Probability and Binomial Distribution Lecture

okay welcome back hopefully the exam wasn't too difficult so today we're going to be uh completing chapter 11 and maybe going into chapter 12 we'll see um just as a reminder uh this coming Thursday is a holiday 4th of July so there will be no classes that's also mentioned in the uh calendar so no lecture for for that day whatever we cover for today tomorrow Wednesday that is all that will be covered for the next exam so let's get started uh section 11.4 so continue with the probability uh binomial probability construct a simple binomial probability distribution apply the binomial probability uh hold on uh construct a simple yeah apply the binomial probability formula for an experiment involving berui trials now beri trials is just fancier name for binomial uh trials named after I can't remember a French or Italian mathematician back in the Renaissance days who studied this uh kind of structure so the B Pro so here's an example the spinner below is spun twice and we are interested in the number of times A2 is obtained so put this in other words think of two as a successful outcome and the one and three as failed outcome so the sample space so coming back to earlier in the chapter the sample space is going to look like this the ACT whatever the number and the first position is this is the results first Spin and the second number is the results Second Spin so two something like 2 three would mean oh we got a two on first spin so for for example Spin and three on Second Spin note the two possible outcomes of success and failure is the basis for the binomial probability the buy means two so we're looking at and this these are the two things we're considering a success and failure now when the outcomes of an experiment are divided into just two categories success and the associated probabilities are called binomial most of the time uh there are other elements in play as we uh look deeper into binomial distribution uh repeated trials of the experiment where probability of success and here's one key factor the probability of success remain remains constant so the probability of the outcome of the first spin is the same as a probability for the Second Spin third spin fourth spin etc etc etc if x so here we have an the variable x if x denotes the and here how it is defined the number of twos curring on each pair of spins so at this point with the probability if you're going to introduce a variable you really need to introduce with it what that variable stands for when X is an example of a random variable the idea of a random variable as opposed to just a variable so a variable in general its values are determined can be determined by some controlled process for example the the formula eal mc^2 uh you may have come across that even lightly in your in your life so e stands for energy m stands for Mass says C is the speed of light e and C is are are constants those are a fixed value irrespective of what E and M are E and M are variables but they are variables whose value are controlled and as far as if you're considering a scenario in physics and you want to determine well how much energy would be in this scenario well you have control over the value of M and once you've fixed that value of M you plug it in you multiply it out with the c^2 and you get the equivalent energy or the other way around I have so much energy but you want to determine how much mass does that represent so you have control over the value of the E you have the a linear expression for a line for drawing a line well you can control what the X values or the Y values is so if you plug in an X you can determine the corresponding y if you plug in a y you can determine the corresponding X so it is a variable in general is controlled by its value is controlled by some process either you are providing the values or a computer or a measuring device is providing the values a random variable on the other hand its values are determined by some random process so like uh this spinning this whole spinning thing the values of X is is limited by the result of the spin of the spinning device in this spinning example in s the sample space the numbers of two are zero and four cases they occur once in four cases and twice and one case say what okay so what this is trying to say here's the sample space the results of all possible outcomes of that spin if that Arrow lands on a two it's counted as a success if it doesn't land on a two that is a one or three then that's a failure so here are all the possible outcomes for the two spins the first spin can be one and then the second can be one or two or three the Second Spin can be a two and the first spin can be a one or two or three the first spin could also be a three and its Second Spin is one or two or three so those are all the possibilities now there's three possible outcomes there two spins so 3^2 equals 9 so we should have nine number outcomes so number of bins or number of uh experiments so there be nine in other words three possible outcomes one two three you square it because you have two spins so that brings a total of nine so when we look at back at this n s the number of twos is zero so this is saying twos do not occur in the spin in four cases so no twos occurring and two spins so if you look at this uh here's one case where you have no two uh that's uh this one no two uh there's a two here a two here two here but here there is node two and at the last no two that's where your four cases comes in X is one and four cases so one so here we're saying X is equal to one in four cases how many outcomes here in the sample space have uh the number two occurring only once that's what the X represents the number of twos occurring on each spare so it can since there two spins you can have no twos showing up or one two showing up or two twos showing up so here you only have one two one two one two and one two three oh yeah yeah yeah and one two so there's your four times where one two appears two so this two means x = 2 so meaning the the outcome two on the spin happens twice x = 2 in only one case which is this so once because the table on the next slide includes all the possible values of X and their probabilities it is an example of a probability distribution in this case it is a binomial probability distribution so here's a table marking off all the probabilities so remember X represents the number of twos that show up after a pair of spins since four of those results satisfy x equals zero because it's a 13 or 31 or something like that or one one 13 this is the probability of getting no twos within two spins four of them had a single two involved so four nths and for the pair of twos that only occurred once so that's one nth now please note here is a check so if you take all of these probabilities and add them up so you get 49ths plus 49+ 1 nth oh looky there we all have common denominators so four + 4 + 1 gives you 9 9 over 9 is essentially 1 whatever probability table you're looking at the probabilities all had a some rules of probabilities number one we're all X from sample space p of X has to be greater than or equal to zero cannot be negative next one so P of X1 plus P of X2 plus blah blah blah blah blah plus x to the N whatever n is all those probabilities have to add up to one so for starters these are some of the probabilities I think they were brought up earlier in the chapter so uh here they are again and if you notice the check 4 9th + 4 9th + 1 9th Give You 9 9ths which is one so hipip raw that works so binomial probability formula so there is a generalized formula and the only thing that generalized means is that you can use it in a multitude of scenarios you just have to apply it in the proper way so in general let N lowercase n be the number of repeated trials so with like the spinning game each trial is just a one spin of of the arrow P Little P is the probability of success On Any Given trial so the probability of success remains fixed now we'll cover some examples of this but I just want it to be driven across the probability of success is going to remain fixed for each trial and we have a second variable called Q so p and Q that's going to be uh 1 minus P so if P was 0.4 that means Q is going to be 1 minus 0.4 so 0.6 and the probability of failure for any given trial so and X the number of successes that occur note all trials are independent remember independent is about influence and part of this means P remains fixed throughout all trials so if you do a trial it's not going to influence the results of the following trials in general act successes can be assigned among and repeated trials and n CX so here is your n choose X your combination uh combination here's the uh formula p is a probability of success Q is a probability of failure with p and Q where Q is 1 minus P remains constant throughout all n trials the probability of exactly X successes is given by this formula wh so these are interchange so these formulas are interchangeable this uh n choose X expands to This n factorial expression here so first example uh tossing or finding the probability of a coin toss so find the probability of now in these examples please be careful about the wording find the probability of exactly three heads and five tosses of a fair coin solution so what are we looking for the probability of obtaining exactly three heads uh this is a binomial experiment so how do I know because number one there's a limit lied number of Trials or so and is not it's not Infinity it's not some you know some doesn't go on and on it's a fixed number of Trials so and five PES three heads in five PES so that is your end uh head so the probability of getting ahead well you know there's only two possible outcomes heads or tails uh each one is equally likely and we're looking at heads so the probability of heads going to be one half and that's going to