 All right, I just jotted this down in a hurry. I can print this up for you afterwards if you want to, so you don't need to write this down unless you feel you learn better by writing this down, or I'll throw it online as well. But some key concepts here. I said, if you can count it, you can solve it. If you know how many happy successes, victories there are divided by how many total outcomes there are, that's the answer. It's the number of successful outcomes divided by the number of outcomes. We did that on the quiz for a card question, I think. OK? Second key concept I've tried to reinforce, or means what? Let's write that in, Mr. Dewick. And means what? Multiply. Or does mean minus the overlap. In fact, that's when I do the formula on the tree, because the way we set up the trees never had an overlap, it took care of itself. But yeah, the formula which we hardly used was it's on your formula sheet, but a or b was the probability of a plus the probability of b minus the overlap, and. The formula hardly used for and was, see, I simplified it to and means multiply, but it's the probability of the first one times the probability of the second one given that the first one has occurred. Brett, what this statement here is saying is, hey, make sure you go down the right branch. Some terminology. Mutually exclusive versus independent. These are two kids get mixed up. So what does mutually exclusive mean? No overlap. It means this, two completely separate groups. What does independent mean? Well, it means that the probability of b occurring if a occurs is the same as the probability of b occurring if a didn't occur. Whether a occurs or not does not affect the odds of b occurring. That was the mathematical definition of independent. What I said was, if you're doing a tree and the second branch matches the second branch over here, you know what, those are independent. It means the odds didn't, the first event had no effect on the second event. We introduced the notion of Venn diagrams, and this is going to give us some of the questions you're going to have to deal with. So I'm just going to make up some examples. Example, in a survey, 250 students are surveyed. 100 like Facebook, 75, Twitter, 50 do both. Done? Done? OK, 50 do both. Hey, what is the probability that a student neither Facebooks nor Twitter's questioned one? And apparently, Facebook is plural, not a word, not a verb yet. Pardon me? Yeah, sure. Now, here's how I would know to tackle this with a Venn diagram. How many different categories are there? Did they give me the overlap or talk about the overlap? Both? Venn is my best approach. You can get there with formulas. I'm not so fond of it. So I would do this, Facebook, Twitter. How would I fill this out? I said you always start in the middle. Now, there were some questions where they did not give you the overlap, and we put a text there. I'll make one up like that in just a second. What kind of questions can I ask you just with this piece of information here? Well, what's going to go right here? It's going to go right here. 50. What's going to go right here? 50. What's going to go right here? And clearly, I made these numbers up because what's going to go right here? 75. Let's see, 150 is 125. I think 125. And I doubt more than half of you. I doubt half of you don't have a Facebook account, but I'm making up numbers. So the probability of neither, which would be not Facebook, or not Twitter, if we wrote it in our symbolism, is which one of these is neither? B, find the probability that a student has a Facebook, but not a Twitter account. Which one of these is Facebook, but not Twitter? This 50 here. I'm just going to write the answer down. 50 out of 250 to make it a probability. 0.2? C, given that a student does not Twitter, what is the probability that the student has a Facebook? I could ask you a conditional probability. How do I know this is a conditional probability, Leslie? What's my trigger word? Given. Never mind. No, not thank you. You should have done this last week. So not thank you. Pardon me? Yeah, back there. If all goes well. Conditional, I said, here is where we use the formula. I didn't put that in my earlier stuff I meant to, and I got interrupted. The one time we use the formula, Dina, is for conditional, but I gave you a way to memorize it. First, you got to translate it. What's the given? What have they told me? What do I know? Given what? Not Twitter. That's going to go right there. Find what? Facebook. That's the translation into the symbols. And then I said, this is going to be the probability of Facebook and not Twitter, both divided by the probability of not Twitter, the second one, the given one. It's both over the given one. Now let's look at my Venn diagram. How many are in the Facebook and not in the Twitter at the same time? Well, these two groups are in the Facebook. But you know what? I think only this one is in the Facebook and not in Twitter at the same time. In fact, it's the same as the answer for B as it turns out. It's going to be 50 over 250 divided by how many are not in the Twitter? 