to be honest, it has been a long time since i taught stats. i plan on making some vids over the summer (too late i know) about stats though. i need to refresh myself first though.
My sisters is in 3+ right now and your video managed to get into her head how to do these problems and I just wanted to say thanks. I had no idea how to do these and you saved my butt. This happened the night before finals for the added stress.
I am in tutoring for the first time in my life because my college xfer from JC to UC depends on this stats class and my gpa (which is fine, except for math). I am really determined to do well in this class, and this video is great. I learned today in psych that if you go home and review the material the same day that you learned it in class, it will help you to retain it- so I can sit here and eat dinner while learning and solidifying my understanding. You're passing me through college!
I was pretty sure I understood the gist of this while working on a problem but was double checking my work. Stumbled here and you clarified it. Thanks!
@Agilulfa Actually, to make the piece of information "It only costs $1 to play" relevant, we would have to change all the variables, to make X stand for the amount of money one would win, but again, (-1, 1) would not be valid. Instead, the values would be: X= {-1, 0, 1, 9}, and P(X)= {1/2, 1/8, 1/4, 1/8}, and the expected value of X would be = (-1)(1/2)+(0)(1/8)+(1)(1/4)+(9)(1/8) = 7/8, which is the same answer given, but justified enough, and for how much one would win, not for...
If a had a teacher like him in college , I would be now a Nobel price winner in maths. Bravo - this is honnestly the way for teaching childs in a pleasant way. Bravo - Patricos Koronos
thank you very much Mr. Patrick your lessons are much much better than my teacher's lessons ! the way you explain the details in a very simple way makes it easy to understand thanks again your student from saudi arabia :)
@BlikeNave there have also been numerous studies done that show if you study/read something right before you go to sleep, it much more readily gets processed into your memory. i used to always look at important stuff for about 10-15 minutes before sleep... you will find you are thinking about it while you lay there, which can be good and bad!
like the video, however if you expect to win $.875 when it cost $1, isnt that a losing game due to not getting your dollar back? or do you mean you win $1.875 total on average for every dollar you spend.
@ThisIsMe812 i was not offended in the slightest. just pointing out that 99% of what you learn in school ends up being useless. people just seem to fixate on math though for some reason (i guess cause they have to think).
Well, I failed stats last year and I am doing everything - even UA-cam - do pass this year. Thank goodness I clicked on your videos. Our professor should watch your videos to see how it should be explained.
@The100mis you miss the point: you win $.875 on average, so after 100 games, you would be up about $87.50. and of course, it IS possible to be down money (try going to a casino)
@ThisIsMe812 and besides, how much of what you learn in school do you use in 'daily living' ? having a job usually means having a skill and having it mastered better than others. if math is not gonna be your skill, that is fine, but for some, this is a skill that makes money! me for example... i make my living understanding math! so to me, it is extremely useful : )
this was very good thank you alot i am majoring in economics im in my first year and statistics has me lost without words but u did a great job thank you alot
I'm looking for creative ways to express the expectancy-value theory of motivation for a graduate class in adult learning. I just might use this! Math is not at all what I initially had in mind, but I love this and think I can spin it to work for my project. Thanks!
At first it seems like getting $0.875 for spending a $1 is a bad choice. Is the right way to interpret this Expected Value problem as: "I'm spending $1 to receive an average of $1.875 per spin, thus netting $0.875?"
Would it not make more sense to subtract one from each of the positive outcomes? While you will definatly lose $1, if you land on $2, you gain $2, so overall you win $1. So there would be a 50% chance you lose $1 overall, 12.5% chance you break even, 25% you win $1 etc. Also, I was taught the probilities should add up to 1.
Could you possibly help me with a stats question??? Mary has a free token to play a game. The probability that Mary will win the game is 0.05, so the probability that she will not win is 0.95. If Mary wins, she will be given $100, while if she loses, she must pay $5. Let X = the amount of money Mary wins (or loses). What is the expected value??