be true for every trial on the first on the second on the third on the fourth on the fifth the probability of getting a heads is one half so this is the probability of success in this scenario success is getting ahead as opposed to a tail you take this result Q = 1 - P = 1 -2 so Q is this this is failure and x equals three so you have five tosses and you want to know the probability of getting a heads or exactly three heads doesn't matter where those heads occur they could her in the first toss second toss fifth it doesn't matter just as long within those five tosses you have three heads which forces the other two to be tails the probability of that is going to be this there's your n there's your X here's your P here's your X here's your Q and this is going to be n- X so 5 - 3 to give you two this is this five choose three that's the same one from chapter 10 so 5 choose three gives you this value uh 12 to the 3 power so that's going to be 12 * 12 * 12 which is 1/8 and then 12 squared so 12 * 12 that gives you this so 10 over 1 * 1 18 * 1 14 uh you can simplify some of this so the 10 on the top eight on the bottom so they have two in common so divide the two into 10 that gives you these with a five a two into the 8 that leaves poent of four and now you can multiply out 5 * 1 * 1 * 1 gives you 5 1 * 4 * 4 gives you 16 so here is your 516 any questions find the problem rolling a dice uh find the probability of finding exactly two threes and six rolls of a faradise so before you jump into it make sure you you understand it properly so the mention of the three is related to the dice the outcome of a dice so I want exactly two outcomes of a three when I roll a die six times this is counted as a success remember this is rolling of dice so the sample space is going to be either one or two or three or four or five or six so the three is counted as a success which means means if you get any of the other numbers those are going to be failures the one the two the four the five six how do I know because find the probability of obtaining exactly two threes so in the context of this question three is counted as a success any other number is counted as a failure now we could change the expression to say well find the probabilities of paining exactly two sixes in six rolles of a diet in that case the six would be a sign of success the one two 3 four five would all be signs of failures that is not getting a six so we have a total of six rolls so this six so well let's break it down solution so it's binomial experiment with n equals 6 so six rows the probability of success is going to be just one out of six so one six because only one number out of the six numbers one outcome out of the six possible outcomes is counted as a success so that's one out of six which means the remaining five possible outcomes are all failures so the probability failure is going to be five out of six and x = 2 so exactly two so x = 2 so think of it this way so you have in the context of six roles you're looking for the likelihood of two of those roles being a three so the six will be your n your total number of Trials your two out possible outcomes will be your X so X will always be less than or equal to and greater than or equal to zero so the probability of X being two is going to be 6 choose 2 so there's your n there's your X six there's your P this will be your x uh 56 there's your q and the four that's going to be your n minus X which is uh 6 - 2 which is 4 so 6 CH 2 is 15 the 16 to the 2 power gives you 1 over 36 and the 56 to the 4th power gives you the 625 over uh 1296 and at this point you know just whip out your calculator and work it out so 15 over one * 1 over 36 * 625 over I mean you could try to reduce it if you want but in this case oh it kind of does reduce uh 12 96 so three goes into 15 but and three goes into 36 so you can have this uh 12 so 5 * 1 * 625 so that's going to be 3,125 all over 12 * 1296 well I'm not entirely sure what that is so 15,552 and when you divide the 15,000 into the 3,000 that will give you a decimal of .201 now this doesn't tell you when those 23es come out it just tells you that within the six roles the likelihood of two of those roles coming out as a three is uh this uh 0201 or 20% chance so if we were to look at other possibilities find likelihood of no threes and six rolls of a fair D the note of the fair Dy all that mean the fair indicates that each outcome is equally likely that's all the fair means here no threes coming out that's going to be X = 0 so the probability of x = 0 is going to be uh 6 chose 0 * 16 to the 0 power time 56 to the 6th power so 6 ch0 that's going to be 1 16 to the power of0 that's going to be 1 and then 56 to the 6 power that's going to be 15,625 over 46,656 and you divide one into the other so 0.335 so roughly onethird chance that out of six rolls none of the outcomes will be a three so likelihood of all six rolls ending up as three so how many threes are we talking about we're talking about six threes why because it's all six rolls ending up as a three so six threes so here here x is going to be six you still have six rolls so this is still your n so the probability of x = 6 is going to be 6 choose 6times 1 16 to 6 power uh times 56 to the 0er power because that's 6 - 6 here's your n here's your X which gives you zero so 626 that's 1 uh 16 to the 6 power that's going to be 1 over 46,656 time 1 which gives you 1 over 46,656 which is roughly 0 .2 or uh two so that's going to be 0.002% chance what's the likelihood of rolling a dice six times and every one of the outcomes being a three. z002 or 0.002% chance so extremely unlikely let me try one more example here so likelihood that's out of four tosses of a fair die we get three ones same scenario but slightly altered instead of looking for threes now we're looking at ones so so now we have a new success so the success is now redefined as getting a one on a us now the probability of that happening has won't change so the probability is still six which means the chance of failure is 56 uh in this case we don't have six tosses we have now four tosses so there's your n three on there's your X X probability of getting three X's so that's going to give you four twos 3 * 1 6 to the 3 power * 56 to the 1 power N - x that gives you 4 - 3 which gives you 1 4 choose three that's going to give you 4 * 16 to the 3 power so that's going to give you 1 over 216 time 56 to the first power so that'll be 56 uh you can simplify a little bit the two and the six they both have twos in common so you can reduce those and so that gives you uh 2 * 5 is 10 and 3 * 216 * 1 uh that'll be uh 64 8 10 over 648 0.015 so 1.5% chance of getting so out of four tosses you get three ones doesn't say three ones in a row they could be it doesn't this doesn't tell you what kind of arrangement these ones are it just tells you that amongst the four toses three of them will be ones we just don't know if there's going to be the first second and third or the first first second or third or the first second and fourth or the first third and fourth or whatever the arrangement is now just to give you a better idea of this let me go over one more example likelihood of getting to even outcomes out of four tosses so here the success is determined not by a single number coming out but by an even number so any outcome that's not even will be uh counted as a failure so two even number so there is our X out of four tosses so there's our n so what is p well here's our sample space what are all the possible outcomes you have 1 2 3 4 five 6 so out of all these outcomes which ones are even two is even three is not four is even five is not six is even and that's it so the failure are any numbers that are not even so the probability how many are even well out of the sample space three are even how many total possible outcomes are there uh six of them so the probability of success is going to be 1/2 which forces Q to be also 1/2 so the probability of xal 2 is going to equal to 4 2 2 to the 1 12 to the 2 power so there's your X there's your x times 12 the first half was your P the next one half is your q and coincidentally that is also two four choose two that's going to be six2 to the 2 power that's going to be 1 14 uh 1 12 to the 2 power that's going to be4 four and the six they both have two in common so three and two so when you multiply this out it's going to give you uh eight which is exactly 625 so out of four tosses you're more likely than not for two of those outcomes to be an even number so hopefully all of this gives you a better idea about how to work these kinds of problems out yeah it goes back to the wording of the exercise now it doesn't have to be something that is exactly find the probability of attaining less than please note less than how many less than two threes in the six rolls of a fair dock same situation as before so solution and please note six rolls so yeah n is still six p is is still one six because we're only looking for one number as an outcome as a successful outcome less than two is now described as X less than two when you're looking at this kind of scenario you need to ask okay these X's uh these are whole numbers you know 0 1 2 3 4 5 6 7 8 9 10 11 12 etc etc etc what values of X satisfy X less than two now remember these are independent so the probability of X being less than two is going to equal the probability of X equaling 0 plus the probability of X equaling 1 these are two independent things so these are not just that but they're they're nonoverlapping so they're mutually exclusive so the probability of X less than two the only numbers that satisfy this are X = 0 and x = 1 so for each individual value that satisfi this inequality you're going to have a separate probability for that individual value 0 and one are the only ones that satisfy this inequality X less than two so we're going to have a one for x equals 