1 00:00:02,200 --> 00:00:06,295 The rule for the probability of conjunctions is a little more 2 00:00:06,295 --> 00:00:09,663 complicated. One reason is that, there're really two 3 00:00:09,663 --> 00:00:14,221 cases we have to keep separate. We're going to look first at the case of 4 00:00:14,221 --> 00:00:19,307 independent events, and we have to begin by saying what it means to call them 5 00:00:19,307 --> 00:00:23,276 independent. To call them independent is simply to say 6 00:00:23,276 --> 00:00:28,798 that the probability of one does not affect the probability of the other. 7 00:00:28,798 --> 00:00:34,622 So, if you flip a coin or roll a die, and get a certain number, and then you pick 8 00:00:34,622 --> 00:00:40,824 up the coin or the die, and you flip it or roll it again, then the second result 9 00:00:40,824 --> 00:00:45,967 is not affected by the first result. So, those are independent events, 10 00:00:45,967 --> 00:00:49,220 and with cards, if you take a deck of cards, 11 00:00:49,220 --> 00:00:54,726 and you pick out a card, two of spades. Then you take the two of 12 00:00:54,726 --> 00:01:01,127 spades, and you put the pack in the deck, you shuffle, it's going to be independent 13 00:01:01,127 --> 00:01:07,215 of whether you get it again, because what you picked up the first tine doesn't 14 00:01:07,215 --> 00:01:10,650 affect what you pick up the second time. But, 15 00:01:10,650 --> 00:01:16,957 take a deck of cards and you pull out the king of spades, and you take that card, 16 00:01:16,957 --> 00:01:21,372 and you throw it away, and don't put it back in the deck, 17 00:01:21,372 --> 00:01:27,837 then it affects all the probabilities of the remaining cards, because now you have 18 00:01:27,837 --> 00:01:32,429 fewer cards than you had before. So that's the difference between 19 00:01:32,429 --> 00:01:37,133 independent events and dependent events. So we're going to look at independent 20 00:01:37,133 --> 00:01:42,700 events first, and look at the rule for the conjunction of independent events. 21 00:01:42,700 --> 00:01:46,540 And the rule for, for this case is pretty simple. 22 00:01:46,540 --> 00:01:53,300 If two events are independent, then the probability that both events will occur 23 00:01:53,300 --> 00:01:58,019 in that order, because we're talking about this one and then that one, we'll 24 00:01:58,019 --> 00:02:03,669 talk a little bit more about that later, but we're assuming that. The product that 25 00:02:03,669 --> 00:02:08,987 they both will occur is the product of the probability of the first event times 26 00:02:08,987 --> 00:02:16,134 the probability of the second event. And notice that the both occur is tied to 27 00:02:16,134 --> 00:02:19,801 the product, or multiplying the two probabilities together. 28 00:02:19,801 --> 00:02:25,113 And the reason for that is that it's less likely that they will both to occur, than 29 00:02:25,113 --> 00:02:29,686 that one of them will occur separately. And when you multiply the two 30 00:02:29,686 --> 00:02:34,742 probabilities between zero and one, you're going to get a lower probability 31 00:02:34,742 --> 00:02:40,008 that they both occur, than for either one of them occurring, you know, all by 32 00:02:40,008 --> 00:02:43,561 itself. We can restate this rule symbolically 33 00:02:43,561 --> 00:02:49,202 using the symbols that we used before for the probability of negation. 34 00:02:49,202 --> 00:02:55,890 Namely, the probability of hypothesis 1 and hypothesis 2 both being true is the 35 00:02:55,890 --> 00:03:01,530 probability of hypothesis 1 times the probability of hypothesis 2. 36 00:03:01,530 --> 00:03:06,678 Now, let's apply it to some cases. Suppose you flip a coin, and you get 37 00:03:06,678 --> 00:03:09,720 heads? And you flip another coin, and you get 38 00:03:09,720 --> 00:03:13,315 heads again? What's the probability of getting heads 39 00:03:13,315 --> 00:03:19,520 twice at two flips of a coin? Well the probability is 5. for the first 40 00:03:19,520 --> 00:03:25,820 flip of getting heads. It's 5. for the second flip of getting 41 00:03:25,820 --> 00:03:33,485 heads and then the probability of getting both is going to be 5. times 5,. or 25.. 