First let me say I wish you were my math teacher! Second thanks for all your videos they are getting me through my advanced math in university! Can you help me with this: your friend bets you $20 that he can pull 2 spades in a row from a deck of 52 cards (which contains 13 spades). What is your expected value from this bet? a. 17.65 b. 17.76 c 18.24 d 18.57
The expected value from this bet it depends from the number of spades your friend will pull out. if he will pull 0 spades --> you will win 20$ if he will pull 1 spade --> you will win 20$ if he will pull 2 spades--> you will LOOSE 20$ GIVEN THAT THERE ARE 2 TRIALS The prabability of pulling out 0 spades is: (13C0 * 39C2) / 52C2 = 741/1326 The probability of pulling out 1 spade is: (13C1 *39C1) / 52C2 = 507/1326 The prabability of pulling out 2 spades is: (13C2 * 39C0) / 52C2 = 78/1326 The expected value of this bet will be: (20$ * 741/1326) + (20$ * 507/1326) + (-20$ * 78/1326) = 17.65 (rounded to the second decimal) hope i helped!
So I think you so do a follow up where you make the bell curve and calculate the Standard deviation. Use a problem simialar to this. or if you want this on since you already have the mean. I wonder would it follow the empirical rule?
@patrickJMT Taking into account my other commentary (9 months later than these ones), I think there's confusion over what X stands for. If it standed for what one would WIN, not HAVE, and the expected value of X = $0.875, then, after 100 games, one would win $87.50, and one would have $187.50. That could have made it clearer for The100mis, but I guess she's not interested anymore...
I agree! I enjoyed the video but was confused when I came up with an Expected Value (EV) of 1.875 and unpaused the video to see he got 0.875. I'm certainly not a math wiz so I am genuinely curious why -1 is considered. It makes sense that each play costs $1.00 but statistically it doesn't make sense that you can have a probability of 200%: The way I calculated expected value for this problem was: P(total) = 1.0 = (0.5 + 0.125 + 0.25 + 0.125) = { [ P(0) = 0.5 = (1/4)*2 ] + [ P(1) = 0.125 = 1/8 ] + [ P(2) = 0.25 = 1/4 ] + [ P(10) = 0.125 = 1/8 ] }, which makes sense and follows the (loosely termed) 'all probabilities must sum to 1.0' guideline/rule. However, the probabilities calculated in the video are: P(total) = 2 = (1.0 + 0.5 + 0.125 + 0.25 + 0.125) = { [ P(-1) = 1.0 ] + [ P(0) = 0.5 = (1/4)*2 ] + [ P(1) = 0.125 = 1/8 ] + [ P(2) = 0.25 = 1/4 ] + [ P(10) = 0.125 = 1/8 ] }, which I do not understand. I do understand that he is accounting for the cost to play of $1.00. However, it seems more natural to do this normalization on the 'back-end' of the problem, where it is more simple to follow. My answer of making an average of $1.875 per game is incorrect, but by subtracting $1.00 from the EV, I would have the correct answer (same as the tutor) of $0.875 per game. Thanks for the video!!
Nope, decided to spin the wheel and landed on a zero, you wouldn't lose $0, you would lose 1 dollar. If you decided to spin the wheel and land on the one, you wouldn't gain a dollar, you would net zero. You have to subtract one at the end there to find the expected value of your net gain.
You can think of it without considering the -1 at first, then you come up with 1.875 expected value. This the average that you can earn from the game. If you could play for free without paying 1 dollar, you got 1.875. But you should pay for each game which cost 1 dollar for each play and after all you should subtract it from 1.875 average, which makes it 0.875 dollar.
Isn't the lower half of the table supposed to be a density function? In which case all the probabilities would have to add up to 1? It seems more logical to leave out the 100% chance of losing a dollar and just subtract it at the end. So the game has an E(X) of 1.875, but it costs a dollar to play so the net gain is 0.875.
I don't see why this game is profitable for the ones playing ? On average you win 0.875 cents per game, but to play the game you need to give 1dollar, so on average you lose money : 0.875 - 1= -0.125cents
I think that X= -1, P(X)= 1 is not a valid variable, because I'll have -$1 if the outcome is 0, as I'll win $1 if the outcome is 2. Two different descriptions cannot stand for the variable X, which is NOT (as stated on the experiment) how much money will I win, but simply what is the outcome that falls on the spinner. Otherwise, we would have to change all the other X variables. So, the expected value would be 15/8 = 1.875, instead of 7/8.