0 and a second one for xal 1 which takes us to The Next Step so this probability of xals 0 expands to this and the probability of xal 1 expands to this which gives us roughly this value around rounds off to this this value rounds off to this and when we put the two together we get 7368 find the probability of hits in baseball so this binomial experiment can be applied to other Arenas uh one arena is like a uh uh baseball so every baseball player has like a a I forgot what it's called um a batting average so a baseball player has a wellestablished batting average of 0.250 in the next series he will bat 10 times find the probability he will get more than two hits probability of more than two hits it's definite that he's going to be batting 10 times within those 10 times we want to know how many of those 10 times he'll get a hit that amounts to more than two hits this 0.25 that is his his probability of success which means this is going to be his P so there's your n there's your p uh Q even if uh you can't calculate it directly you can always calculate indirectly because no matter what p is Q will always be one less than that. 25 so that g gives you this value here and X more than two hits is expressed as X greater than two remember there are times when calculating a valued directly can be quite involved when we look at just the P the probability of X greater than two what are all the values of x that satisfy this inequality remember you can't go beyond 10 because you only have 10 hits total so X it could be well it can't be two because we're not we're we're asking more than two so the smallest X can be is three uh it could also be four five 6 7 8 9 and and 10 so 10 is the max so it maxes out at n notice how many numbers are involved here so if we were to do this directly then we would be asking okay what is the probability of xal 3 plus the probability of x = 4 plus the probability of x = five and keep on going up and up and up until you get xal 10 and this is a lot of work there is an alternative instead of looking at so instead of X greater than two look at its complement the complement of X greater than two is going to be X less than or equal to two so just as a reminder if you have this inequality you want to look at the complement notice that the inequality changes directions and where there was no equals involved now there is so if there was an equals involved now you're going to take it away but here there's no equals involved so it gets introduced now because this is the complement we're going to treat it as a complement so if this were your your event e this will be your events e Prime so the probability of e plus the probability of its complement is going to equal one so subtract the complement from both sides you get this now the reason why we bring this up is because how many values of X satisfy this well this is uh X can be 0 one or two to do it directly you're going to have there's going to be like eight values to calculate here we only have three values so which one is more work now if you have a good calculator this is not going to be any different but if you're going to do it by hand then you know this kind of stuff these complement brings down the work significantly and it gives you the exact same result no changes to the answer it's not an approximation it's not a guess it's an exact value but in this scenario with less so 1 minus P the probability of X being 0 one or two so continued this expression can be expanded to this so there's your uh probability of x equal Z here's your probability of x = 1 here's your probability of x = 2 and notice that it is all inside the brackets here because we want to maintain that this all the sum of all of these is equal to this uh probability and the higher in the earlier expression so this is to maintain this expression the value in the expression in the earlier step so because of the bracket the parentheses this means we're going to be doing we're going to add up all of these terms together before we subtract anything so this becomes this so all of this now becomes this all of this now becomes this and and this becomes that so we add them up all together so after you sum them all up now you can subtract it from one and you get this so this is the probability that the baseball player will uh get more than two hits out of 10 chances at at the that what does this mean so let's uh remind ourselves of what it what the uh like we h of a batter getting more than two hits in 10 chances at bat and probably success remember was 0.25 his batting average was uh 0.25 so the probability that we calculated was 0.47 44 there are two ways to look at probabilities there's a singular way and there is a collective way now remember our scenario or the context of this exercise is is 10 chances at bat along with this being his batting average so the singular perspective of probability so the probability oops 0.474 4 if uh the batter went through one round of 10 chances that's the context of this 10 chances at bat their likelihood to hit more than twice is 47.443840 slightly less than even so more likely than not that he will not hit more than twice now this doesn't tell you which of the batting will be a hit it just tells you that uh within the 10 uh chances at bat the number of hits will be at least three or more than two so that's the singular perspective just looking at one opportunity of 10 chances at B the other way to look at this is like is a collective perspective this is just a way of viewing these probabilities suppose the batter only has 10 chances per season I know batters don't are not necessarily limited to that this is just a way of viewing uh probabilities in uh within say 10 seasons so this is not the same 10 as 10 chances at bat that's a that that 10 chances at bat determines the context of the exercise this is one season he gets he gets 10 chance another season he gets 10 chance another season he gets 10 chance so out of within those 10 seasons uh how many seasons will he likely hit more than two times within a season don't you're going to take that 44 4744 you're going to multiply it by this 10 and that's going to give you 4744 and then you're going to round it off to the nearest unit in about five seasons we don't know which one we just guessing we're estimating that amongst those 10 seasons about five seasons will be he'll be hitting more than twice and by the way the more than twice can mean three times six times eight times it doesn't matter it's more than twice more than twice probably not going to live this long but suppose he had he lives 100 seasons and he's able to maintain his uh uh he's able to maintain his uh batting average So within 100 Seasons you take the 0.44 I'm sorry 4744 you multiply it by the 100 and that gives you 47.443840 expects he hits more than twice in a season in each season so you take this you round off you get point4 you take this number you round off okay so that ends holy cow 11 4 uh I don't think I'm going to finish 115 so we'll see okay so expected value and simulation so basic concepts determine expected value of a random variable and expected net winnings in a game of chance uh determine if a game of chance is a fair game use expect value to make business and insurance decisions use simulation and genetic processes such as determination of flower color and birth gender so expected value so children in third grade were surveyed and told to pick the number of hours that they play electronic games each day the probability distribution is given below so the First Column represents the number of hours that a child play uh the corresponding row or the the the the value in the second row tells you the corresponding probability of that happening first you would take a survey and then you find out all the data you know child won how many hours so I do so many hours I do so many hours I do so many hours you know you add up all these hours and then you uh you know a lot of these uh it's probably not going to be exact values oh I don't I do zero hours I do one hour you know probably some do half an hour or three4 of an hour or you know two and a half hours or whatever the case may be so when you're doing these surveys you have to determine how are you going to count all these these that don't these cases that don't follow specifically an hour but here's the example so zero hours that probably is 30% 1 hour probably 40% two hours 20% and 3 hours 10% and notice the uh here you have X here you have X here you have P of x now what we're going to do is do something like uh we're going to take the zero multiply it by3 one * 4 2 * 2 3 * .1 give us this and that is uh when you do all the arithmetic you get 1.1 so this 1.1 hours is the expected value of the quantity of time times spent playing electronic games so expected value this is a weighted average so it's tied to the idea of an average but not entirely it is a weighted average and the idea is okay we have all these kids we took the survey we figured out all these probabilities so now I can go back and say oh if I pick a random kid out of that third grade what's the expected number of hours of they're playing the games oh it the expected number of hours is 1.1 hours or it could work the other way suppose a new kid is introduced to this third grade uh class what's the expected number of hours that they play uh games oh we expect them to play 1.1 hour so here's a more concise thing for expected value so if a random variable can have any of the values X1 X2 X3 through xn and the corresponding probabilities of these values occurring are XP probability of X1 probability of X2 probability of x3 up through Pro probability xn then the expected value of x is given by this so we take each value of x we take the corresponding probabilities we take the products we're repeat that for every other expected value and then we add up all those products that is