 All right, let's start. Lesson six. So it says this, for experiments with only a few draws from a deck of cards or a pot or whatever, it's usually easiest to calculate probabilities using a probability tree. Tyler, my rule of thumb is this. Two cards tree, three cards maybe, anything more than three today's lesson. As the number of draws gets larger, the number of branches in the tree increases dramatically and the tree method becomes very tedious. So what we're gonna try and use is some of our permutation combination tricks from last unit. Suppose two cards are drawn without replacement from a shuffle deck of 52 cards. Calculate the probability that they are both aces. Well, and what we're gonna do here, Jesse, is we're gonna do it for two cards. We're gonna see if we can get the same answer using combinations and then we'll try and extend that using combinations for three, four, or five cards. We'll do five cards, five card poker. You can do the odds if you want to for seven card Texas Holden. It's gonna be the same idea. It just gets a bit unwieldy. How many aces are there in the deck? Four out of 52, 48 out of 52. Down this branch, we picked an ace. How many aces are there in the deck now? Out of, how many non aces are there in the deck? 48 out of 51. Hopefully use the compliment, it didn't count. Down this branch, we did not get an ace. So Amy, how many aces are still left in the deck? Out of 51, nice. And 47 out of 51. So they want us to calculate the probability that they are both aces. They want us to calculate the probability of ace, ace, which branches that? Four out of 52, and three out of 51. And I know that's 12 over 26, 52, but this time I want you to get your calculators out and reduce that, because when we use the combinatorics approach, we're gonna get answers in lowest terms anyhow. What is 12 out of 26, 52 in lowest terms? One out of what? One out of 2, 21, okay? Now I wrote it this way, if you wanted to use the formula, this is what the formula would look like. The probability of both cards aces equals the probability of ace one times the probability of ace two given ace one. That number times that number. I find the tree much simpler. What if we wanted to use combinations? Let's see if we can get this same answer without a tree. Now I still like a tree because Ian, it's visual to me, and I have all those built in error checks along the way. But for five cards, my tree would have so many, forget it. In fact, even with the homework that Pat asked, a three level tree was getting a bit unwieldy already. So the sample space is the list of all possible equally likely ways to get two cards from 52. How many ways can you pick two unordered cards from 52 cards? How many unordered ways? 52, choose two. How do I know what to choose and not a permutation? Because they said unordered order doesn't matter. What is 52 choose two? Six thousand, what? Oh, what is it? One, three, two, six. And then we have the number of unordered ways to select two aces. How many aces are there in the deck? Four, choose, how many aces do I want to choose? What is four choose two? Six, so here's what I'm gonna say. Remember, Victoria, I said for probability, if you can count it, you can solve it. Do you know how many ways there are to pick two cards? The number of ways to pick two cards is that many. Do you know how many ways there are to pick two aces? The number of ways to pick two aces is that many. So the odds of getting two aces are the total number of ways of success divided by the total number of outcomes. Does that reduce to one over 221? Does that reduce to that? Yes, you can also get there with permutations. However, I almost never use permutations. I'm gonna show it to you and then I'll show you where we'll use permutations, but it's almost never. The number of ordered ways to get two cards from 52 cards, why that's 52 P2, which is what? With your hair, it's too tough for me to tell. I wanna trust you, I think that I don't trust, so I have to do the ogre program. What is 52 P2? 2,652, strangely enough, is 2652 the denominator that we had earlier? Ah, how many ways can you select in order two aces? Well, that's for P2, is that 12, is it? So you could also go 12 over 2652, which is one over 221. Now, right now, I think all I've really done is confuse you. Let's make things a little clearer. Ari, all I've tried to do on this first page here is to say it seems that we can get the same answer using chooses without a tree. What I wanna give you guys now is here is Mr. Dewick's handy-dandy approach. Example one, here is how I do card questions. When you're picking cards, when you have a poker hand, do you care about the order that you pick the cards or do you just care whether they're in your hand or not? If you're in your hand or not, card questions are chooses. All right, how many cards are we picking? Five. So A says we wanna find the probability that the five cards are all spades. And although it wants to walk us through it, I'm gonna use a tool that we've already used. Do you remember what we drew for choose questions last unit? Then draw a table. Bucket. Draw a bucket. And our bucket is gonna have two categories for this question. Spades and others, except now instead of writing others, I'm gonna write not spades because now you know that shortcut. A little easier to write. Matt, how many spades are in the deck? How many not spades are in the deck? How many spades does this question want us to pick? Five. How many not spades does this question want us to pick? Zero. Okay. Put your pencils down and look up. The total number of ways to get five spades is 13 choose five and 39 choose zero. Don't write this down yet. That was last unit. That was how many ways to get five spades. Here we want to calculate the probability, which means we're gonna have to divide by something. What are we gonna divide by? How many cards are we picking grand total? 