If you win a dollar, you just get your dollar back, so that's (0*.125) or one-eighth of the spinner. Zero times anything is zero so you can leave that off if you want.The chance that you would lose your dollar is fifty percent, or expressed as a probability is .50 since the zero takes up half the spinner (-1*.50). The 10 dollar space pays off only 9 dollars since you don't get your dollar back, and that is one-eighth of the spinner, or .125 as a probability (9 *.125) The 2 dollar pays only 1 dollar since you don't get your original dollar back or (1*.25) (0 *.125)+(-1*.50)+(9*.125)+(1*.25)=.875
@yoshiyambao Taking into account my other commentary, I think there's confusion over what X stands for. If it standed for what one would WIN, not HAVE, and the expected value of X = $0.875, then, after 100 games, one would win $87.50, and one would have $187.50. That could have made it clearer for The100mis...
how about one where we are looking for the expected profit. example insurance policy is 600 amount it cost the company $1200 amount for each policy holder hospitalized. furthermore the have estimated that 85% of policy holders will not be hospitalized, 10% will be hospitalized once a year and no one will be hospitalized more than twice.
HELP! If you know that 35% of university students require vision correction (glasses or contact lenses) and assume that enrolment in classes is random, (a) How many students would you expect to require vision correction in a class of 40? (b) What distribution would the number of students in a class of 40 that require vision correction follow? Why? (c) What is the probability that less than 10 students in a class of 40 require vision correction?
@ThisIsMe812 i bet almost every decision you make in life is really some variation of expected value. for example i mean, why not drop out of school? because the probability of getting a decent job goes up with an education. does one have to go to school to get a good job? of course not...
How do you do this if you aim to win the 10 dollars in 4 tries? Then what will your expected outcome be? I can do this basic stuff, but I just recently got hit with an intense version about shooting darts and I can't figure it out for the life of me.
Dear Patrick- Thanks for the video as always helpful. However, I believe that E(X) is independent of the number of times the game is played so whether you play once or 100 or 1000 times the E(X) will remain 0.875.
Can I have a question.please do answer me Suppose that you are playing a game with single die assumed fair. In this game, you win Rs 20 if a 3 turns up. Rs 40 if a 5 turns up, lose Rs 30 if a 6 turns up. You neither win nor lose if any other faces turn up. What is the expected sum of money you win?
A winning straight ticket earns $5,000 and the winning box ticket wins $2,000 for Cash-3, what is the expected value for each type of bet if each ticket costs $1 and you either win or you lose? Anyway you could help me out with this?
+Raunchy_bullet there will be two probabilites either win or lose and so given that the price of ticket $1 and winning is 5000-2000 = 3000 and losing will get you nothing . outcome = $1 $3000 $0 probabilities = -1 1/2 1/2 so expected value = (-1 * 1) + (1/2 * 3000 )+ (1/2 * 0) = $1499 per game that is the answer considering idk what is cash-3.
The pie example through me off, I thought since 0 was an option twice it would be 2/5 which would give you a probability of .4, and everything else a probability of .2
thanks for ur video..but main problem in practical life is how can we find expected output and how can we assign weight? because Random number are not capable to bring real expected return. Most tough thing is finding expected outcome and giving them probability weight......................... please tell us for that
Can some one assist me in this question ? Question 1 Using the same game as in the video, but with $100 as the highest amount rather than $10, what is the expected value for playing the game only once, to the nearest whole dollar?
Dear Patrick, thank you for this video! I've one question . suppose you are required to pay 1$ each time you plan ( not paying one time 1$) how you describe this on that Outcome&Probability table?
He included a probability of loosing a dollar as one. Meaning you are 100% gaurenteed to loose a dollar every time you play. You could also have calculated just your expected winnings (which would give you a higher number of course), and then subtract $100 from that as the cost to play 100 times. Because it's the exact same math but with a slightly different logical motivation.
to be honest, it has been a long time since i taught stats. i plan on making some vids over the summer (too late i know) about stats though. i need to refresh myself first though.
You're my favorite math tutor on UA-cam. I've passed my math classes thanks to you. I just wish you had more stats videos!
Thank you!!
I have a final in 6 hours and you always save me, thank you so much!
ur hot
good luck!
So you already have your bachelor ?
My sisters is in 3+ right now and your video managed to get into her head how to do these problems and I just wanted to say thanks. I had no idea how to do these and you saved my butt. This happened the night before finals for the added stress.
04:24
The sum of all probabilities must be one. ( Is it according to kolmogorov's system of axioms?) The way you have written it, the sum is 0.
care to show your thought process ?