our expected value for that uh scenario find the expected number of boys for a three child family assuming girls and boys are equally likely so solution number one what are the possible outcomes remember we have three child family how many boys can we have well we can have none of them can be boys one of them can be a boy two of them can be a boy all of them can be void what's the probability well that depends on these sample space so we look at this here the girlo girl okay how many boys are there none and the second one Gore boy so that's going to be one boy this is going to be one boy two boys here one boy here two boys here two boys here and three here so these are all the expected outcomes and now we look at the probability so how many of these have none coming out oh well it's only this one outcome that have no boys so one out of eight uh one boy well that's going to be this one this one and this one so there's three scenarios three outcomes where there's only one boy so three ofes two boys so that's going to be this one this one and this one so three outcomes where you have two boys so again three a and there's only one outcome where you would have three boys so one over8 so now you're going to take the product of each one of these so 0 * 1/8 gives you 1/8 1 * 38 gives you 38 2 * 38 gives you 68 notice I'm not reducing anything because we're going to have to add these all together no reducing Until the End uh 3 * 1 18 is 38 so now we're going to take the sum of all of these numbers so the expected value is the sum of the third column the third column was this gives you this so when you add them all up you get 128 which reduces to 3 Hales or 1.5 so what's the expected number of boys and a three child family it's going to be 1.5 a player plays $3 to play the following game sells out $3 he rolls a die and receives $7 if he chooses if he tosses a six uh he gets $1 back if he gets anything else find the players expected net winnings so keyword is net winnings for the game so for the privilege of playing the game he pays $3 so he's already lost $3 if he uh toss a die there are only two outcomes one outcome is rolls a what is it toses a six so if he gets a six then a player receives $7 if it is not A6 player receives $1 so what satisfies this the not a six well it'll be a one a two a three a four four and a five so solution so the information for the game is displayed below uh D outcome if it's a one two three four or five the payoff is $1 so the net gain for the player so remember the player lost $3 they had to sell out $3 in order just to for the privilege to play this game uh it's not just rolling the die it's to get some money back whether it's low or high so that's the game in full so remember he had to play he had to pay $3 to get in so this uh $1 minus the $3 gives you a -2 with respect to the player now if it rolls a six then you know she gets $7 back but remember she still had to pay out $3 to play the game so one or seven minus three so ultimately in this scenario she only made $4 so make $4 or lose $2 ultimately so this is our X so we'll take the -2 we'll multiply it by 56 and that will give us negative uh 106 uh for the uh for the winning the four multip by six gives you 46 of a dollar the sum of these numbers because these are the only two possible scenarios either they lose or they win so we take the sum of those two and that gives us uh $6 over6 which reduces to1 so what does this mean this is not a short-term calculation in the long run the player is expected to lose on average $1 per game if you were to Mo uh play this multiple times and kept track of all the winnings and losses and you average all that out you know once you get to a large number of of plays I mean large number I mean like 20 or 30 or 40 or it should start to approach uh a loss of $1 now notice that I said up here net gain for player had we switched the context to the game maker context of game maker so this person gains $3 and then for the play there's two possible outcomes if the player rolls a six then pay $7 to fler but remember this is a context of so if I'm the game maker I get $3 back so if this person win rolls of six I have to sell out $7 so for the game maker the $3 is a plus the $7 is negative player not six so here I have to pay out $1 to player so again I pay out so that means it's negative one for me so player win which is a six that means my um X my net gain well $3 in for me but I lost $7 so this is going to be4 uh probability of this happening is 16 so X of PX is going to be -46 player lose okay that's a one two three four five $3 from the player to me but they lose so I still have to Shell out money but not as much as course so negative one so that means a $2 gain the probability of that happening is 56 so multiply these two together you get plus uh 106 so when we add these up together the sum is is positive 66 so in the long run the game maker is expected to make $1 an average of $1 per game you know these expectations are depending on the context uh and the original perspective it was from the uh perspective of the player the player players sheld out money got money back in the perspective of the game maker it's the everything's the other way around a game in which the expected net winnings are zero is called a fair game a game with negative expected winnings is unfair against the player so negative expected winnings unfair against the player which makes sense our previous example uh had a negative expectation so the in the long run all the players in the long run were going to lose on average a dollar per game so unfair against the player a game with positive expected net winnings is unfair in FA in favor of the player most of the time it is with respect to the player so if the player gets a negative expected winnings then it's unfair against if it's a positive expected winnings then it's unfair because it's not zero but in favor of the player is now they're in the long run they're the expectations that they make money uh on average per game what should the game in the previous example cost so that it is a fair game so if we look back at the previous example can we adjust the payin fee originally was $3 but can we adjust that $3 so that the expectation was uh uh would be zero so solution because the cost of $3 resulted a net loss of $1 we can conclude that the $3 was $1 to high a fair cost to play the game would be $3 minus $1 give you $2 so if we're to look at that in this new scenario if we lose meaning we rolled a one two three four or five then we lost $2 but we got $1 back so minus $2 plus the $1 back gives us a net of1 U multiply it by its probability of 56 and that gives us negative 56 a win means we rolled a six well the payoff was $7 but remember we had to pay $2 for the P privilege of playing the game so the $2 I had to pay in plus the $5 uh I get back so that makes it $5 net uh $5 * 56 gives us positive 5 six so when we add these two together we get zero now if the game was more complicated then the adjustment may may need a little bit more uh and this case you know the $3 in expected negative $1 so so that was the $3 was $1 too high so if we dropped it by a dollar then that should give us less of a loss how much less of a loss well we calculated it and it comes out to even things out so this would be a perfectly fair game so no one the player or the game maker is not expected to win out in the long run one simple type of roulette is played uh with an i Ivory ball wheel set in motion the wheel contains 38 compartments uh well in America yes the 38 compartment uh 18 of the compartments are black 18 are red one is labeled zero and one is labeled double zero uh these last two the zero and double zero are neither black nor red usually well here in America it's often Green in this case assume the player places $1 on either red or black if the player picks the correct color of the compartment in which the ball finally lands the payoff is $2 otherwise the payoff is $0 find the expected net winnings Solution by the expected uh value formula uh expect the net winnings are Point net winnings so remember they put in $1 uh for the privilege to play so in the context of the player that's negative $1 right off the bat if they win so this is the scenario of winning they get $2 back so the $2 plus the1 that I had to sell out to play the game gives us a net income of $1 what was the likelihood of that happening but remember this is where if you're playing Reds or blacks there's 38 possible slots 18 of them are red 18 of them are black so no matter how you cut it 18 over 38 is a probability of winning now if you don't get a red or black you have $0 coming in plus the $1 you had to Shell in so that's going to give you negative one that's where that comes in remember that 18 slots are red 18 slots are so if you picked red which means 18 of the slots are red how many are not not red well there's 18 blacks plus the zero and0 so that's 20 so if you picked red there's 18 red that's where that comes in and 20 that are not red that's where that comes in 18 over 38 - 20 over 38 gives you -2 over 38 which simplifies to -1 over1 19 so here you have a NE expected loss here of one over1 19 of a dollar or about 5.3 cents per game in the long run so you might win for a period of time but you play long enough the house will win okay so I'm gonna stop here at because I ran out of time start on this tomorrow