52 choose five. That's how you can do five cards without a tree. And now get your calculators out and crunch the numbers. And I think this time when you get a decimal, it's gonna stay as a decimal. It won't let you change it to a fraction. That's a decimal you can try. And you need to practice this on your calculator, so all of you try typing this in and see what you get. By the way, what is 39 choose zero? One, I'm not gonna bother typing that into my calculator. I don't think it'll turn that to a fraction. No, it won't. You get 4.95 times 10 to the negative four. 0.000495. 4.95 times 10 to the negative four. Well, Troy, what do you call a hand that has all spades? A flush, is that a good hand? Yes, should the odds be high? No, should be pretty small. There is a built-in error check, and I'll show it to you in a second because I wanna see if you can spot it. B, B, Stacey, what does B want us to make a hand of? So I'm gonna draw a bucket. It's gonna have three compartments. Spades, hearts, and others. Stacey, how many spades are there in the deck? How many hearts? How many others? 26. How many spades do they want me to choose? How many hearts? How many others? Trick question, none. Okay. The probability of three spades and two hearts is going to be 13 hearts, choose three, and 13 hearts, choose two, and 26 hearts, choose none, divided by, divided by what? 52 cards, choose five. Look up for a second. Look up. Here, Michelle, is your built-in error check. Do you see that the first number plus the first number add to the first number? And the last number plus the last number add to the last number? Does the first number plus the first number plus the first number add to the first number? Does the last number plus the last number plus the last number add to the last number? Now, that only works, Matt, if you put the choose zero in, and I always put the choose zero in in my work, I just never include it on my calculator because I know that choose zero always works out to what? One. But that is a handy-dandy built-in error check if you're trying to figure out Jesse, have you missed something? First plus first plus four equals first. Or if you're not sure what goes on the bottom? First plus first plus four. Oh, there's the first number, choose. Last plus last plus. Oh, that's what the last number goes. What's the answer? Point zero zero eight five eight. What are the odds of getting three spades in two hearts? Not very big. The odds are pretty good. You'll get more than two colors, more than two suits when you have five cards. The odds of only having two suits, not that likely. See, there are three spades and two non-spades, bucket. My bucket's gonna have spades and non-spades. I don't think there's a third compartment because I think everything is either a spade or not a spade. So I don't have to come up with a third compartment. Miguel, how many spades in the deck? How many non-spades? How many spades do they want us to pick? How many non-spades? 13, choose three and three spades. And 39, choose two, two non-spades divided by 52, choose five. What do you get? Point zero eight one five. Is that correct? There are only two spades in the five card hand. I think that's gonna be almost identical to this. I think my categories, Ian, are gonna be spades and not spades. But how many, oh, 13, 39. How many spades do they want us to get this time, Ian? Two. How many non-spades? Three, 13, choose two and 39, choose three divided by 52, choose five. What do you get, Victoria? Sorry, I can't hear. Point two seven four three, round it off properly. The odds, picking five cards of getting two spades, no more, no less, about one and four. Let's see, I just cut the deck. One, two, three, four, five. Heart, one spade, two spades. Oh, three spades. This would be the C question. The odds of that happening are pretty slim. But not the D question. E, here's a good poker hand. Two pair, two nice high pair. Two aces, two kings and one other card of a different type. How many cards are we picking, Matt Grand Total? Five, three? No, for two cards? Yeah, for three, maybe. For five, bucket. And this time we're gonna have three compartments. Aces, kings and others. Kyle, how many aces are there in the deck? How many kings are there in the deck? How many others do the math? 44, compliment. How many aces do they want us to choose? Two, how many kings do they want us to choose? Two, how many others do they want us to choose? One. So it's gonna be four choose two and four choose two and 44 choose one divided by, divided by what? 52 choose five. What are the odds of getting an ace king pair? This should be low because it's a good hand. Sorry, zero, zero, zero. Your calculator said 6.095 times 10 to negative four. Is two pair a good hand? Yep. Is it as good as a flush? So the odds of a flush should have been slightly less, were they? Yeah, there was a four there instead of a six. Example two, three prizes are awarded in a raffle. 100 people hold one ticket each. What is the probability that Alice, Ben and Conchetta win first, second and third prizes respectively? Tyler, what? Why is this a permutation? Respectively means in that order, order matters, but I'm gonna be honest. I'm not gonna solve this with a permutation. I'm going to first solve this with a three level tree. I'm gonna go like this, A, B, C. And I'm gonna say, how many people are there in the raffle? 