42 00:03:33,485 --> 00:03:41,675 So the probability of getting two heads in a row is 25. on two flips of a fair 43 00:03:41,675 --> 00:03:45,235 coin. A little more complicated example uses 44 00:03:45,235 --> 00:03:48,660 cards. Here we have a standard deck of cards. 45 00:03:48,660 --> 00:03:57,060 Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King. 46 00:03:57,060 --> 00:04:04,644 So there are 13 different types of cards, and there are 4 different suits, spades, 47 00:04:04,644 --> 00:04:10,090 hearts, diamonds and clubs. So we can remember that. 48 00:04:10,090 --> 00:04:16,069 The probability of having one of these types of cards chosen, assuming that 49 00:04:16,069 --> 00:04:22,903 they're shuffled properly, is one out of thirteen, and the probability of getting 50 00:04:22,903 --> 00:04:31,104 a particular suit, say a club, is one out of four, because there are 13 clubs out 51 00:04:31,104 --> 00:04:35,280 of 52 cards. Clubs are one suit out of four, so the 52 00:04:35,280 --> 00:04:39,010 probability of picking a club is one in four. 53 00:04:39,010 --> 00:04:44,896 and the probability of picking a particular card, say a 3, is one in 54 00:04:44,896 --> 00:04:47,466 thirteen. Again, a deck of cards. 55 00:04:47,466 --> 00:04:52,726 Suppose I pull out a 4, and then I pull out a club. 56 00:04:52,726 --> 00:04:59,885 Now in a standard deck of cards, there's thirteen cards, we're assuming no jokers. 57 00:04:59,885 --> 00:05:04,360 So the odds of getting a four are one in thirteen. 58 00:05:05,640 --> 00:05:11,096 There're four suits, spades, hearts, diamonds and clubs. 59 00:05:11,096 --> 00:05:17,171 So the odds of picking a card that's a club is one in four. 60 00:05:17,171 --> 00:05:23,863 What's the probability of picking a four, and then picking a club? 61 00:05:23,863 --> 00:05:34,760 Well it's going to be 1 in 13 times 1 in 4 And that means, this can be 1 in 52. 62 00:05:35,900 --> 00:05:43,505 So the odds of that combination occurring, right, in that order, is going 63 00:05:43,505 --> 00:05:50,800 to be, 1 over 13 times 1 over 4 equals 1 in 52. 64 00:05:50,800 --> 00:05:56,077 That example shows you how to calculate the probability of a conjunction with 65 00:05:56,077 --> 00:06:01,151 independent events, but what if the events are not independent? That is, what 66 00:06:01,151 --> 00:06:06,091 if the probability of one effects the probability of the other? Here is a 67 00:06:06,091 --> 00:06:09,842 question. Joe can run a mile in less than five 68 00:06:09,842 --> 00:06:14,626 minutes, about half the time. When he runs and he's timed, half of the 69 00:06:14,626 --> 00:06:19,480 time he's under five minutes, half of the time he's over five minutes. 70 00:06:19,480 --> 00:06:25,179 So what's the probability he can run a mile in less than five minutes on a given 71 00:06:25,179 --> 00:06:28,696 occasion? .5. But, what's the probability he can run 72 00:06:28,696 --> 00:06:34,254 one mile in under five minutes, and then run another mile in under five minutes 73 00:06:34,254 --> 00:06:40,441 right after that? Well, if you did this calculation you 74 00:06:40,441 --> 00:06:45,480 will have 5. times 5.. And that means 25,. but you know that's 75 00:06:45,480 --> 00:06:49,361 not right, because he's going to be dead tired after the first mile. 76 00:06:49,361 --> 00:06:54,104 There's no way that the probability of running the second mile in under five 77 00:06:54,104 --> 00:06:57,677 minutes is just as high, whether or not he would have just 78 00:06:57,677 --> 00:07:01,373 finished running a mile. So those events are not independent, 79 00:07:01,373 --> 00:07:06,486 and that example shows you why you need a new formula to calculate the probability 80 00:07:06,486 --> 00:07:09,505 of a conjunction, when events are not independent, 81 00:07:09,505 --> 00:07:12,770 but instead their probabilities depend on each other. 