 Well, welcome to Episode 37. This is the last episode of the new material in our course. We have one more episode after this that will be a review for the final exam. And let's look at our list of objectives here for Episode 38. First of all, we want to look at terminology for probability, and then we'll look at various applications of probability. We'll look at a formal definition for probability, and we'll introduce several probability formulas, and then we'll look again at some more applications. Okay, if we go to the very first to the next graphic, we'll introduce some fundamental terminology here for probability. First of all, we want to define an experiment, and an experiment is an activity that has observable outcomes. So in other words, rolling dice, drawing a card from a deck of cards, using a number out of a box, something like that, those would be the observable outcomes that we could have. Flipping a coin, I see Jeff back here signaling flipping a coin, of course. As a matter of fact, that's what our first example is going to be about. Now if you take the set of all outcomes, that's referred to as the sample space, the sample space S, and an event is any subset of the sample space. So for example, if you flip a coin, the sample space would be a head or a tail, and an event might be just the head. That is, could you flip a head with a coin? Okay, now if we have two different events, E1 and E2, those are said to be independent if they have nothing in common. So for example, if you roll a die, and if you flip a coin, those are independent events because rolling the die has no effect on the outcome of flipping a coin. And finally, the way that we calculate the probability is if we have equally likely outcomes, then the probability of an event, we'll call that P of E, is the ratio of the number of elements in the event divided by the number of elements in the sample space. So I've abbreviated that as N of E divided by N of S. Okay, let's go to the next graphic and we'll take a closer look at these terms in a specific experiment. Okay, suppose we have an experiment in which we want to flip a coin twice. Well the sample space, which is the set of all outcomes, would be that we could get two heads, H, H, or we could get a head and a tail, that is H, T, or we could get a tail and then a head, T, H, or two tails. Now you notice that we're counting H, T, and T, H separately because if they're two coins, the first coin could be heads and the second coin tail, or the first coin could be tail and the second coin head. So we count those twice, we don't count that just as one outcome. Now an event is a subset of the sample space and let's describe our event in words by saying we're going to flip a coin twice and we want to flip at least one heads. Well if I look at the members of the sample space, there are only three members in which we get at least one heads. That would be H, H, H, T, and T, H. Two heads, a head and a tail, or a tail and a head. And the probability is going to be three over four. Now you notice I call this the theoretical probability P of E is equal to three over four. This is the number of elements in the event. There are three elements in the event divided by four elements in the sample space. Now I call this the theoretical probability and the reason is that if you actually flip a coin four times, there's no guarantee that you'll, or rather if you flip a coin twice, and if you do this four times in a row, there's no guarantee that you'll ever get a head in all of those flips. But theoretically in the long run, that is if you did this for thousands or hundreds of thousands of times, that three out of four times that you flip a coin twice, you would expect to see at least one head. Now compare that with what I call the experimental probability right here. What is the experimental probability of flipping at least one head based on the following observations? And that is, suppose that we get two heads, a head and a tail. Let's go back to that graphic so we can see that. A head and a head, a tail and a tail, a tail and a head, and a head and a tail. So in other words, if we actually flipped a coin twice and we did this six times, what if these were the observations that we made? Then based on this experiment, the probability of us getting at least one head would be, let's see, one, two, three, four, four out of six. So if we go to the green screen, I would say that the experimental probability would be, let's see, out of six observations, four of those did I get at least one head and that turns out to be two thirds. Whereas we said the theoretical probability was three out of four. And so there's no reason to expect that based on that experiment of only six times that we flipped a coin twice. There's no reason to expect that we would see exactly three fourths. As a matter of fact, if we flipped a coin twice six times, there's no way I could make that ratio be three over four if I have an integer over an integer. But this is what I see based on the experiment and this is what I theoretically would find overall. Now, for example, if you were a bowler and someone asked you what's the probability that on the next roll of the ball that you'll roll a strike, well, the way you would estimate that probability would be based upon your previous experiment and that would be something like looking at these six outcomes here. And there's no way that you could theoretically predict what is the probability on your next roll that you would get a strike except to be based on a sample of data like we just had there in the experimental probability. Okay, in this example, we ask what is the probability of flipping a penny, a nickel and a dime once each and getting two heads and one tail? Well, let's see. First of all, let's try working this out by listing the individual cases and then we'll see if there's a shortcut for computing this. I'm thinking that to begin with, if we have a penny, a nickel and a dime, let's say the penny, the nickel and the dime. Okay, so I guess we could start off by getting heads, heads, and heads. Or we could get heads, heads, and then tails on the dime. Now, let's see, that's one tail. Suppose I move that tail through this list. Suppose we have heads, tails, heads. So now I still have one tails, but it's for the nickel. And then I have tails, heads, heads. So we've covered the case where we have all heads and we've covered the three cases where we have one tail. Now, the next, I guess