100. What are the odds that Alice gets picked? How many Alice's are there in the deck? See, I think it's one out of 100. How many Ben's are there in the deck? One, but not out of 100. How many people are left in the deck now? 99 and one out of 98. I think that's one way to do it. The other way to do it is to do it as a permutation to say, look, from 100 people, permutate groups of three and out of all those groups of three, only one of them is the one that you want. Is 100 P3 the same as 100 times 99 times 98? Turns out it is. But I'll be honest, Victoria, I would have done this with a tree because it's the same thing happening over and over. I can visualize a single branch of a tree for four or five levels, even, as long as they're not changing the event. 100 P3, one divided by that, one over 970,200. One over 970,200. In fact, when we were doing trees, the order did matter and they were technically permutations. B, what's the difference between B and A? Or it doesn't matter, it's the choose. You know what we're gonna do? We're gonna do a bucket. And our bucket is gonna contain two groups, winners and losers. I'll call them non-winners. How many winners are there in this group? How many people do they want to win? Three. How many losers are there in this group? 97. How many winners do we want to choose? All three. How many losers do we want to choose? It's a trick question. None. Can you see what the equation's gonna be though? It's gonna be three choose three and 97 choose zero divided by, what's the bottom gonna be? Oh yeah, add the first, that's the first. Add the last, that's the last. 100 choose three. By the way, what is three choose three? One, what's 97 choose zero? One, I'm gonna go one over. Find out what 100 choose three is please. What is it? I want as a fraction, what is 100 choose three? 161700. This is gonna allow us to do 649. So in the BC 49 draw, six different numbered balls are chosen from numbered one to 49. They're randomly selected. So before the draw, you select six numbers. Find the probability that all three of your numbers are drawn. You never know. Sorry, all six of your numbers are drawn. Does the order matter? No, they'll put them in order Jesse to display them, but all I need to know is whether my number was picked or not. I don't care whether it was picked last or for long as it was picked, that means bucket. Our bucket is gonna have winners and losers. How many winning numbers are there? Six, how many losing numbers are there? 43, if I wanna win the grand prize of the winning numbers, how many do I need to choose? All six, how many losing numbers do I want to choose? Zero, what are the odds of winning? Six, 49. Six, choose six and 43, choose zero divided by, divided by what? 49, choose six. The top works out to one because six, choose six and 43, choose zero is one times one. What is 49, choose six? One, three, comma. Chance of winning, one in 14 million, basically. Oh, sorry, that's the chance of winning the grand prize. You can also win a prize if you get five, right? You'd have five winners, one loser. Or, oh, and you'd add them because or means add. You can also win a prize if you get four, right? You'd have four winners and two losers. I don't think you can win a prize if you get three, right? Maybe you can. Now, I have my tree, my bucket right here. Let's see, without redrawing the bucket, can we get the equation for exactly three? From six winners, choose how many this time? Three and from 43 losers, choose how many? Three divided by 49, choose six. Let's try that one. Still pretty small, 0.0177. Put your pencils. Turn on your workbook, please, to page 400 and 73. 473, 473. And I'd like you to look at example three. Read example three. How many marbles are we picking? Three, tree, I could. Let's try using combinatorics. Is this question saying anything about first or second or third in part A? No, so part A is gonna be a bucket. Part A I'm gonna have blue and not blue because those are the only two conditions they gave me. How many blue marbles are there grand total? Six, how many non blue marbles are there grand total? Eight, how many blue marbles do they want me to choose? For part A, two, how many non blue will I choose then? One, what's the equation gonna be? Six blues, choose two and eight others, choose one divided by, divided by what? From 14 marbles, choose three. What do you get? Are you guys okay in typing these in? Then to save time, I'm just gonna get the equation. I'll let you type them in on your own in your homework. B, at least one is blue. What would that mean? What does that least one mean? What cases? One or, or, oh okay, let's do this. And I have the bucket over here. Let's just see if I can modify it. So from six marbles, if one is blue, how many of the eight remaining do I have to choose? Two or, or means that. From six marbles, if two are blue, eight choose one. Or, from eight marbles, choose, sorry, from eight marbles. From six marbles, from six blue, choose, how many Tyler? Three and eight choose none. All divided by, 14 choose three. Let's see, first plus first, last plus last works. First plus first, last plus last works. First plus first, last plus last works. Built in a way of making sure you haven't missed anything. You would do the whole top line, get an answer, do the bottom line, and then you have your fraction. C, does the order matter in C? Yeah? Then I'm gonna do it by visualizing the tree. So I'm gonna write this as red one, comma, green two, comma, blue three. How many, on the first pick, how many red marbles are in the bag? Three, how many red marbles are in the bag? Five out of, sorry, 14. We've just picked a red marble. How many green marbles are left in the bag? Three out of, 13. We've just picked a green marble. How many blue marbles are left in the bag? Six out of, 12. Permutations are, trees are permutations. I can visualize a three branch, fairly basic tree, okay. D, one is red, one is green, and one is blue. Hmm, what would that look like? Does the order matter here? Bucket, now my bucket is gonna have red, green, and blue. How many reds in the bag