okay welcome back hopefully the exam wasn&#39;t too difficult so today we&#39;re going to be uh completing chapter 11 and maybe going into chapter 12 we&#39;ll see um just as a reminder uh this coming Thursday is a holiday 4th of July so there will be no classes that&#39;s also mentioned in the uh calendar so no lecture for for that day whatever we cover for today tomorrow Wednesday that is all that will be covered for the next exam so let&#39;s get started uh section 11.4 so continue with the probability uh binomial probability construct a simple binomial probability distribution apply the binomial probability uh hold on uh construct a simple yeah apply the binomial probability formula for an experiment involving berui trials now beri trials is just fancier name for binomial uh trials named after I can&#39;t remember a French or Italian mathematician back in the Renaissance days who studied this uh kind of structure so the B Pro so here&#39;s an example the spinner below is spun twice and we are interested in the number of times A2 is obtained so put this in other words think of two as a successful outcome and the one and three as failed outcome so the sample space so coming back to earlier in the chapter the sample space is going to look like this the ACT whatever the number and the first position is this is the results first Spin and the second number is the results Second Spin so two something like 2 three would mean oh we got a two on first spin so for for example Spin and three on Second Spin note the two possible outcomes of success and failure is the basis for the binomial probability the buy means two so we&#39;re looking at and this these are the two things we&#39;re considering a success and failure now when the outcomes of an experiment are divided into just two categories success and the associated probabilities are called binomial most of the time uh there are other elements in play as we uh look deeper into binomial distribution uh repeated trials of the experiment where probability of success and here&#39;s one key factor the probability of success remain remains constant so the probability of the outcome of the first spin is the same as a probability for the Second Spin third spin fourth spin etc etc etc if x so here we have an the variable x if x denotes the and here how it is defined the number of twos curring on each pair of spins so at this point with the probability if you&#39;re going to introduce a variable you really need to introduce with it what that variable stands for when X is an example of a random variable the idea of a random variable as opposed to just a variable so a variable in general its values are determined can be determined by some controlled process for example the the formula eal mc^2 uh you may have come across that even lightly in your in your life so e stands for energy m stands for Mass says C is the speed of light e and C is are are constants those are a fixed value irrespective of what E and M are E and M are variables but they are variables whose value are controlled and as far as if you&#39;re considering a scenario in physics and you want to determine well how much energy would be in this scenario well you have control over the value of M and once you&#39;ve fixed that value of M you plug it in you multiply it out with the c^2 and you get the equivalent energy or the other way around I have so much energy but you want to determine how much mass does that represent so you have control over the value of the E you have the a linear expression for a line for drawing a line well you can control what the X values or the Y values is so if you plug in an X you can determine the corresponding y if you plug in a y you can determine the corresponding X so it is a variable in general is controlled by its value is controlled by some process either you are providing the values or a computer or a measuring device is providing the values a random variable on the other hand its values are determined by some random process so like uh this spinning this whole spinning thing the values of X is is limited by the result of the spin of the spinning device in this spinning example in s the sample space the numbers of two are zero and four cases they occur once in four cases and twice and one case say what okay so what this is trying to say here&#39;s the sample space the results of all possible outcomes of that spin if that Arrow lands on a two it&#39;s counted as a success if it doesn&#39;t land on a two that is a one or three then that&#39;s a failure so here are all the possible outcomes for the two spins the first spin can be one and then the second can be one or two or three the Second Spin can be a two and the first spin can be a one or two or three the first spin could also be a three and its Second Spin is one or two or three so those are all the possibilities now there&#39;s three possible outcomes there two spins so 3^2 equals 9 so we should have nine number outcomes so number of bins or number of uh experiments so there be nine in other words three possible outcomes one two three you square it because you have two spins so that brings a total of nine so when we look at back at this n s the number of twos is zero so this is saying twos do not occur in the spin in four cases so no twos occurring and two spins so if you look at this uh here&#39;s one case where you have no two uh that&#39;s uh this one no two uh there&#39;s a two here a two here two here but here there is node two and at the last no two that&#39;s where your four cases comes in X is one and four cases so one so here we&#39;re saying X is equal to one in four cases how many outcomes here in the sample space have uh the number two occurring only once that&#39;s what the X represents the number of twos occurring on each spare so it can since there two spins you can have no twos showing up or one two showing up or two twos showing up so here you only have one two one two one two and one two three oh yeah yeah yeah and one two so there&#39;s your four times where one two appears two so this two means x = 2 so meaning the the outcome two on the spin happens twice x = 2 in only one case which is this so once because the table on the next slide includes all the possible values of X and their probabilities it is an example of a probability distribution in this case it is a binomial probability distribution so here&#39;s a table marking off all the probabilities so remember X represents the number of twos that show up after a pair of spins since four of those results satisfy x equals zero because it&#39;s a 13 or 31 or something like that or one one 13 this is the probability of getting no twos within two spins four of them had a single two involved so four nths and for the pair of twos that only occurred once so that&#39;s one nth now please note here is a check so if you take all of these probabilities and add them up so you get 49ths plus 49+ 1 nth oh looky there we all have common denominators so four + 4 + 1 gives you 9 9 over 9 is essentially 1 whatever probability table you&#39;re looking at the probabilities all had a some rules of probabilities number one we&#39;re all X from sample space p of X has to be greater than or equal to zero cannot be negative next one so P of X1 plus P of X2 plus blah blah blah blah blah plus x to the N whatever n is all those probabilities have to add up to one so for starters these are some of the probabilities I think they were brought up earlier in the chapter so uh here they are again and if you notice the check 4 9th + 4 9th + 1 9th Give You 9 9ths which is one so hipip raw that works so binomial probability formula so there is a generalized formula and the only thing that generalized means is that you can use it in a multitude of scenarios you just have to apply it in the proper way so in general let N lowercase n be the number of repeated trials so with like the spinning game each trial is just a one spin of of the arrow P Little P is the probability of success On Any Given trial so the probability of success remains fixed now we&#39;ll cover some examples of this but I just want it to be driven across the probability of success is going to remain fixed for each trial and we have a second variable called Q so p and Q that&#39;s going to be uh 1 minus P so if P was 0.4 that means Q is going to be 1 minus 0.4 so 0.6 and the probability of failure for any given trial so and X the number of successes that occur note all trials are independent remember independent is about influence and part of this means P remains fixed throughout all trials so if you do a trial it&#39;s not going to influence the results of the following trials in general act successes can be assigned among and repeated trials and n CX so here is your n choose X your combination uh combination here&#39;s the uh formula p is a probability of success Q is a probability of failure with p and Q where Q is 1 minus P remains constant throughout all n trials the probability of exactly X successes is given by this formula wh so these are interchange so these formulas are interchangeable this uh n choose X expands to This n factorial expression here so first example uh tossing or finding the probability of a coin toss so find the probability of now in these examples please be careful about the wording find the probability of exactly three heads and five tosses of a fair coin solution so what are we looking for the probability of obtaining exactly three heads uh this is a binomial experiment so how do I know because number one there&#39;s a limit lied number of Trials or so and is not it&#39;s not Infinity it&#39;s not some you know some doesn&#39;t go on and on it&#39;s a fixed number of Trials so and five PES three heads in five PES so that is your end uh head so the probability of getting ahead well you know there&#39;s only two possible outcomes heads or tails uh each one is equally likely and we&#39;re looking at heads so the probability of heads going to be one half and that&#39;s going to be true for every trial on the first on the second on the third on the fourth on the fifth the probability of getting a heads is one half so this is the probability of success in this scenario success is getting ahead as opposed to a tail you take this result Q = 1 - P = 1 -2 so Q is this this is failure and x equals three so you have five tosses and you want to know