@@vegancog4628what is unclear?
I seriously wish you were my STAT prof. Your videos teach more than my prof ever could.
i have been doing some search on this topic for a while, and this is probably the best explanation i have seen. good job
I am in tutoring for the first time in my life because my college xfer from JC to UC depends on this stats class and my gpa (which is fine, except for math). I am really determined to do well in this class, and this video is great. I learned today in psych that if you go home and review the material the same day that you learned it in class, it will help you to retain it- so I can sit here and eat dinner while learning and solidifying my understanding. You're passing me through college!
I was pretty sure I understood the gist of this while working on a problem but was double checking my work. Stumbled here and you clarified it. Thanks!
@Agilulfa Actually, to make the piece of information "It only costs $1 to play" relevant, we would have to change all the variables, to make X stand for the amount of money one would win, but again, (-1, 1) would not be valid. Instead, the values would be: X= {-1, 0, 1, 9}, and P(X)= {1/2, 1/8, 1/4, 1/8}, and the expected value of X would be = (-1)(1/2)+(0)(1/8)+(1)(1/4)+(9)(1/8) = 7/8, which is the same answer given, but justified enough, and for how much one would win, not for...
If a had a teacher like him in college , I would be now a Nobel price winner in maths.
Bravo - this is honnestly the way for teaching childs in a pleasant way. Bravo - Patricos Koronos
@Stirfry0 no, this is not ap calc
@yuyupb sure, but i would just draw pictures and not actually answer anything
thank you very much Mr. Patrick
your lessons are much much better than my teacher's lessons !
the way you explain the details in a very simple way makes it easy to understand
thanks again
your student from saudi arabia :)
@BlikeNave there have also been numerous studies done that show if you study/read something right before you go to sleep, it much more readily gets processed into your memory.
i used to always look at important stuff for about 10-15 minutes before sleep... you will find you are thinking about it while you lay there, which can be good and bad!
like the video, however if you expect to win $.875 when it cost $1, isnt that a losing game due to not getting your dollar back? or do you mean you win $1.875 total on average for every dollar you spend.
@ThisIsMe812 i was not offended in the slightest. just pointing out that 99% of what you learn in school ends up being useless. people just seem to fixate on math though for some reason (i guess cause they have to think).
I wish my stats teacher could explain stuff simple like you. Great vid, helped me out a lot
Well, I failed stats last year and I am doing everything - even UA-cam - do pass this year. Thank goodness I clicked on your videos. Our professor should watch your videos to see how it should be explained.
@patrickJMT so you were saying, if I play 100 times, after invested $100, I will have $87.50 in my pocket?
Thank you. This helped me understand expected value the day before my final.
Im with u....
One step closer to not failing my Alg2 class. Thank you very much :o
This is the best explanation of this concept I've found.
glad it helped :)
Never really paid attention in stat class..... exam in about a week....No problem I got patrickJMT to help me :D
Ur a great teacher everytime I don't get something I go to you! Thank you.
i swear if i hadnt seen this video i would of failed thanks i like how you dont make it sound confusing this vid should have a million views
Thank you for giving an example verse all other math teachers who just give equations and your expected to just understand it.
Thanks for making a statistics video that isnt the most boring thing on the planet
@The100mis you miss the point: you win $.875 on average, so after 100 games, you would be up about $87.50.
and of course, it IS possible to be down money (try going to a casino)
@ThisIsMe812 and besides, how much of what you learn in school do you use in 'daily living' ? having a job usually means having a skill and having it mastered better than others. if math is not gonna be your skill, that is fine, but for some, this is a skill that makes money! me for example... i make my living understanding math! so to me, it is extremely useful : )
this was very good thank you alot i am majoring in economics im in my first year and statistics has me lost without words but u did a great job thank you alot
@laucherhan I think the 1 dollar you spend in each game is already factored in calculating the E(X)
THANK YOU! The videos my instructor posted were so confusing and this method puts it into a perspective that make sense to me.
glad i could help :)
I'm looking for creative ways to express the expectancy-value theory of motivation for a graduate class in adult learning. I just might use this! Math is not at all what I initially had in mind, but I love this and think I can spin it to work for my project. Thanks!
Thanks a lot. I've watched your videos for pre-calc and everything. I'm taking AP stats right now and having a lot of troubles.