to start out with? Five, how many greens? Three, how many blues? Six, how many reds do they want us to choose? One, how many greens? One, how many blues? One, Miguel, can you see the equation? What's it gonna be? Five, choose one, and? No, three choose one, and? Six choose one divided by 14, choose three. Does that make sense, Eric? Okay, again, if they give me the order, you'll notice I fall back on the tree if they give me a permutation. You can get there with permutations, ah, I'm not so fond of it. Next page, example four. City council consists of nine men and six women. Three representatives are chosen at random to form an environmental subcommittee. What's the probability that, I guess, a male mayor and two females are chosen? Well, here my bucket, first of all, does order matter? Does it say anyone has to pick first? No, then I'm gonna use a bucket, and my bucket's gonna have this, three categories. I'm gonna have mayor Jim, I'll use a letter J for Jim, and then males and females, or men and women. How many mayor Jims are there in the bag? One, how many males, don't say nine, because I've had to break this group up. How many males are they if I put mayor Jim in a separate category? Eight, how many females? Six, that's how I would divide this bucket up. Let's start. How many mayor Jims do I wanna pick? One, how many of the remaining males do I wanna pick? None. How many of the females do I wanna pick? Two, can you see the equation? One, choose one, and eight, choose zero, and six, choose two, all over, all over what? Plus, plus, plus 15, choose three people. B says, if you know the mayor is on the committee, what are the odds that two women are chosen? So if we already know the mayor is on the committee, if we know this guy is on the committee, my bucket's gonna get modified just a little bit. It's gonna be males and females. How many males, eight, how many females, six? The mayor is on the committee. He's guaranteed to be on the committee, so since I know he's on the committee, I'm not gonna include that in my calculation. I'm gonna say, then I want two, and none. It's gonna be eight, choose zero, and six, choose two, all over. Since I know the mayor is on, there's only 14 remaining to choose, and out of them, I'm picking two. See the slight difference in wording, okay? We've done some card questions. I'm gonna leave that instead. Let's look at example six. Example six. The famous birthday question. This is the one that I threw at you, and I wanna make sure I get this right, so I'm actually gonna look at my notes because I did these wrong ones. Yeah. In a class of 30 students, calculate the probability to the nearest hundred that they all have different birthdays. Assume that we're not in a leap year. Okay. Well, let's see if we can do this kind of using the fundamental counting principle. How many different birthdays does the first student have to choose from? 365. Now if I want the next student to have a different birthday, how many birthdays do I have to choose from? 364. And if I want the third student to have a different birthday, 363. Now I would keep going for 30 students. I think I would get 365 times 364 times 363 times 36, I do that 30 times. I think that that's actually 365 permutate 30. How many ways can I make sure everybody has a different birthday? That many ways. Okay. Many birthdays. Well, for each blank, there's 365 choices, 365 choices, 365 choices, 365 to the 30th, that's the total number of birthdays possible in the class. 30 kids, 365 days to choose from. Here's how many are different. Here's how many there are grand total. Does that make sense? So the answer is gonna be 365 P 30 divided by 365 to the 30th. That's the odds of 30 students all having different birthdays. Can you crunch that for me please? 0.293, 0.294. That's the odds if I have a class of 30 of them all having different birthdays. So here's the question that I asked you guys. What are the odds of at least two having the same birthday? Or let me say it this way, what are the odds of none of them having different birthdays? I heard it, what'd you say? Compliment, 1 minus 0.294. If I had 30 students in this class, what are the odds of all of them having different birthdays? This is the odds of two or more having the same. What is it? 0.706, about a 70% chance. Now, how many did we have in this class? We had 24. So Irwin, I'm gonna redo this to see whether I should have made the bet with you or not really quickly. It would be 365 P 24 divided by 365 to the 24th. One minus that, yeah, I should have made the bet. Still, the odds were 54% in my favor. I'll take that every time, but once in a while I'll lose. Slightly less than half the time though, which means slightly more than half the time I'll win. I'll take that. Now, example six I did for its nerdiness. It's not going to be on your test, but it is a neat probability question, the birthday question. I won't test you on it. What's your homework? Sorry, it went a bit longer. We started a bit later. Number three. Now, number three is two marbles. You could do a tree. See if you can do it with combinatorics. The only one, B, the order matters. I would do a mental tree for B. I would go red one, comma, yellow two, and I think you can just walk your way down the branch. Four A, four B, four C, you know what, all of number four. Now, I like number nine. Number nine, how many marbles are we choosing? Number nine, how many marbles are we choosing? Five. Tree? No. Skit B. Skit B. 12 girls, 12 boys, 10 girls. Sure, 10 is good too. Okay. Using shoes. On your test, I'm going to give you a card question or a marble question, because those are the easiest to type up, but there's some nice applications here. And most of you have watched enough gambling to be interested in the card questions anyways. Sorry.