to continue this list, what if we had two tails? And if we had two tails, that would mean one head. So suppose we were to put tails, tails, and one head on the end. And then I move that head through, tails, heads, tails. And then I move it further through, heads, tails, tails. So now these last three, you notice we have two tails or equivalently one heads. So let's see, we have three heads. Then we did these three where we had two heads. Then we did these three where we had one head. And then finally, what if we had no heads at all? That would be all tails, tails, tails, and tails. So I think this covers all the possible outcomes that we could have when we flip a penny, a nickel, and a dime. So it looks like there's a total of eight outcomes. And I would say these are all equally likely. I don't see that any one of those would be any more likely than any other. So it looks like our total number of outcomes is eight. So that would be the number of elements in the sample space appears to be eight. Now the event was that we were going to get two heads and one tails. And I think those are listed right here in this group, two heads and one tails. So the number of elements in the event is equal to three. And therefore the probability for this event to occur would be the number of elements in the event divided by the number of elements in the sample space. And that would be three over eight. So what we've determined here is that theoretically the probability of flipping a penny, a nickel, and a dime in getting two heads and a tail is three over eight. And to interpret this in another way, if I were to flip, if I were to do this eight times, flip a penny and a nickel and a dime, we would generally expect that three of those eight times we would have two heads and one tail. Okay, what would be the probability of flipping all heads in this case? One out of eight? One out of eight in this case, right. Now, you know, I might abbreviate that one by just calling it H, H, H. And you notice that for this event, three heads, there's only one way that can happen. That would be on this first line. So that would be one over and once again in the sample space there were eight outcomes, so one out of eight in that case. Exactly. Okay, now there is some abbreviation for some of the notation that I'm using and it comes from some information you've seen previously in other courses on unions and intersections. Let's go to this next graphic and we can just review what that information is about. You may recall that, let's see if you can show my hand on this screen. Okay, you may recall that if you put this symbol, the union symbol between two sets A and B, that the union of A and B, the union of A and B is the set of all elements X such that X belongs to A or X belongs to B. And so if this circle represents A and this circle represents B and they possibly overlap right here in the middle, then if I were to shade in A union B, I would shade in the whole thing. That's why this entire, this entire graph here is shaded in as red because it represents the union. And I think what's significant here is the word, the symbol, the symbol and the word for union go with the word or. X is in A or X is in B. If X is in A or if X is in B, then X is in this shaded region A union B. Now on the other hand, if I invert that symbol, this refers to the intersection of the two sets. And the intersection of A and B is the set of all X such that X belongs to A and X belongs to B. So X is in the, is in the intersection such as the intersection of two streets. And if this is set A and if this is set B, then the intersection is what they have in common. And I want to point out that the symbol for intersection is associated with the word and. So if the U opens up, if the symbol opens up, the union will use an R. And if it opens down, we'll use an N. Now you'll see this in some of the problems that come up in just a moment. Okay, let's go to the next graphic and consider a new problem. And I'll use some of the union and intersection notation as I explain this. It says a ball is drawn at random from a Grecian urn containing, I don't know why it has to be Grecian, but it just seemed appropriate, containing six balls. Now there's one red ball, two green balls, and three yellow balls. So if the ball selected is green, then we want to know what is the probability, what is the probability that the ball selected will be green? I guess I should say it, the probability it will be green. Now I think we can assume that if we're drawing these balls at random, what that means is that the outcomes of choosing one of the red balls, or the red ball, or one of the two green balls, or one of the three yellow balls, they're all equally likely to come up. Now if you look at the green balls in total, it's more, it's more likely that you'll pick a green ball than a red one. But if you look at the green balls separately, they, we could choose one of those just as likely as we would choose the red ball. So we want to know what is the probability that when you do this at random you'll choose a green ball? Well let's see, it looks like there are six possible outcomes. So we'll put a six in the denominator. And there's only one outcome in which a ball is, no there are two outcomes where the ball is green. So that'll be two over six, or that reduces to be one third. Now what that means is that in the long run, if I did this experiment many times, selecting a ball out of the urn and then replacing it, I would expect that one out of every three times I would pick a green ball, more or less. And now we may actually perform this experiment and we may choose a ball three times out of the six and then replace it, mix them up and choose another ball again. And we may never pick a green ball, or we may always pick a green ball, but in the long run we think that one out of three times we should pick a green ball. Okay now, suppose we want to know the probability that the ball selected is not green. And the way I'll abbreviate this I'll put a little negative sign in front of the green, but I've, I've made it sort of wiggly here so it doesn't look like a subtraction sign or a true negative sign. But if you see a mark such as this, like in our textbook, that means that it's not going to be green. You know there's another way to represent this and that would be to put a bar over it and in some textbooks you'll see, you'll see a bar drawn above, drawn above that. And if you see a bar over the the term