At first it seems like getting $0.875 for spending a $1 is a bad choice. Is the right way to interpret this Expected Value problem as:
"I'm spending $1 to receive an average of $1.875 per spin, thus netting $0.875?"
Would it not make more sense to subtract one from each of the positive outcomes? While you will definatly lose $1, if you land on $2, you gain $2, so overall you win $1. So there would be a 50% chance you lose $1 overall, 12.5% chance you break even, 25% you win $1 etc. Also, I was taught the probilities should add up to 1.
Could you possibly help me with a stats question???
Mary has a free token to play a game. The probability that Mary will win the game is 0.05, so the probability that she will not win is 0.95. If Mary wins, she will be given $100, while if she loses, she must pay $5. Let X = the amount of money Mary wins (or loses).
What is the expected value??
Excellent. Very clearly explained and simple to understand.
First let me say I wish you were my math teacher! Second thanks for all your videos they are getting me through my advanced math in university! Can you help me with this:
your friend bets you $20 that he can pull 2 spades in a row from a deck of 52 cards (which contains 13 spades). What is your expected value from this bet?
a. 17.65
b. 17.76
c 18.24
d 18.57
The expected value from this bet it depends from the number of spades your friend will pull out.
if he will pull 0 spades --> you will win 20$
if he will pull 1 spade --> you will win 20$
if he will pull 2 spades--> you will LOOSE 20$
GIVEN THAT THERE ARE 2 TRIALS
The prabability of pulling out 0 spades is: (13C0 * 39C2) / 52C2 = 741/1326
The probability of pulling out 1 spade is: (13C1 *39C1) / 52C2 = 507/1326
The prabability of pulling out 2 spades is: (13C2 * 39C0) / 52C2 = 78/1326
The expected value of this bet will be:
(20$ * 741/1326) + (20$ * 507/1326) + (-20$ * 78/1326) = 17.65 (rounded to the second decimal)
hope i helped!
@@katsilianouzafeiria6037 brilliant!
I started EV chapter today and i was lost..it helps me a lot..thanks.
thank you so much for your help on this! This makes so much more sense to me now!
So I think you so do a follow up where you make the bell curve and calculate the Standard deviation. Use a problem simialar to this. or if you want this on since you already have the mean. I wonder would it follow the empirical rule?
Very simple explanation to a seemingly tough concept! Thanks!!
Thanks for the help - these vids really do help. Question for you, how much WOULD you pay to play this game...assuming a risk:reward ratio of 1:3?
so if I played 100 times spending $100, would ! win $87.50 back but lose $13.50, or would i win back the $100 plus $87.50?
If you started with $100 you'd come out with $187.50.
@@johnbrewer7221 awesome.!
Great example, you made this so much easier to understand. Thanks.
yo you here from a homeschooling website, particularly discoveryk12 aswell?
Patrick, can you explain the Proof for the E(X)?
@patrickJMT Taking into account my other commentary (9 months later than these ones), I think there's confusion over what X stands for. If it standed for what one would WIN, not HAVE, and the expected value of X = $0.875, then, after 100 games, one would win $87.50, and one would have $187.50. That could have made it clearer for The100mis, but I guess she's not interested anymore...
Thank you very much!!! I really appreciate this video!! It's helping me out with a different kind of example. Thanks again!
THis helps a LOT! THanks:)
lucky that I didnt listen to the professor
And so the day before the midterm, not only did I learn that there was a God, but that he posted youtube tutorials on statistics.
I guess this is the best explanation!! thank you!
my teacher can`t teach.. I just realised why it makes sense. this video is so helpful...thx
this is not only teach us how to calculate the expexted value for exams its also explain the logic behind the lottery games in general 👍
@patrickJMT i believe what ur saying is.. gambling is a loosing game..unless u got lucky..but this game...a "low risk game" ?
Very nice video. But I don't understand why -1 has to be considered. Isn't the notion of losing $1 captured by the sector having $0?
I agree! I enjoyed the video but was confused when I came up with an Expected Value (EV) of 1.875 and unpaused the video to see he got 0.875. I'm certainly not a math wiz so I am genuinely curious why -1 is considered.