the probability of getting a heads or exactly three heads doesn&#39;t matter where those heads occur they could her in the first toss second toss fifth it doesn&#39;t matter just as long within those five tosses you have three heads which forces the other two to be tails the probability of that is going to be this there&#39;s your n there&#39;s your X here&#39;s your P here&#39;s your X here&#39;s your Q and this is going to be n- X so 5 - 3 to give you two this is this five choose three that&#39;s the same one from chapter 10 so 5 choose three gives you this value uh 12 to the 3 power so that&#39;s going to be 12 * 12 * 12 which is 1/8 and then 12 squared so 12 * 12 that gives you this so 10 over 1 * 1 18 * 1 14 uh you can simplify some of this so the 10 on the top eight on the bottom so they have two in common so divide the two into 10 that gives you these with a five a two into the 8 that leaves poent of four and now you can multiply out 5 * 1 * 1 * 1 gives you 5 1 * 4 * 4 gives you 16 so here is your 516 any questions find the problem rolling a dice uh find the probability of finding exactly two threes and six rolls of a faradise so before you jump into it make sure you you understand it properly so the mention of the three is related to the dice the outcome of a dice so I want exactly two outcomes of a three when I roll a die six times this is counted as a success remember this is rolling of dice so the sample space is going to be either one or two or three or four or five or six so the three is counted as a success which means means if you get any of the other numbers those are going to be failures the one the two the four the five six how do I know because find the probability of obtaining exactly two threes so in the context of this question three is counted as a success any other number is counted as a failure now we could change the expression to say well find the probabilities of paining exactly two sixes in six rolles of a diet in that case the six would be a sign of success the one two 3 four five would all be signs of failures that is not getting a six so we have a total of six rolls so this six so well let&#39;s break it down solution so it&#39;s binomial experiment with n equals 6 so six rows the probability of success is going to be just one out of six so one six because only one number out of the six numbers one outcome out of the six possible outcomes is counted as a success so that&#39;s one out of six which means the remaining five possible outcomes are all failures so the probability failure is going to be five out of six and x = 2 so exactly two so x = 2 so think of it this way so you have in the context of six roles you&#39;re looking for the likelihood of two of those roles being a three so the six will be your n your total number of Trials your two out possible outcomes will be your X so X will always be less than or equal to and greater than or equal to zero so the probability of X being two is going to be 6 choose 2 so there&#39;s your n there&#39;s your X six there&#39;s your P this will be your x uh 56 there&#39;s your q and the four that&#39;s going to be your n minus X which is uh 6 - 2 which is 4 so 6 CH 2 is 15 the 16 to the 2 power gives you 1 over 36 and the 56 to the 4th power gives you the 625 over uh 1296 and at this point you know just whip out your calculator and work it out so 15 over one * 1 over 36 * 625 over I mean you could try to reduce it if you want but in this case oh it kind of does reduce uh 12 96 so three goes into 15 but and three goes into 36 so you can have this uh 12 so 5 * 1 * 625 so that&#39;s going to be 3,125 all over 12 * 1296 well I&#39;m not entirely sure what that is so 15,552 and when you divide the 15,000 into the 3,000 that will give you a decimal of .201 now this doesn&#39;t tell you when those 23es come out it just tells you that within the six roles the likelihood of two of those roles coming out as a three is uh this uh 0201 or 20% chance so if we were to look at other possibilities find likelihood of no threes and six rolls of a fair D the note of the fair Dy all that mean the fair indicates that each outcome is equally likely that&#39;s all the fair means here no threes coming out that&#39;s going to be X = 0 so the probability of x = 0 is going to be uh 6 chose 0 * 16 to the 0 power time 56 to the 6th power so 6 ch0 that&#39;s going to be 1 16 to the power of0 that&#39;s going to be 1 and then 56 to the 6 power that&#39;s going to be 15,625 over 46,656 and you divide one into the other so 0.335 so roughly onethird chance that out of six rolls none of the outcomes will be a three so likelihood of all six rolls ending up as three so how many threes are we talking about we&#39;re talking about six threes why because it&#39;s all six rolls ending up as a three so six threes so here here x is going to be six you still have six rolls so this is still your n so the probability of x = 6 is going to be 6 choose 6times 1 16 to 6 power uh times 56 to the 0er power because that&#39;s 6 - 6 here&#39;s your n here&#39;s your X which gives you zero so 626 that&#39;s 1 uh 16 to the 6 power that&#39;s going to be 1 over 46,656 time 1 which gives you 1 over 46,656 which is roughly 0 .2 or uh two so that&#39;s going to be 0.002% chance what&#39;s the likelihood of rolling a dice six times and every one of the outcomes being a three. z002 or 0.002% chance so extremely unlikely let me try one more example here so likelihood that&#39;s out of four tosses of a fair die we get three ones same scenario but slightly altered instead of looking for threes now we&#39;re looking at ones so so now we have a new success so the success is now redefined as getting a one on a us now the probability of that happening has won&#39;t change so the probability is still six which means the chance of failure is 56 uh in this case we don&#39;t have six tosses we have now four tosses so there&#39;s your n three on there&#39;s your X X probability of getting three X&#39;s so that&#39;s going to give you four twos 3 * 1 6 to the 3 power * 56 to the 1 power N - x that gives you 4 - 3 which gives you 1 4 choose three that&#39;s going to give you 4 * 16 to the 3 power so that&#39;s going to give you 1 over 216 time 56 to the first power so that&#39;ll be 56 uh you can simplify a little bit the two and the six they both have twos in common so you can reduce those and so that gives you uh 2 * 5 is 10 and 3 * 216 * 1 uh that&#39;ll be uh 64 8 10 over 648 0.015 so 1.5% chance of getting so out of four tosses you get three ones doesn&#39;t say three ones in a row they could be it doesn&#39;t this doesn&#39;t tell you what kind of arrangement these ones are it just tells you that amongst the four toses three of them will be ones we just don&#39;t know if there&#39;s going to be the first second and third or the first first second or third or the first second and fourth or the first third and fourth or whatever the arrangement is now just to give you a better idea of this let me go over one more example likelihood of getting to even outcomes out of four tosses so here the success is determined not by a single number coming out but by an even number so any outcome that&#39;s not even will be uh counted as a failure so two even number so there is our X out of four tosses so there&#39;s our n so what is p well here&#39;s our sample space what are all the possible outcomes you have 1 2 3 4 five 6 so out of all these outcomes which ones are even two is even three is not four is even five is not six is even and that&#39;s it so the failure are any numbers that are not even so the probability how many are even well out of the sample space three are even how many total possible outcomes are there uh six of them so the probability of success is going to be 1/2 which forces Q to be also 1/2 so the probability of xal 2 is going to equal to 4 2 2 to the 1 12 to the 2 power so there&#39;s your X there&#39;s your x times 12 the first half was your P the next one half is your q and coincidentally that is also two four choose two that&#39;s going to be six2 to the 2 power that&#39;s going to be 1 14 uh 1 12 to the 2 power that&#39;s going to be4 four and the six they both have two in common so three and two so when you multiply this out it&#39;s going to give you uh eight which is exactly 625 so out of four tosses you&#39;re more likely than not for two of those outcomes to be an even number so hopefully all of this gives you a better idea about how to work these kinds of problems out yeah it goes back to the wording of the exercise now it doesn&#39;t have to be something that is exactly find the probability of attaining less than please note less than how many less than two threes in the six rolls of a fair dock same situation as before so solution and please note six rolls so yeah n is still six p is is still one six because we&#39;re only looking for one number as an outcome as a successful outcome less than two is now described as X less than two when you&#39;re looking at this kind of scenario you need to ask okay these X&#39;s uh these are whole numbers you know 0 1 2 3 4 5 6 7 8 9 10 11 12 etc etc etc what values of X satisfy X less than two now remember these are independent so the probability of X being less than two is going to equal the probability of X equaling 0 plus the probability of X equaling 1 these are two independent things so these are not just that but they&#39;re they&#39;re nonoverlapping so they&#39;re mutually exclusive so the probability of X less than two the only numbers that satisfy this are X = 0 and x = 1 so for each individual value that satisfi this inequality you&#39;re going to have a separate probability for that individual value 0 and one are the only ones that satisfy this inequality X less than two so we&#39;re going to have a one for x equals 0 and a second one for xal 1 which takes us to The Next Step so this probability