that means that it's not green rather than that it is green. And finally in some textbooks you see a little prime, a little apostrophe mark behind the G or after the G, and that means that rather than being green it's not green. That's sort of the the negation of the G. Well let's see, now what's the probability that the ball selected is not green? Well there are six possible outcomes and it looks like there are four ways I can choose a ball that's not green. There's one red and there's three yellow. So there are four ways I can choose a ball that's not green and they're all equally likely so I can take the ratio of these two and I get two-thirds. By the way what's the sum of the two answers that I've gotten here? Well one-third and two-thirds adds up to be one. So what that tells me, what that suggests is another way to determine that the ball is not green is that I could just take one minus the probability that it is green. And so one minus a third would have been two-thirds. So we could have worked the problem in that form as well. Okay finally there's a third question here and this says suppose we want to know the probability that the ball chosen is both green and yellow. So the probability that it's both green and yellow. Now you notice the way this has been abbreviated here is I'm putting in an intersection sign G for green, Y for yellow and the intersection sign is associated with the word and. So G intersect Y is represents the possibility that the ball could be green and yellow. Now once again there are six possible outcomes but in no case can I pick one ball that's both green and yellow so there are zero possibilities there and therefore here's an experiment here's a here's a here's an event in which the probability is zero and we'll interpret that to mean it's just impossible. Okay so some of the ideas that we brought up here are summarized in four laws on probability that we bring up in the next graphic. Okay law number one says that the probability that event E does not occur is one minus the probability that event E does occur or if you put these two together if you add P of E and P of negative E these two add up to be one. Now the largest that a probability can be is one and that means that it's a sure thing and if I subtract the probability of E then I'll have the probability that E does not occur. Law number two the probability of E is zero means what? It's impossible. Means it's impossible yeah if we're talking about an experiment which there are finitely many outcomes if the probability is zero then it's impossible such as choosing a ball that's both green and yellow and if the probability of E is one we just mentioned this a moment ago what does that mean? Sure thing. Means it's a sure thing yeah so for example uh in the six balls in the urn if we ask what's the probability that when you choose a ball it will be red green or yellow well that's a sure thing because that's the only kind of balls that were in the urn and finally for any event E the probability of E has to be greater than or equal to zero or less than or equal to one so we know that if we ever come up with a probability that's negative or if we come up with a probability that's more than one we know that there's something wrong because all of these answers are going to be between zero and one. Okay let's go to the next graphic and we'll look at another probability problem. Suppose we draw a card from an ordinary deck of 52 cards and we ask several questions about what may happen. Now for those of you who aren't familiar with the deck of cards let me just explain uh what a deck of cards looks like and uh what some of the terms mean now here I have um here I have a deck of uh an ordinary deck of cards and that would be 52 cards and there are four suits so for example uh that's the four of what what suit is that? Clubs okay and this is the two of spades okay and here is a five of diamonds yes and now here is another two that's the two of hearts okay these are the four suits and we have 13 cards in each suits in in each of the suits for example uh let's take um the suit of clubs for example the smallest club is the uh two of clubs let me see if I can find the two of clubs here and uh it's the very last card wouldn't you know it okay that's usually considered the smallest card and then the biggest card would be the ace of clubs so I'll put the ace over here and the cards go two three four five six seven eight nine ten and then we come to the three face cards there's uh there's the jack there's the the queen and there's the king those are called the three face cards now you know there are some games in which the ace is considered a one and you put it over here it's the smallest value card but I think in most games ace is considered a high card rather than a low card so uh we have 13 cards in each suit and these three cards in the suit are referred to as the face cards so let's see let me just kind of summarize some of this information here uh we said that there were four suits there were four suits there are the the clubs there are the diamonds there are the hearts and there are the spades and on the club you see a symbol sort of like this to represent the club on the diamond the diamonds are red by the way clubs are black for the hearts you see a shape like this and for the spade it's sort of an inverted heart but it's also black so this one is black the clubs are black the spades are black the diamonds are red and the hearts are red now I point this out because in some of your homework problems they'll ask something about uh perhaps the probability of drawing a red card or a red face card something like that okay now within each suit we have the arrangement of cards that goes like this there's the two the three the four five six seven eight nine ten and then we come to the three face cards the jack the queen the king and then there's the ace isn't considered a face card it doesn't have a face on it and uh you know this I think the the reasoning behind the jack the queen and the king comes from medieval uh uh courtly life there was the king the queen and then the jack was the heir apparent to the throne so those were the those were the three face cards uh originally as they were designated okay now let's go back to our problem and answer some questions about a deck of cards suppose we draw a card it ran in from an ordinary deck of cards and you see the ordinary deck of 52 and you see 52 is four times 13 we have four suits 13 