It makes sense that each play costs $1.00 but statistically it doesn't make sense that you can have a probability of 200%:
The way I calculated expected value for this problem was:
P(total) = 1.0 = (0.5 + 0.125 + 0.25 + 0.125) = { [ P(0) = 0.5 = (1/4)*2 ] + [ P(1) = 0.125 = 1/8 ] + [ P(2) = 0.25 = 1/4 ] + [ P(10) = 0.125 = 1/8 ] }, which makes sense and follows the (loosely termed) 'all probabilities must sum to 1.0' guideline/rule.
However, the probabilities calculated in the video are:
P(total) = 2 = (1.0 + 0.5 + 0.125 + 0.25 + 0.125) = { [ P(-1) = 1.0 ] + [ P(0) = 0.5 = (1/4)*2 ] + [ P(1) = 0.125 = 1/8 ] + [ P(2) = 0.25 = 1/4 ] + [ P(10) = 0.125 = 1/8 ] }, which I do not understand. I do understand that he is accounting for the cost to play of $1.00. However, it seems more natural to do this normalization on the 'back-end' of the problem, where it is more simple to follow.
My answer of making an average of $1.875 per game is incorrect, but by subtracting $1.00 from the EV, I would have the correct answer (same as the tutor) of $0.875 per game.
Thanks for the video!!
Nope, decided to spin the wheel and landed on a zero, you wouldn't lose $0, you would lose 1 dollar. If you decided to spin the wheel and land on the one, you wouldn't gain a dollar, you would net zero. You have to subtract one at the end there to find the expected value of your net gain.
You can think of it without considering the -1 at first, then you come up with 1.875 expected value. This the average that you can earn from the game. If you could play for free without paying 1 dollar, you got 1.875. But you should pay for each game which cost 1 dollar for each play and after all you should subtract it from 1.875 average, which makes it 0.875 dollar.
It is consider to put -$1 have the prob. Of 100% for every turn because it is part of the game.
Think of it as winning $0.
Which translates to nothing.
Isn't the lower half of the table supposed to be a density function? In which case all the probabilities would have to add up to 1? It seems more logical to leave out the 100% chance of losing a dollar and just subtract it at the end. So the game has an E(X) of 1.875, but it costs a dollar to play so the net gain is 0.875.
Wow, you actually make statistics fun! My teacher is hopeless
I don't see why this game is profitable for the ones playing ? On average you win 0.875 cents per game, but to play the game you need to give 1dollar, so on average you lose money : 0.875 - 1= -0.125cents
not everyone will see this.... it is taught in just about any statistics class, and sometimes in a probability/discrete math class.
Thanks!
thanks for the donation!
One million subs, congrats!
thank you! ;)
I think that X= -1, P(X)= 1 is not a valid variable, because I'll have -$1 if the outcome is 0, as I'll win $1 if the outcome is 2. Two different descriptions cannot stand for the variable X, which is NOT (as stated on the experiment) how much money will I win, but simply what is the outcome that falls on the spinner. Otherwise, we would have to change all the other X variables. So, the expected value would be 15/8 = 1.875, instead of 7/8.
If you win a dollar, you just get your dollar back, so that's (0*.125) or one-eighth of the spinner. Zero times anything is zero so you can leave that off if you want.The chance that you would lose your dollar is fifty percent, or expressed as a probability is .50 since the zero takes up half the spinner (-1*.50). The 10 dollar space pays off only 9 dollars since you don't get your dollar back, and that is one-eighth of the spinner, or .125 as a probability (9 *.125) The 2 dollar pays only 1 dollar since you don't get your original dollar back or (1*.25)
(0 *.125)+(-1*.50)+(9*.125)+(1*.25)=.875
@yoshiyambao Taking into account my other commentary, I think there's confusion over what X stands for. If it standed for what one would WIN, not HAVE, and the expected value of X = $0.875, then, after 100 games, one would win $87.50, and one would have $187.50. That could have made it clearer for The100mis...
God bless u. The explaination is as simple as it should be
Patrick you're just so funny and amazing all at once. So grateful 4 ur vids :D
Thank you so much!:) I'm taking liberal arts and a lot of it is expected value.This helped me!!
THANK YOU!!!!!! This was a great, easy explanation.
you sir. Have earn't yourself a new subscriber.
how about one where we are looking for the expected profit. example insurance policy is 600 amount it cost the company $1200 amount for each policy holder hospitalized. furthermore the have estimated that 85% of policy holders will not be hospitalized, 10% will be hospitalized once a year and no one will be hospitalized more than twice.