82 00:07:12,770 --> 00:07:18,538 The same distinction between independence and dependence applies to our old friend, 83 00:07:18,538 --> 00:07:21,736 cards. So just to simplify matters let's start 84 00:07:21,736 --> 00:07:26,323 with a very limited deck. It's just got four aces It's got the ace 85 00:07:26,323 --> 00:07:28,269 of spades, the ace of hearts, 86 00:07:28,269 --> 00:07:31,119 the ace of diamonds, and the ace of clubs. 87 00:07:31,119 --> 00:07:36,748 And notice that the ace of hearts and the ace of diamonds are red, and the ace of 88 00:07:36,748 --> 00:07:43,064 spades and the ace of clubs are black. So, knowing that, out of these cards, if 89 00:07:43,064 --> 00:07:50,703 we shuffle them, and don't look at them, what's the probability of picking a red 90 00:07:50,703 --> 00:07:55,120 ace? Well, 1 in 2, right, because 2 of them 91 00:07:55,120 --> 00:08:02,340 are red out of 4. Now. Sorry, Oh, 92 00:08:02,340 --> 00:08:05,191 a red one, I've picked the red one. Now, 93 00:08:05,191 --> 00:08:09,393 let's put it back in the deck and shuffle them together, 94 00:08:09,393 --> 00:08:12,619 okay. What's the probability now of picking 95 00:08:12,619 --> 00:08:15,750 another red Ace? Again 1 in 2. 96 00:08:15,750 --> 00:08:22,048 So, if you look at the possibilities all laid down, the first pick is in the 97 00:08:22,048 --> 00:08:25,843 columns, right? So the leftmost column is if you pick an 98 00:08:25,843 --> 00:08:30,529 ace of spades on your first pick, second column is ace of hearts on your 99 00:08:30,529 --> 00:08:33,848 first pick. Third column is ace of diamonds on your 100 00:08:33,848 --> 00:08:37,036 first pick, fourth column is ace of clubs on your 101 00:08:37,036 --> 00:08:40,290 first pick. And then your second pick is this rows, 102 00:08:40,290 --> 00:08:43,804 and again it's either spades, hearts, diamonds or clubs. 103 00:08:43,804 --> 00:08:49,372 So there are 16 possible combinations of picks here, in order, and how many of 104 00:08:49,372 --> 00:08:52,999 them are picks where you have two red aces? 105 00:08:52,999 --> 00:08:57,806 Well, this is one, this is one, this is one, and this is one. 106 00:08:57,806 --> 00:09:04,976 So, the 4 possibilities in the middle are the possibilities where you've picked an 107 00:09:04,976 --> 00:09:11,723 ace that's red, both on your first pick and also on your second pick. So the odds 108 00:09:11,723 --> 00:09:19,061 of doing that are 4 out of 16 or one out of four, or 25,. just like we calculated 109 00:09:19,061 --> 00:09:23,370 using the formula. But, what if there's no independence? 110 00:09:23,370 --> 00:09:27,011 Let's start with the same four aces. Okay, 111 00:09:27,011 --> 00:09:30,966 and what's the probability of picking a red ace? 112 00:09:30,966 --> 00:09:35,251 Still one in two. So let's suppose I pick the ace of 113 00:09:35,251 --> 00:09:36,981 hearts, throw it away, 114 00:09:36,981 --> 00:09:42,666 do not put it back in the deck. And now I've only got three aces left. 115 00:09:42,666 --> 00:09:48,186 What's the probability of picking a second red ace without looking? 116 00:09:48,186 --> 00:09:54,201 Well it's going to be one in three, because one of them is red and the other 117 00:09:54,201 --> 00:09:57,332 two are not. There are three cards left, 118 00:09:57,332 --> 00:10:01,856 one of them is a red ace. We can also look at it this way. 119 00:10:01,856 --> 00:10:07,492 There are sixteen possibilities. How many of these include a red ace on 120 00:10:07,492 --> 00:10:11,698 the first pick and also a red ace in the second pick? 121 00:10:11,698 --> 00:10:18,027 Well, those are the four in the middle, but if we got rid of the ace of hearts by 122 00:10:18,027 --> 00:10:23,619 throwing it out and not putting it back in the deck, then, we've left out this 123 00:10:23,619 --> 00:10:27,178 whole row. And we know that we are in this column 124 00:10:27,178 --> 00:10:30,809 because we picked an ace of hearts the first time. 125 00:10:30,809 --> 00:10:36,474 And only one out of the three is picking a red ace the second time, so we know 126 00:10:36,474 --> 00:10:41,630 that there's one out of three. So when it's dependent like this, because 127 00:10:41,630 --> 00:10:45,770 you don't put the card back, you need a different formula. 