cards in each suit what is the probability that the heart that the card is a heart well let's see um would you say that the that the choices of picking a card at random uh you think it's equally likely that you'd pick a heart or a spade or a club yes yeah so these are all equally likely uh outcomes and therefore if our event is to pick a heart I should put the number of hearts divided by the number of cards in the deck now how many hearts are there 13 13 and how many cards are there 52 so we'll put 13 over 52 let's come to the green board so I can write that down so uh the probability that we could draw a heart is 13 hearts to choose from 52 cards in the deck now that reduces to be what one fourth one over four now you notice in this case what I have is one suit divided by four suits and the the suits come up equally likely there's one suit out of four suits are there 13 hearts out of 52 cards uh I think when you're working the problem this is probably the way you would interpret it but then you'd want to reduce your answer to be one fourth you wouldn't want to leave it as 13 over 52 when you could get a such a nice answer as one fourth okay let's go back to the graphic and look at the next question now we want to know in part b uh what is the probability that the card is a face card that is a jack a queen or a king well let's see the probability that we get a face card I'll just write the word face in there to abbreviate that event uh is going to be something over 52 now let's see there are how many jacks four jacks there are four queens there are four kings so how many face cards are there 12 okay so let's let's go back to the green screen okay so that'll be 12 over 52 and that reduces also now what will divide 12 and 52 four uh four will and that gives me three over 13 now let me ask you this if you had a choice of from a from a shuffle deck of cards picking your card what do you think would be more likely that you would pick a heart or that you would pick a face card a heart a heart yeah Stephen why do you say a heart because um one fourth or 13 over 52 is greater than 3 13s or 12 over 52 exactly right there are 13 ways out of 52 a heart can be drawn only 12 ways a face card can be done but you know those are fairly close together aren't they they're almost the same but in the long run you would probably pick a heart a little bit more often than you would pick a face card now of course if you sat at home and you you did this three times and you picked a card at random put it back into the deck shuffled it pick a card at random put it back in the deck it could be that in the in the short haul as they say you might pick a face card more often than a heart but in the long run the hearts are going to win out now you know this is exactly how gambling games work say if you go to Las Vegas or Atlantic City or to a casino it's possible that in the short term you could win at the game if you're playing blackjack or craps or playing a slot machine but in the long run the shall we say the law of averages takes over and in the long run you generally don't win at those games but in the short term you could you could have a lucky streak or you could be really unlucky and never win a cent but but in the long run I think the casino figures that they're they're going to they're going to win some money from you okay let's go back to the graphic look at part c suppose that the card you choose is a heart and a face card at the same time what's the probability that that would happen now i'm going to abbreviate this on the green screen as h intersect f of course h is for heart f is for face card what does the intersection sign mean and means the word and yes if you think back to our venn diagrams and our discussion of unions and intersections a moment ago so this means heart and face card now if you prefer you could write this as h and f but i'm pointing this out because in the textbook you're going to see this symbol on some problems okay now a heart and a face card well let's see there are 52 choices still now in how many ways can i pick a heart and a face card at the same time three three ways yeah now someone may say well Dennis you can only pick one card they can't both happen sure they can what if you pick the jack of hearts the queen of hearts or the king of hearts you could pick a heart and a face card at the same time but there are only three ways that can happen and three and 52 don't reduce i'll have to leave it like that so you see this is considerably less likely than just picking a heart or just picking a face card that both of those would happen okay and one more question on the graphic okay now if you look at line d it says the card this time we want to know what is the probability the card is either a heart or a face card well let's see the way i'm going to abbreviate that is to say p of h union f because you remember union is associated with the word or intersection is associated with the word and and so in this case let's see there are 52 choices now in how many ways can i choose a heart well let's see there are 13 ways i can choose a heart and in how many ways can i choose a face card 12 12 so if i add those two together a person might say well then Dennis there must be 25 ways you can pick a heart or a face card but you know we've actually counted some things twice here because there are some hearts that are also face cards so rather than counting them twice we better subtract off the duplications how many times do i pick a card that's a heart and a face card at the same time there's three three times okay so what i have is 13 plus 12 and then i remove the duplicates and this gives me 22 over 52 which is 11 over 26 now in this last example let me just show you how this leads to a law of probability and it says that if i write this is 13 over 52 that's the probability of choosing a heart and if i add on 12 over 52 that's the probability that i choose a face card then i have to subtract off the probability that i choose a heart and a face card h intersect f yeah let's see 13 over 52 that's the probability of picking a heart 12 over 52 is the probability of picking a face card but i have to remove the duplication which means i'm removing the that is the probability of choosing a heart and a face card so the probability of the h r f is the probability of h plus the probability of f minus the probability of h and f we're going to see this now in the next graphic when we state two more laws of probability in