --> 00:07:13,745 you add it to the probability of this one plus this one and that gives you 0.5 of 86 00:07:13,745 --> 00:07:19,820 getting heads on the second flip. But then, if you add them together to get 87 00:07:19,820 --> 00:07:25,456 the probability of getting heads on either the first flip or the second flip, 88 00:07:25,456 --> 00:07:29,482 you've double counted the one in the upper left in red. 89 00:07:29,482 --> 00:07:35,278 You counted it in the probability of the first flip and also, in the probability 90 00:07:35,278 --> 00:07:39,325 of the second flip. And you don't want to double count it, 91 00:07:39,325 --> 00:07:45,180 because then you'll get the wrong answer. So, I hope this chart illustrates why in 92 00:07:45,180 --> 00:07:50,384 order to generalize the formula for disjunction, we need to subtract the 93 00:07:50,384 --> 00:07:53,564 probability that both the events will occur. 94 00:07:53,564 --> 00:07:59,419 Here's another example. Out of a standard deck of cards, what's the probability of 95 00:07:59,419 --> 00:08:03,800 picking either a six or a club? Well, 96 00:08:03,800 --> 00:08:09,621 we know that there're 4 sixes and 13 different types of cards, so the 97 00:08:09,621 --> 00:08:13,298 probability of picking a six is one in 13. 98 00:08:13,298 --> 00:08:20,810 What's the probability of picking a club? Well, there are 13 clubs and 4 suits, so 99 00:08:20,810 --> 00:08:25,357 the probability of picking a club is one in four. 100 00:08:25,357 --> 00:08:36,314 And if we add those probabilities together, we'll get 4 + 13 makes 17 out 101 00:08:36,314 --> 00:08:46,617 of 52 cards, but that's not right because you've just double-counted the six of 102 00:08:46,617 --> 00:08:51,905 clubs. You counted it as a club and you also 103 00:08:51,905 --> 00:08:57,328 counted it as a six. So, in order to get the probability of 104 00:08:57,328 --> 00:09:04,441 picking either a six or a club, you need to add the probability of getting a six, 105 00:09:04,441 --> 00:09:11,376 one in thirteen, to the probability of getting a club, one in four, and subtract 106 00:09:11,376 --> 00:09:18,310 the probability of them both happening namely, getting the six of clubs, one in 107 00:09:18,310 --> 00:09:24,090 52, because otherwise, you'll be double-counting that six of clubs. 108 00:09:24,090 --> 00:09:30,151 So now, I hope you understand the rationale behind this generalized version 109 00:09:30,151 --> 00:09:34,001 of the rule for probabilities and disjunctions. 110 00:09:34,001 --> 00:09:39,080 And I hope you are able to do a few exercises using that rule.