175, did you add them up or did you complement? I went 75 minus 250. Did you add them up? I think it's 250 minus 75. Or you can go that and that. But those are not Twitter, right? So 175 over 250. How do I divide by a fraction? Flip it and multiply. This is going to end up being 50 over 250 times 250 over 175. Can I cancel? Yes, I can. These are factored this time. And I get 50 over 175. And I'm curious, lowest terms, two out of seven kids. If you know they don't, Twitter, based on my made up statistics, which are terrible, there's a two in seven chance that they have a Facebook account. So I could give you a Venn diagram question and ask you to do something with it. The other type of Venn diagram question that I can give you requires you to start on the outside sort of. So something like this. I'm making this up and it may end up being impossible because I'm trying to do the math in my head. We may have to adjust this. So you might want to hold off and run this down until we know the Venn diagram actually works. Oh, sure. 87% of the students in a school are cheering for the Canucks. 22% of the students in a school are cheering for the Maple Leafs. I apologize for those of you for whom I just swore. But Maple Leafs, if he says desperately trying to do some math in his head, 31% cheer for no team. What percent cheer only for the Canucks? I'm making this one up. There's a chance this one might not work. Let's find out. How do I know this is a Venn diagram question, Brett? First of all, two categories, Maple Leafs, Canucks. Also, often percentages. I didn't do percentages on the previous one. You're going to make it? Because you're sure? OK, come forward further if you need to. Always can. OK, I'm going to start out with my Venn, where we have Canucks and Maple Leafs. And I said, whenever possible, we want to start out in the middle with the overlap, with the both, with the and. Did this question tell me the middle, the overlap, the both, the and? No. You know what, then? I'll say the overlap is x. What percent cheer for the Canucks, 87? What percent only cheer for the Canucks? What am I going to put here? 87 minus x. What am I going to put here? 31, Mr. Dewick, read the question properly. 22 minus x. What am I going to put here? 31, hoping desperately x ends up being positive. Otherwise, I've got a problem. Because I also know one more thing. You know what, this plus this plus this plus this have to add to? 100%. That's also why this works nice. So this is going to be 87 minus x plus x plus 22 minus x plus 31 equals 100. Hey, one of my x's cancels. Kind of cool. 87 plus 22 plus 31, 87 and 22 is 117. Oh, heck, Mr. Dewick. Don't make a sloppy math mistake at this juncture. And I get 140 minus x equals 100. What's x? And now I can fix this. I can say, hey, that's 40. Hey, that's 37. Hey, that's, there's my problem. This is what I was worried about occurring. Dope, negative 18. Well, it's the Leafs fans. Oh, no. So I've made up a bad example. But do you understand the concept of if they don't give you the overlap solving your way there? And on the test, it will work out. I'm just going to write that example. Sorry. But this was our first couple of days. We used OR, we used Venn diagrams. Then we moved on to the notion of a tree. So we could do something like this. Two cards are drawn without replacement. What is the probability that you get at least one face card? How many cards are we drawing? Two, tree. I mean, draw a tree. I said that last time. And someone said, no, two. I said tree. No, two. I said, no, draw a tree. I can say three, believe me. So what's my event going to be? I think face card or not, face card or not. And I usually put a subscript. So that's face card on the first, face card on the not on the first, face card on the second, face card not on the second, face card on the second, face card not on the second. How many face cards are there in the deck? 12 out of 52, 40 out of 52. OK, down this branch, you pick the face card. How many face cards are left in the deck? 11 out of 51, 40 out of 51. Here you have 12 out of 52, 39 out of 51, Mr. Duick, 51, Mr. Duick, 51, Mr. Duick. And that's why I like that built-in error check to always make sure they add to one that caught me right there. Oh, they don't add to one. I've moved somewhere. By the way, Dina, are these two different from these two? These are dependent, not independent. What does at least one face card mean? One or two. Now, I could go this branch or this branch or this branch, or as a faster way, I could use the complement, one minus this branch, because one minus none is at least one. Neither of those would be just fine. I'm going to go just to reinforce the idea of the complement. So I'm going to do this. Boom, boom, boom. But I'm going to say, hey, one minus the probability of none, which is one minus 40 out of 52 and 39 out of 51. One minus 40 out of 52 and 39 out of 51. Math, enter, enter, 7 out of 17. Mr. Duick, what answers will you accept? I would accept a decimal, let's say the four decimal places, rounded off properly. I prefer the fraction either not reduced or in lowest terms. I don't care which. This makes more sense to me than that. What if it was three cards? Well, oh, if it was three face cards, so not at least two, but three exactly, I could do that single branch of the tree, probably without even drawing the tree. It would be 12 out of 52 times 11 out of 51 times 10 out of 50. Done, three face cards. But really, anything more than that, I'm going to use bucket and combinatorics. What if it's a permutation? Actually, most of our trees were naturally built in permutations, because this has face card on the first, followed by face card on the second. The permutations kind of carry themselves. So let's do one with a bucket. Example, jar A contains four blue, comma, three red, comma, and five green marbles. Jar B contains about three blue, two red, and five green, seven. I don't want to add it to the same. What have we got? Four plus three is seven. 