the first case it says law number five says that if events will call them e and f are independent events remember that means they have nothing in common such as a rolling a die and flipping a coin so that those are independent the outcome of one has no effect on the other then the probability of p and f happening is simply the probability of e times the probability of f that's a that's a product there so we're multiplying two probabilities together now the law that i've just mentioned is the probability that e r f happens is the probability of e plus the probability of f minus the probability of e and f happening at the same time and the reason we subtract that off is because this gets rid of any duplication now suppose e and f have no duplication suppose these are independent events then what i would be doing here is well excuse me not independent but suppose they don't overlap then the probability of their intersection is zero and in that case it's just the probability of e plus the probability of f okay i think it's time to look at something other than cards let's go to another experiment and hey why don't we go to rolling dice there's another interesting problem so the question is what is the probability of rolling a sum of seven with a pair of dice and then i ask some other questions after that but let's consider the first one at seven now you know on a die on a on a pair of dice one one die has how many faces six there are six faces yeah so like if this represents one die if there's a little dot there of course that means one if i put two dots over here that means two and if i put three dots diagonally across there that means three so in other words on the back sides would see four five and six uh i don't know if you're aware of this but the numbers on opposite faces of a die total seven so the number on the back side directly behind the one should be a six what number would you guess what number of dots do you think would be on the bottom of the die four four yeah because the top and the bottom numbers always had to be seven and on the left and the right side they had to be seven so there must be a five over on this side so there are six outcomes on the die uh and they're all equally likely we're assuming this is an unloaded die so when you roll it it's equally likely that you get one through one through six now the question is what is the probability of rolling uh a pair of dice and getting a sum of seven well now with a pair of dice the outcomes of the total are not equally likely let's see what they could be i could roll uh a whoops let's see too high there a one and a one or a one and a two that'd be a total of three a one and a three one and four one five one and six by the way when you roll a one and a one what do they call that in dice snake eyes yeah it's supposed to look like two bdi is looking looking at you i guess out of a a hole in the ground or a tunnel now you could also roll a two and a one a two and a two two and a three two and a four two five and two six now you notice i'm counting two one differently from one two because we have two dice and let's say one's green and one's red this one might say the first the green one comes up one the red one two this one says the green one comes up to the red one one so those are considered different outcomes now we also have three one three two three three three four three five three six this is where the first die comes up with a three on it obviously and then four one four two four three four four four five four six so when you come up with four and six that means you've rolled a total of 10 then there's five one five two etc i'll just write the other possibilities down here and then there's six one six two all the way down to now what's the number what's the name we give to a pair of sixes what do they call that 12 well they do they very good they do call that 12 sometimes and there is another name kind of like a pair of ones is called snake eyes a pair of sixes sometimes called box cars i guess because the sixes the way the dots are lined up they look like the top and bottom of a of two box cars on a on a drain okay now how many outcomes are there when you roll a pair of dice 36 how many distinct outcomes here we have 36 outcomes and i've listed these separately so that these will all be equally likely there's no more likelihood that you'd roll a pair of fives from rolling a four and a six although if you want to roll a 10 there there are three ways to roll a 10 there's a four six a five five and there's a six four now in how many ways can you roll a total of seven with a pair of dice well let's see here's a seven right here a one into six but there's also a two and a five and there's also a three and a four and a four and a three and a five and a two and a six and a one how many ways can i roll a seven with a pair of dice six six so if i want to compute the possibility the probability that i roll a sum of seven not rolling an individual die to get seven that's going to be six out of 36 or one over six not very likely but you know what seven is the most likely thing you can roll with a pair of dice because take sixes for example and how many ways can you roll a six there's one and here's one all together how many ways can you roll a six five only five ways five out of 36 so you notice that the longest diagonal has six in it there are five ways to roll a six there are four ways to roll a five there are three ways to roll a four there are two ways to roll a three and of course there's only one way to roll a total of two so seven is the most likely is the most likely roll okay now there's another question though up here and it says what is more likely rolling a sum of either seven or eleven or rolling a sum of either two three or twelve now some people might say oh i would pick the two three or twelve because you have more possibilities you have two three or twelve you have three different possibilities over here there are only two possibilities but of course that wouldn't be that wouldn't be very sound reasoning what is the probability of rolling a seven or eleven well let's see how many ways can i roll a seven six ways six ways how many ways can i roll an eleven two two ways that's a total of eight ways by the way those don't overlap do the you can't roll a seven and an eleven at the same time those would be independent events so there are there are what six plus two out of 36 ways to roll a seven or eleven that's going to be eight out of 