128 00:10:45,770 --> 00:10:51,380 The probability is one out of two times one out of three equals one out of six, 129 00:10:51,380 --> 00:10:56,919 instead of the one out of four that we had when they were independent and you 130 00:10:56,919 --> 00:11:00,612 put the card back. So we need a different formula to 131 00:11:00,612 --> 00:11:05,020 generalize for this case where there is dependence, okay. 132 00:11:05,020 --> 00:11:10,383 And this formula's going to use the conditional probability, which is just a 133 00:11:10,383 --> 00:11:14,052 fancy way of saying what I've just mentioned to you. 134 00:11:14,052 --> 00:11:17,369 Right? The conditional probability of picking a 135 00:11:17,369 --> 00:11:22,803 red ace on the second pick given that I pick a red ace on the first pick and 136 00:11:22,803 --> 00:11:27,813 threw it away, is one in three. It's how many times does x occur picking 137 00:11:27,813 --> 00:11:34,245 a red ace on the second pick out of the cases where I pick a red ace on the first 138 00:11:34,245 --> 00:11:38,817 pick and threw it away and that's going to be your one in three. 139 00:11:38,817 --> 00:11:42,960 So that's the conditional probability, and now the formula. 140 00:11:42,960 --> 00:11:48,175 The probability of both of two of events occurring is the product of the 141 00:11:48,175 --> 00:11:53,675 probability of the first event occurring times the conditional probability of the 142 00:11:53,675 --> 00:11:57,890 second event occurring given that the first event occurred. 143 00:11:57,890 --> 00:12:00,634 That's where you use the conditional probability. 144 00:12:00,634 --> 00:12:04,722 Now notice, when the events are independent by conditional probability of 145 00:12:04,722 --> 00:12:08,698 the second event occurring, given that the first event occurred, is just 146 00:12:08,698 --> 00:12:12,450 going to be the same as the probability of the second event occurring, 147 00:12:12,450 --> 00:12:15,924 because the first event doesn't affect the second event. 148 00:12:15,924 --> 00:12:20,764 So that first formula for an independence is just an instance of this formula, 149 00:12:20,764 --> 00:12:25,542 where the conditional probability of the second, given the first, is identical 150 00:12:25,542 --> 00:12:30,258 with the probability of the second. So that's why it's a generalized formula 151 00:12:30,258 --> 00:12:35,159 and actually includes the first formula, but when they're independent it's just 152 00:12:35,159 --> 00:12:37,890 simpler to think of the first formula alone. 153 00:12:37,890 --> 00:12:43,534 We could also symbolize this rule. We can say that the probability of 154 00:12:43,534 --> 00:12:50,896 hypothesis 1 and hypothesis 2 is equal to the probability of hypothesis 1 times the 155 00:12:50,896 --> 00:12:57,112 probability of hypothesis 2, given hypothesis 1, and that straight up and 156 00:12:57,112 --> 00:13:02,675 down line in this formula is what indicates conditional probability. 157 00:13:02,675 --> 00:13:08,483 This little formula here means the probability of hypothesis 2 given 158 00:13:08,483 --> 00:13:12,563 hypothesis 1. So you can use this rule to calculate the 159 00:13:12,563 --> 00:13:17,590 probability of any old conjunction, whether they're independent or not, 160 00:13:17,590 --> 00:13:22,511 but when they are independent, you might want to use that simple rule, which is 161 00:13:22,511 --> 00:13:25,540 the first version of probability of conjunction.