of xals 0 expands to this and the probability of xal 1 expands to this which gives us roughly this value around rounds off to this this value rounds off to this and when we put the two together we get 7368 find the probability of hits in baseball so this binomial experiment can be applied to other Arenas uh one arena is like a uh uh baseball so every baseball player has like a a I forgot what it&#39;s called um a batting average so a baseball player has a wellestablished batting average of 0.250 in the next series he will bat 10 times find the probability he will get more than two hits probability of more than two hits it&#39;s definite that he&#39;s going to be batting 10 times within those 10 times we want to know how many of those 10 times he&#39;ll get a hit that amounts to more than two hits this 0.25 that is his his probability of success which means this is going to be his P so there&#39;s your n there&#39;s your p uh Q even if uh you can&#39;t calculate it directly you can always calculate indirectly because no matter what p is Q will always be one less than that. 25 so that g gives you this value here and X more than two hits is expressed as X greater than two remember there are times when calculating a valued directly can be quite involved when we look at just the P the probability of X greater than two what are all the values of x that satisfy this inequality remember you can&#39;t go beyond 10 because you only have 10 hits total so X it could be well it can&#39;t be two because we&#39;re not we&#39;re we&#39;re asking more than two so the smallest X can be is three uh it could also be four five 6 7 8 9 and and 10 so 10 is the max so it maxes out at n notice how many numbers are involved here so if we were to do this directly then we would be asking okay what is the probability of xal 3 plus the probability of x = 4 plus the probability of x = five and keep on going up and up and up until you get xal 10 and this is a lot of work there is an alternative instead of looking at so instead of X greater than two look at its complement the complement of X greater than two is going to be X less than or equal to two so just as a reminder if you have this inequality you want to look at the complement notice that the inequality changes directions and where there was no equals involved now there is so if there was an equals involved now you&#39;re going to take it away but here there&#39;s no equals involved so it gets introduced now because this is the complement we&#39;re going to treat it as a complement so if this were your your event e this will be your events e Prime so the probability of e plus the probability of its complement is going to equal one so subtract the complement from both sides you get this now the reason why we bring this up is because how many values of X satisfy this well this is uh X can be 0 one or two to do it directly you&#39;re going to have there&#39;s going to be like eight values to calculate here we only have three values so which one is more work now if you have a good calculator this is not going to be any different but if you&#39;re going to do it by hand then you know this kind of stuff these complement brings down the work significantly and it gives you the exact same result no changes to the answer it&#39;s not an approximation it&#39;s not a guess it&#39;s an exact value but in this scenario with less so 1 minus P the probability of X being 0 one or two so continued this expression can be expanded to this so there&#39;s your uh probability of x equal Z here&#39;s your probability of x = 1 here&#39;s your probability of x = 2 and notice that it is all inside the brackets here because we want to maintain that this all the sum of all of these is equal to this uh probability and the higher in the earlier expression so this is to maintain this expression the value in the expression in the earlier step so because of the bracket the parentheses this means we&#39;re going to be doing we&#39;re going to add up all of these terms together before we subtract anything so this becomes this so all of this now becomes this all of this now becomes this and and this becomes that so we add them up all together so after you sum them all up now you can subtract it from one and you get this so this is the probability that the baseball player will uh get more than two hits out of 10 chances at at the that what does this mean so let&#39;s uh remind ourselves of what it what the uh like we h of a batter getting more than two hits in 10 chances at bat and probably success remember was 0.25 his batting average was uh 0.25 so the probability that we calculated was 0.47 44 there are two ways to look at probabilities there&#39;s a singular way and there is a collective way now remember our scenario or the context of this exercise is is 10 chances at bat along with this being his batting average so the singular perspective of probability so the probability oops 0.474 4 if uh the batter went through one round of 10 chances that&#39;s the context of this 10 chances at bat their likelihood to hit more than twice is 47.443840 slightly less than even so more likely than not that he will not hit more than twice now this doesn&#39;t tell you which of the batting will be a hit it just tells you that uh within the 10 uh chances at bat the number of hits will be at least three or more than two so that&#39;s the singular perspective just looking at one opportunity of 10 chances at B the other way to look at this is like is a collective perspective this is just a way of viewing these probabilities suppose the batter only has 10 chances per season I know batters don&#39;t are not necessarily limited to that this is just a way of viewing uh probabilities in uh within say 10 seasons so this is not the same 10 as 10 chances at bat that&#39;s a that that 10 chances at bat determines the context of the exercise this is one season he gets he gets 10 chance another season he gets 10 chance another season he gets 10 chance so out of within those 10 seasons uh how many seasons will he likely hit more than two times within a season don&#39;t you&#39;re going to take that 44 4744 you&#39;re going to multiply it by this 10 and that&#39;s going to give you 4744 and then you&#39;re going to round it off to the nearest unit in about five seasons we don&#39;t know which one we just guessing we&#39;re estimating that amongst those 10 seasons about five seasons will be he&#39;ll be hitting more than twice and by the way the more than twice can mean three times six times eight times it doesn&#39;t matter it&#39;s more than twice more than twice probably not going to live this long but suppose he had he lives 100 seasons and he&#39;s able to maintain his uh uh he&#39;s able to maintain his uh batting average So within 100 Seasons you take the 0.44 I&#39;m sorry 4744 you multiply it by the 100 and that gives you 47.443840 expects he hits more than twice in a season in each season so you take this you round off you get point4 you take this number you round off okay so that ends holy cow 11 4 uh I don&#39;t think I&#39;m going to finish 115 so we&#39;ll see okay so expected value and simulation so basic concepts determine expected value of a random variable and expected net winnings in a game of chance uh determine if a game of chance is a fair game use expect value to make business and insurance decisions use simulation and genetic processes such as determination of flower color and birth gender so expected value so children in third grade were surveyed and told to pick the number of hours that they play electronic games each day the probability distribution is given below so the First Column represents the number of hours that a child play uh the corresponding row or the the the the value in the second row tells you the corresponding probability of that happening first you would take a survey and then you find out all the data you know child won how many hours so I do so many hours I do so many hours I do so many hours you know you add up all these hours and then you uh you know a lot of these uh it&#39;s probably not going to be exact values oh I don&#39;t I do zero hours I do one hour you know probably some do half an hour or three4 of an hour or you know two and a half hours or whatever the case may be so when you&#39;re doing these surveys you have to determine how are you going to count all these these that don&#39;t these cases that don&#39;t follow specifically an hour but here&#39;s the example so zero hours that probably is 30% 1 hour probably 40% two hours 20% and 3 hours 10% and notice the uh here you have X here you have X here you have P of x now what we&#39;re going to do is do something like uh we&#39;re going to take the zero multiply it by3 one * 4 2 * 2 3 * .1 give us this and that is uh when you do all the arithmetic you get 1.1 so this 1.1 hours is the expected value of the quantity of time times spent playing electronic games so expected value this is a weighted average so it&#39;s tied to the idea of an average but not entirely it is a weighted average and the idea is okay we have all these kids we took the survey we figured out all these probabilities so now I can go back and say oh if I pick a random kid out of that third grade what&#39;s the expected number of hours of they&#39;re playing the games oh it the expected number of hours is 1.1 hours or it could work the other way suppose a new kid is introduced to this third grade uh class what&#39;s the expected number of hours that they play uh games oh we expect them to play 1.1 hour so here&#39;s a more concise thing for expected value so if a random variable can have any of the values X1 X2 X3 through xn and the corresponding probabilities of these values occurring are XP probability of X1 probability of X2 probability of x3 up through Pro probability xn then the expected value of x is given by this so we take each