36 okay now let's see well this still fit on the screen i think it will what's the probability of rolling a two a three or a twelve well once again is going to be something over 36 how many ways can you roll a two one way one way i'll put a box around that one how many ways can you roll a three two ways two ways yeah and how many ways can you roll a 12 one only one way over here so there are only four ways i can roll a two three or twelve so that's four out of 36 okay so let's see four out of 36 now let's just reduce these further the eight out of 36 reduces to be two over nine the four out of 36 reduces to be one over nine so it looks like it's twice as likely that you roll a seven or eleven as opposed to a two three or twelve now do you know where this question comes from actually you know why i'm asking about these particular numbers if you go to vegas right and if you roll dice if you shoot craps as they say on the first roll of the dice if you roll a seven or eleven you win and if you roll a two three or twelve you lose you say well hey i love that game i've got twice as good a chance of winning as losing but you notice what is the sum of two nights and one night it's three nights or one third so what's going to happen the other two thirds of a time or what's going to happen the other two thirds of all of you the times you roll well you don't actually lose but what happens is you wrote you don't roll a seven or eleven or a two three or twelve and what happens in the game of craps is whatever you roll other than seven eleven two three or twelve that's called your point and after that you get the dice back and you have to roll again and you have to keep rolling until you roll your point again so for example if your point is a five you keep rolling the dice until you roll a five or you roll a seven and now seven is a killer and if you roll a seven you lose you don't win so seven's a winner on the first roll and eleven's a winner two three and twelve are losers but most of the time what happens is you don't roll any of these numbers you roll some other number and that's called your point and then you get the dice back and you roll some more until you roll your point again or you roll a seven and now it's more likely that you roll a seven than anything else so on the second and subsequent rolls the probability is really poor now that you're going to win after that so i think you're enticed into shooting craps thinking that on the first roll you have twice as good a chance of winning is losing but that's only on the first roll on the rolls after that the probability is significantly poorer that you're going to win okay so much for that let's go to the next graphic here's sort of a geometrical look at probability what is the probability that we hit a bull's eye that's that that inner gray circle when we randomly hit the target shown to the right now what i've shown you there is the radius the inner circle has radius one and the outer circle has radius three and we want to know the probability that assuming you hit the target at random that you will land in the inner circle now of course if you're throwing darts at a dart board things aren't random number one you don't always hit the dart board at least i don't and number two if you're guiding the dart then hopefully you have a little bit of influence on where the dart lands but if you were to just say toss a dart out there at random and it hits the board what's the probability that you'd hit the target well you know in this case you could say well now Dennis there are actually infinitely many different places you could land inside the circle there's infinitely many different points and in the big circle there are infinitely many points that where you could land so how can you take a ratio of the number of points on the inside versus the number of points possible in the entire circle well i agree in this case you can't actually do this by counting the number of outcomes but instead we do it based on area so in a problem like this we would say the probability of hitting a bull's eye will be determined by the area of the bull's eye divided by well what would you guess the area of the whole the area of the entire target right exactly so you see what i'm doing is i'm reducing the problem to area rather than trying to count points because there's no way you can count up infinitely many points across there well let's see this means we have to know a formula for the area of a circle let's just put that over here let's see the area of a circle is let me move that in a little bit the area of a circle is pi r squared right pi r squared pi times the radius squared so the area for the target would be pi times three squared because that's its radius and the area for the bull's eye is pi times one squared and if i take this ratio the pi's cancel off and what i have then is one over nine so the probability that if you just randomly throw a dart at the board and it hits the board only one out of nine times where you get a bull's eye you know another way of thinking about this is what if this were what if this were a tile on the floor and what if you were just tossing pennies or stones or whatever and they just bounce into the target at random now that's that's the important thing is they arrive at random then the probability that the coin or the stone would ally in the inner circle is one in nine the probability is one out of nine that it would land there uh steven kind of like if you set the tiger if you set the target out and raindrops fell on it exactly yeah that'd be a good example right so uh if if you were if it was getting ready to rain and a raindrop landed on the target uh what's the probability the first drop would land in the bull's eye or in the inner circle that would be one out of nine exactly okay well you know as you can see i think from the variety of these examples that the probability is a very broad subject and um we're just looking at the rudiments of probability and we're trying to do so by looking at a variety of problems that i think are interesting um and uh uh that we are considering here let's go to the next problem what is the probability that two of the members of this class uh that's including me have the same birthday the probability of the same birthday well now this this is uh getting a bit more complicated um i think the way we should handle this is decide what's the probability that no two of us have the same birthday and then if we look