HELP!
If you know that 35% of university students require vision correction (glasses or contact lenses) and assume that enrolment in classes is random,
(a) How many students would you expect to require vision correction in a class of 40?
(b) What distribution would the number of students in a class of 40 that require vision correction follow? Why?
(c) What is the probability that less than 10 students in a class of 40 require vision correction?
A) 35% of 40
The rest sounds like binomial distribution stuff
+MonniBelle A) 35% * 40 = 14 student
C) 10/40 * 14 = 3.5% it is less than 3.5%
B) don't understand is there a rest of the question
@ThisIsMe812 i bet almost every decision you make in life is really some variation of expected value. for example i mean, why not drop out of school? because the probability of getting a decent job goes up with an education. does one have to go to school to get a good job? of course not...
Can we calculate E(1/x) ?
i.e., can we find expected value of a function given in a fraction?
This is very useful for my Rust gambling base.
How do you do this if you aim to win the 10 dollars in 4 tries? Then what will your expected outcome be? I can do this basic stuff, but I just recently got hit with an intense version about shooting darts and I can't figure it out for the life of me.
Thanks Mr.Patrick!
Dear Patrick- Thanks for the video as always helpful. However, I believe that E(X) is independent of the number of times the game is played so whether you play once or 100 or 1000 times the E(X) will remain 0.875.
@Agilulfa ...which would be the outcome on the spinner.
Please, let me know if I'm wrong and why, because that (-1, 1) makes it confusing.
Can I have a question.please do answer me
Suppose that you are playing a game with single die assumed fair. In this game, you win Rs 20 if a 3
turns up. Rs 40 if a 5 turns up, lose Rs 30 if a 6 turns up. You neither win nor lose if any other faces
turn up. What is the expected sum of money you win?
A winning straight ticket earns $5,000 and the winning box ticket wins $2,000 for Cash-3, what is the expected value for each type of bet if each ticket costs $1 and you either win or you lose?
Anyway you could help me out with this?
+Raunchy_bullet there will be two probabilites either win or lose and so given that the price of ticket $1 and winning is 5000-2000 = 3000 and losing will get you nothing .
outcome = $1 $3000 $0
probabilities = -1 1/2 1/2
so expected value = (-1 * 1) + (1/2 * 3000 )+ (1/2 * 0) = $1499 per game
that is the answer considering idk what is cash-3.
The pie example through me off, I thought since 0 was an option twice it would be 2/5 which would give you a probability of .4, and everything else a probability of .2
Amazing video sir!
thanks for ur video..but main problem in practical life is how can we find expected output and how can we assign weight? because Random number are not capable to bring real expected return. Most tough thing is finding expected outcome and giving them probability weight.........................
please tell us for that
@Srjuanando thanks : )
Can some one assist me in this question ?
Question 1
Using the same game as in the video, but with $100 as the highest amount rather than
$10, what is the expected value for playing the game only once, to the nearest whole
dollar?
lol im taking this quiz rn
goooooooooooooood luck! : )
Gorgeous video sir
Where did the 1/4 come from?
1/4 of the circle represents 2 and there are two 0's each representing 1/4 of the circle. Thus, 1/4 + 1/4 = 1/2 = Probability of getting 0.
Can u explain me how does this even work
wow ur better as a techer than mine i like how you make it easy to understand
Dear Patrick, thank you for this video!
I've one question . suppose you are required to pay 1$ each time you plan ( not paying one time 1$) how you describe this on that Outcome&Probability table?
He included a probability of loosing a dollar as one. Meaning you are 100% gaurenteed to loose a dollar every time you play. You could also have calculated just your expected winnings (which would give you a higher number of course), and then subtract $100 from that as the cost to play 100 times. Because it's the exact same math but with a slightly different logical motivation.
so you would be down $12.50 after 100 plays?
SAME! all the people in bc calc class said the same =\ hopefully the curve will be very generous this year
Thank you very much, you explained it perfectly
Glad it was helpful!
thanks!
What does it indicate when the expected value is 0?
You played 100 games
It's a fair game
If the expected value is 0 it means that it is a fair game
Nice lecture! Thanks!
Thanks great video math exams tomorrow for me last second studying 😰