value of x we take the corresponding probabilities we take the products we&#39;re repeat that for every other expected value and then we add up all those products that is our expected value for that uh scenario find the expected number of boys for a three child family assuming girls and boys are equally likely so solution number one what are the possible outcomes remember we have three child family how many boys can we have well we can have none of them can be boys one of them can be a boy two of them can be a boy all of them can be void what&#39;s the probability well that depends on these sample space so we look at this here the girlo girl okay how many boys are there none and the second one Gore boy so that&#39;s going to be one boy this is going to be one boy two boys here one boy here two boys here two boys here and three here so these are all the expected outcomes and now we look at the probability so how many of these have none coming out oh well it&#39;s only this one outcome that have no boys so one out of eight uh one boy well that&#39;s going to be this one this one and this one so there&#39;s three scenarios three outcomes where there&#39;s only one boy so three ofes two boys so that&#39;s going to be this one this one and this one so three outcomes where you have two boys so again three a and there&#39;s only one outcome where you would have three boys so one over8 so now you&#39;re going to take the product of each one of these so 0 * 1/8 gives you 1/8 1 * 38 gives you 38 2 * 38 gives you 68 notice I&#39;m not reducing anything because we&#39;re going to have to add these all together no reducing Until the End uh 3 * 1 18 is 38 so now we&#39;re going to take the sum of all of these numbers so the expected value is the sum of the third column the third column was this gives you this so when you add them all up you get 128 which reduces to 3 Hales or 1.5 so what&#39;s the expected number of boys and a three child family it&#39;s going to be 1.5 a player plays $3 to play the following game sells out $3 he rolls a die and receives $7 if he chooses if he tosses a six uh he gets $1 back if he gets anything else find the players expected net winnings so keyword is net winnings for the game so for the privilege of playing the game he pays $3 so he&#39;s already lost $3 if he uh toss a die there are only two outcomes one outcome is rolls a what is it toses a six so if he gets a six then a player receives $7 if it is not A6 player receives $1 so what satisfies this the not a six well it&#39;ll be a one a two a three a four four and a five so solution so the information for the game is displayed below uh D outcome if it&#39;s a one two three four or five the payoff is $1 so the net gain for the player so remember the player lost $3 they had to sell out $3 in order just to for the privilege to play this game uh it&#39;s not just rolling the die it&#39;s to get some money back whether it&#39;s low or high so that&#39;s the game in full so remember he had to play he had to pay $3 to get in so this uh $1 minus the $3 gives you a -2 with respect to the player now if it rolls a six then you know she gets $7 back but remember she still had to pay out $3 to play the game so one or seven minus three so ultimately in this scenario she only made $4 so make $4 or lose $2 ultimately so this is our X so we&#39;ll take the -2 we&#39;ll multiply it by 56 and that will give us negative uh 106 uh for the uh for the winning the four multip by six gives you 46 of a dollar the sum of these numbers because these are the only two possible scenarios either they lose or they win so we take the sum of those two and that gives us uh $6 over6 which reduces to1 so what does this mean this is not a short-term calculation in the long run the player is expected to lose on average $1 per game if you were to Mo uh play this multiple times and kept track of all the winnings and losses and you average all that out you know once you get to a large number of of plays I mean large number I mean like 20 or 30 or 40 or it should start to approach uh a loss of $1 now notice that I said up here net gain for player had we switched the context to the game maker context of game maker so this person gains $3 and then for the play there&#39;s two possible outcomes if the player rolls a six then pay $7 to fler but remember this is a context of so if I&#39;m the game maker I get $3 back so if this person win rolls of six I have to sell out $7 so for the game maker the $3 is a plus the $7 is negative player not six so here I have to pay out $1 to player so again I pay out so that means it&#39;s negative one for me so player win which is a six that means my um X my net gain well $3 in for me but I lost $7 so this is going to be4 uh probability of this happening is 16 so X of PX is going to be -46 player lose okay that&#39;s a one two three four five $3 from the player to me but they lose so I still have to Shell out money but not as much as course so negative one so that means a $2 gain the probability of that happening is 56 so multiply these two together you get plus uh 106 so when we add these up together the sum is is positive 66 so in the long run the game maker is expected to make $1 an average of $1 per game you know these expectations are depending on the context uh and the original perspective it was from the uh perspective of the player the player players sheld out money got money back in the perspective of the game maker it&#39;s the everything&#39;s the other way around a game in which the expected net winnings are zero is called a fair game a game with negative expected winnings is unfair against the player so negative expected winnings unfair against the player which makes sense our previous example uh had a negative expectation so the in the long run all the players in the long run were going to lose on average a dollar per game so unfair against the player a game with positive expected net winnings is unfair in FA in favor of the player most of the time it is with respect to the player so if the player gets a negative expected winnings then it&#39;s unfair against if it&#39;s a positive expected winnings then it&#39;s unfair because it&#39;s not zero but in favor of the player is now they&#39;re in the long run they&#39;re the expectations that they make money uh on average per game what should the game in the previous example cost so that it is a fair game so if we look back at the previous example can we adjust the payin fee originally was $3 but can we adjust that $3 so that the expectation was uh uh would be zero so solution because the cost of $3 resulted a net loss of $1 we can conclude that the $3 was $1 to high a fair cost to play the game would be $3 minus $1 give you $2 so if we&#39;re to look at that in this new scenario if we lose meaning we rolled a one two three four or five then we lost $2 but we got $1 back so minus $2 plus the $1 back gives us a net of1 U multiply it by its probability of 56 and that gives us negative 56 a win means we rolled a six well the payoff was $7 but remember we had to pay $2 for the P privilege of playing the game so the $2 I had to pay in plus the $5 uh I get back so that makes it $5 net uh $5 * 56 gives us positive 5 six so when we add these two together we get zero now if the game was more complicated then the adjustment may may need a little bit more uh and this case you know the $3 in expected negative $1 so so that was the $3 was $1 too high so if we dropped it by a dollar then that should give us less of a loss how much less of a loss well we calculated it and it comes out to even things out so this would be a perfectly fair game so no one the player or the game maker is not expected to win out in the long run one simple type of roulette is played uh with an i Ivory ball wheel set in motion the wheel contains 38 compartments uh well in America yes the 38 compartment uh 18 of the compartments are black 18 are red one is labeled zero and one is labeled double zero uh these last two the zero and double zero are neither black nor red usually well here in America it&#39;s often Green in this case assume the player places $1 on either red or black if the player picks the correct color of the compartment in which the ball finally lands the payoff is $2 otherwise the payoff is $0 find the expected net winnings Solution by the expected uh value formula uh expect the net winnings are Point net winnings so remember they put in $1 uh for the privilege to play so in the context of the player that&#39;s negative $1 right off the bat if they win so this is the scenario of winning they get $2 back so the $2 plus the1 that I had to sell out to play the game gives us a net income of $1 what was the likelihood of that happening but remember this is where if you&#39;re playing Reds or blacks there&#39;s 38 possible slots 18 of them are red 18 of them are black so no matter how you cut it 18 over 38 is a probability of winning now if you don&#39;t get a red or black you have $0 coming in plus the $1 you had to Shell in so that&#39;s going to give you negative one that&#39;s where that comes in remember that 18 slots are red 18 slots are so if you picked red which means 18 of the slots are red how many are not not red well there&#39;s 18 blacks plus the zero and0 so that&#39;s 20 so if you picked red there&#39;s 18 red that&#39;s where that comes in and 20 that are not red that&#39;s where that comes in 18 over 38 - 20 over 38 gives you -2 over 38 which simplifies to -1 over1 19 so here you have a NE expected loss here of one over1 19 of a dollar or about 5.3 cents per game in the long run so you might win for a period of time but you play long enough the house will win okay so I&#39;m gonna stop here at because I ran out of time start on this tomorrow

Transcript for:Probability and Binomial Distribution Lecture

Transcript for:
Probability and Binomial Distribution Lecture