at the complement of that that'll be the probability that two of us do have the same birthday okay so to begin with i'm going to ask this question uh what is the probability that we have different birthdays now there are five students in the room plus me there's six of us all all together and we want to know the probability that we have different uh birthdays well let's see now uh let's begin with me i have my my birthday is august the 18th now what's the probability that your birthday is different than mine well there are 365 days a year i'm i'm going to leave out leap year let's let's avoid that issue um february 29th so what is the probability that your birthday out of 365 days is different than mine seems like it'd be 364 out of 365 okay so um i'm going to write down 364 out of 365 so that's the probability that if i ask david here david what's your birthday is it different than mine of course it's very likely his birthday is different than mine by the way david what is your birthday 26th november ah he's not even close to my birthday august the 18th well of course we sort of expected it would be different didn't we it's 364 over 365 now uh what's the probability that matt's birthday will be different from davids or mine well let's see that'll be something over 365 how many choices do we have for matt's birthday if he's going to be different from us 363 363 okay 363 by the way matt what is your birthday october 14 october the 14th okay so so far we're we're everything's falling in line okay now let's go to uh steven steven let's see the probability that your birthday is different from mine or davids or matt's is some number over 365 what do you think that should be on top 362 362 sure because every every time we include a new birthday we have to reduce the pool by one by the way what is your birthday steven december 31st december 31st hey a new year's eve kid okay congratulations uh okay now if we go to jeff back there jeff i think you're going to be 361 over 365 and of course we have to know what your birthday is march 29 march 29th okay so um of course it's very likely that all of us have different birthdays so you notice this product is still very big and we come to susan susan what do you think i should put in the numerator of this next fraction 360 360 okay and by the way what is your birthday march 26 march i don't think we did have another march birthday here i think we didn't mean but not the 26 okay now if i multiply these numbers together this gives me the probability that all of our birthdays are different now what if i take one minus that this would be the probability that our birthdays are not all different and that would be the probability i'll just call it p that at least two of us have the same birthdays now if you multiply out all these fractions and subtract it from one you have the probability that two of us share the same birthday now that number would be of course very small the likelihood of the six of us having the same any two of us having the same birthday would be very small uh steven's multiplying that out i think right now steven did you get an answer now there are pretty big numbers in fractional form do you want me to put it in oh can you convert it to a decimal yeah sure okay tell me we'll come back to steven in just a moment what i want to point out is this what if we put one more person in the room for example we have tony the camera operator over here tony would put one more fraction on here wouldn't we and it'd be 359 over 365 and what do you think that's going to do to the product if i put one more fraction in there will it make this product larger or smaller smaller because we're putting in a fraction that's still smaller than one so what will it do to the difference if i put an extra fraction in there it makes this fraction smaller but when i take one minus that it'll make it bigger because i'm taking away less it'll make it bigger now here's the surprising result if there had been 23 people in the room and not just six of us well tony when we went to tony there were seven if there were 23 people or more this expression this difference will be more than 0.5 which means that if you have 23 people or more in the room there's more than a 50-50 chance of two of you having the same birthday so if you're at a party this weekend and there are you look around and there are hey 23 people in the room or more you might say someone hey i'll bet you two of us have the same birthday and most people say no i don't think that's going to happen there's only 23 of us but in the long run if you do this repeatedly more than half the time two of you will have the same birthday and of course there could be lots of duplications you could even have three people with the same birthday but if there are 23 people or more then the likelihood is better the probability is more than one half of two of you having the same birthday now how would you actually prove that after this court after this episode is over how would you actually prove this to yourself that 23 is all you need you'd take one minus and then how many of these fractions would you write down 23 well not 23 because you see there were six people but i only listed five fractions that's because the very first person i chose me i just i just had my birthday at random and we didn't introduce the first fraction till we went to the second person so if you write down 22 of these consecutive fractions subtract it from one you'll find out that this number will end up being a little bit more than a half and of course as you put more people in the room you put in more fractions which makes this product smaller which makes this difference bigger now you know most people i think if you just estimate random how many people do you need in the room for two of us to have the same birthday that say well let's see half of 365 is what 182 183 but actually if you have 180 some odd people in the room there's a fantastically high chance that two of you have the same birthday but 23 is the is the halfway point okay so try that at your next party let's go to the next graphic um let's see in this case it says what is the probability that you correctly name three cards in order is this if they are turned over from a well shuffled deck of cards hey i think we got to try this okay let me let me make sure this is well shuffled okay no peeking here let's let's come back to the green board and i'm going to turn over three cards one at a time i think the students