12 in the first one. I'm going to have 10 in the second one. Green marbles. Oh, what's a jar? Good point. Container, with a lid, technically. Usually, it's a clear container made out of glass. So this is, that's a jar. Thank you, I'm going, what are you looking at? Currently, I have one. Good point. A, let's see. What do I want to do here, Mr. Dewey? A jar is randomly chosen. Find the probability, a jar is randomly chosen, and a marble is taken from the jar. Find the probability that it is, oh heck, green. How many events are going on here? Two, tree. My first inclination here would be a tree. So I'm going to draw a tree for this, and then we'll do, hey, what if we're picking more marbles? A, B, and it looks like it's 1 half, 1 half. Brett, on your test, instead of just picking the jar randomly, I may say roll a dice, and if it's a 1 or a 2, pick jar A. If it's a 3, 4, 5, or 6, pick jar B, in which case, Dean, it would be, instead of a half a half, 2 out of 6, 4 out of 6. And then we have blue, red, green, blue, red, green, 12, so 4 out of 12, 3 out of 12, 5 out of 12 for the first jar, 3 out of 10, 2 out of 10, 5 out of 10. Here or here, I think it's going to be 1 half times 5 out of 12 or 1 half times 5 out of 10. And here I don't see a common denominator. I would whip out and go to my graphing calculator right away here, Leslie. 0.5 times 5 out of 12 plus 0.5 times 5 out of 10, math, enter, enter, 11 out of 24. Almost but not quite 50-50 given that a blue marble is chosen. What is the probability, probability that it came from jar A? What's the given? What did they tell you? No, read. What's the given? And this is the key because you get this backwards. I'm sorry, you have to get the wrong answer then. What's the given? What did they tell you? It's blue. By the way, what level on the tree is that? Bottom, what do they want you to find Leslie? And that's going up the tree. This is our classic conditional, okay? Here's the, we did the Venn diagram one, which worked, but here is where we really use this. We said, okay, this is going to be the probability of given that it's blue came from jar A. And because I have a letter B here and a letter B here, I'm going to use BL for blue. So I don't foolishly get confused and get them mixed up, which is going to be on my formula sheet. You know what? It's both over the given one. And I'm hoping when you glance at the conditional formula, you can see that, right? It is, look at it on the right, that is both of them divided by the given one, right? That's how I translate that one, Dina, because the letters get all mixed up. So I'm just going to say this is jar A and blue divided by blue. Look at my tree. What's A and blue? I think that's that branch, isn't it? Half 412s divided by what's blue. Well, I think that branch or that branch, I think in the bottom, it's going to be a half 412s or a half 310s. Can I cancel? No. I'm going to work out the top. 0.5 times 412s equals math, enter, enter. Oh, this works out to 1, 6 divided by, oh, 0.4 times 412s plus 0.5, that's a 0.4, 0.5 times 412s plus 0.5 times 310s, math, enter, enter. Nineteen sixtieths, and now I can go, how do I divide by fractions and multiply? Or I can go to my calculator. I mean, I can go bracket, one sixth divided by bracket, 19 over 60. I think that's dividing two fractions, but I would have to put the top and the bottom in brackets for this to work. Enter, yeah, apparently 10, 19th C. Actually, let's just do a different example, because here I'll do a combinatorics one. A jar contains five blue marbles, six red and three green marbles. If five and six is 11 plus three is 14, if six marbles are chosen without replacement, what is the probability that two are blue, two are red, and two are green? How many marbles am I picking? Tree? No, I don't want to do a sixth level tree. How many marbles? You're right, six bucket or some other approach. And here I think it's independent because we're moving without replacement. So I'm going to draw a bucket. And in my bucket, I have red, blue, and green, and there are five red, sorry, five blue, Mr. Do it, get this right, sheesh, six red, five blue, and three green. And for some reason, I went red, blue, and green, even though in the question, I went blue, red, and green. But as long as your bucket is correct with the right numbers, you're fine. How many blues do I want to pick here, Dina? How many reds? How many greens? So what's my equation? It's going to be, did you write blues first and you follow me and do reds first for some strange reason? Okay, so it's going to be from six reds, choose two, and from five blues, choose two, and from three greens, choose two, divided by, what, 14 marbles in total, choose six at a time. Worth practicing typing that in right now, folks, because that's one of the skills I think. And this one, it leaves a decimal. What do you get? Point, sorry, louder. Give it to me to four decimal places, please. One, four, nine, nine. Yeah. How did you know to do the bucket combinatorics approach, Leah? Because anything more than two, I probably don't want to do a tree. Now, unless they had said, pick six marbles, all six red, that I could do as a tree, because it would be six out of 14 times five out of 13 times four out of 12 times, I could visualize that branch, no problem. Could I do a conditional here? Probably, given that two are red, what's the probability that at least three are red, at least would mean three red or four red or five red or six. That'd be a lot of work. I would not give you one quite that much or means, but we could handle it. I'm trying to think, what did I ask you for the combinatorics? I think I will ask you, and at least or at most, so a good part B would be, what's the probability of getting at least four red, which would mean four or five? It would be four or five, but not six. Why would it be four or five, but not six, and only five red marbles? I'm going to pause for a second, because I want to look at my test, so I need to freeze this. We're back to live. I think the last thing is binome PDF and CDF, or are you guys comfortable with that? So on the written, on the written, there'll be a question, something like this. The probability that my car starts is, let's pretend I'm driving a junky car, 72%. Over the next 10 days, what is the probability that it starts exactly eight times out of 10? Now, how can I recognize that this is not a tree and not a bucket? First of all, how many trials are we doing? 10, so I don't want to do a tree, but Leah, this is different from the one that you asked me, because the odds never change. What are the odds on day two, 72%, what are the odds on day three, 72%, what are the odds on day four, 72%? So it's not a tree where the branches change each level. If I want to find exactly eight times out of 10, the long way, and you'll find this on your formula sheet, is 10 shoes, 8.728 starts, 0.182 non-starts, or on your calculators, it's binome PDF of 10, 0.72, 8, which is what, to four decimal places, 0.2548, basically a one in four chance of starting exactly eight times, be at least eight times out of 10. What does at least eight times mean? Eight or, or, and I could find each of those individually, and it's only three of them, and I could add them up on your test. I'm probably going to give you an at least or an at most, where there's more than three, where it's worth using the CDF shortcut. Now what was the CDF shortcut? Don't write this down because this is wrong. A lot of students, a lot of students will want to go like this. They'll go, oh, binome, they'll say at least a CDF of 10, 0.72, 8. That does not give you eight or nine or 10. That gives you 8, 7, 6, 5, 4, 3, 2, 1, 0. So how can I get 8, 9, or 10? Compliment, 1 minus binome CDF of 10, 0.72, 7. Because if I remove 7, 6, 5, 4, 3, 2, 1, 0, Leslie, you know what's left? 8, 9, or 10. That'll be the odds of this car starting up eight or more times. Which is what? 0.5622? That seems low to me, 0.4378. I figured because if the odds are 72% that it starts. Is that right? Yeah. Or 40, 44% if I make that a percent different. Okay. How do I know it's a binome and not a bucket or a choose? Because there'll be a high number of trials and the odds will never change. I think I put two like this on the multiple choice and it'll even be really blatant. It's the last question on the written because it's the last thing that we did. Okay. I might even make fun of Brett as part of it. Yeah, sure. All right. Now that's kind of a quick overview of what we did. Or and trees to do more detailed and conditional to go backwards along the tree or look at subgroups and Venn diagrams. Choose buckets when we had more than two things. Binome. Now, are there any from the review that you would like me to go over that you've looked at? Remember the review answer key is completely online. There are some weird-ish ones, but I tried to explain online. What's what? Any you want me to go over? You guys are good? So this one here is online. And this is a pretty good indication up to a point. You'll notice if you look at this review, Brett, when you get to around number 20, that's when I ran out of BC provincial exam questions and started using some of the tougher Alberta exam questions. They're still fine, but these ones are a little bit of overkill in my mind. Well, 25 is still good, 26 is still good, and 27 is a conditional. But like this mouse question, for example, that's kind of nerdily cruel, but it's a little overkill. Certainly, I would say you want to make sure you can comfortably handle 119. And then the rest are good curveballs and tougher questions. And I think most of you, if you look at my work, you'll be able to figure out what the heck I did. I tried very hard to show you exactly what I had done. We're good? Who would like a copy of this printout? Hang on one second here.