can see this on the monitor if you go to the green screen uh let's see uh david i'd like for you to name the first card i'm going to turn over let's see if he's psychic four clubs seven of clubs hey it wasn't too far away okay well let's try one another card what do you think the next now see this time you could say four clubs again you just don't want to say seven of clubs um four hearts four hearts oh ten of diamonds well you know we really didn't expect this going to happen let's try somebody else matt what do you think the next card's going to be ace of spades ace of spades oh it's the do so hard so he was a long way away well of course it wasn't very likely that david and matt were going to be able to name three cards in order as they were turned over so how can we figure out what is the probability that this could have happened and we expect this would be a very small number okay so uh i'm just going to call this p for probability and on top i want to list the number of ways to name these these cards correctly but i'll try to economize on words here the number of ways that we can name these cards correctly and then we'll divide by the number of ways um three cards can be shown now you notice this is in order so let me put in order below that now that becomes very important in the in the solution of the problem because we have to be able to name the cards in order so when i work this out you know there's only one way that you can name the cards correctly as they come over there's you have to name the first card correctly the second and the third but on the bottom uh would you say this is a permutation or a combination problem permutation permutation it's going to be permutations right so uh this is going to be p of 52 three because there are 52 objects to choose from and we're choosing three and the order makes a difference now if i had if i had put combinations then this would mean just turn three cards over together and see if somebody can name them but in no particular order so we're doing a permutation problem this is one over now let's see we have a formula for permutations that's 52 factorial divided by the difference factorial that's 49 factorial and that's going to be 49 factorial over 52 factorial when i invert and multiply and that will be one over now let's see when i cancel what numbers will be left in the denominator 52 times 51 times 50 52 times 51 times 50 now you know 50 times 50 times 50 is about 125 000 50 cubed is 125 000 this is going to be one over some number a little bit larger than 125 000 and uh steven you're working this on your calculator what did you get one over 132 600 one over 32 600 now what this means is if we tried this um if we tried this 132 600 times we would expect that only once would somebody be able to name the cards correctly in order okay i think we just have to try this don't you let's just let me shuffle these and let's just try this so susan if you're going to name a card what would you name four diamonds four diamonds oh queen of diamonds we got the suit right okay and uh jeff seven of clubs okay now what if i what if instead of taking it off the top what if i pull it out of the middle do you think that makes a difference yeah okay why do you think it would make a difference is is it any more likely that jeff will be able to name the top card as opposed to the middle card not really yeah so actually it doesn't matter someone says oh can i pick the one in the middle well uh as long as they're doing this randomly it wouldn't matter but i'll turn over the top card by the way you said what seven of clubs seven of clubs ten of diamonds hey we're getting closer to your diamond susan and uh okay steven uh show you name a card three of clubs three of clubs deuces spades oh sorry no no cigar um and uh so we didn't get any of them right and of course that is the most likely outcome that we wouldn't be able to name any of those cards correctly okay uh let's go back to that same graphic let's go back to the same graphic about probability and there was a second question on here it says what is the probability that the card if the cards are not necessarily named in order well i think the answer there is instead of taking permutations you'd take combinations in the denominator so it'd be one over c 52 3 instead of one over p 52 3 and it ends up being slightly larger but don't hold your breath it's not a whole lot larger okay let's go to the next problem the last problem and i think we only have time to briefly talk about this because we're almost out of time so rather than working it out let's just discuss the answer it says if you randomly guess answers on a 10 question true false exam and passing a 70 percent what is the probability that you pass the exam well you know would have to put a ratio of two numbers on the bottom i'd put the number of ways that i could answer a 10 question true false test and we've actually talked about this problem before do you remember what that would be it'd be two to the tenth power which is one thousand twenty four yes a thousand twenty four right well let's see that marker is not behaving very well so let's try a different one here okay now um in how could you pass the exam well you have to get 70 percent and this is a 10 question exam so you have to get seven out of ten correct or eight out of ten correct or nine out of ten correct or ten out of ten correct okay now in how many ways can you can you can you randomly pick the right answers to seven out of ten questions well i'll give you a hint this is either a permutation or a combination problem do you think you'd want to choose permutations or combinations combinations combinations because the order doesn't make a difference so what we would do would be to pick the number of combinations combinations on ten things taken seven at a time plus the number of ways you could pick in ten things correctly pick eight at a time or pick um ten things nine at a time or the combination on ten things ten at a time of course there's only one way to do to do that that would be them all right so you'd have to figure how many ways can you get seven right how many ways can you get eight right nine right or ten right where the order doesn't make a difference you're just picking seven things out of ten not in any particular order okay well if you add up those numbers and divide it by two to the tenth you'll have the probability of passing the exam i'll see you next time for the review for the final exam