This is by far the VERY BEST math instruction on probability that I have ever seen. ABSOLUTELY the very best. I KNOW understand probability. THANK YOU! THANK YOU!
Both of these probability videos were WONDERFUL!! I had NO clue how to explain either of these skills to a 9 y/o (esp. a visual learner) these will be perfect. Clear, simplified and easy to pay attentions to. We really appreciate it, THANK YOU!!!
Thanks, I really enjoy how structured this video is, it's clearly thought out and everything is explained right down to its core. (What I've been browsing through TH-cam looking for for ages!!) Even though I already know all this (watching for tutoring ideas cause my friend's struggling in gr. 7) I find it easy to follow along, as a beginner or a person with some previous knowledge. I'm definitely subscribing and using your videos as my future studying materials. Just thought you'd like to know. Keep up the good work!!! :D
Great video. I've been using this to calculate the probability of winning games on the price is right. I just watched a woman play a perfect game on "Three Strikes" which is essentially 4 green jelly beans (numbers) and 3 orange beans (strikes) Then, which each number drawn, the contestant has to decide which of the four spaces the number belongs in. I actually thought it would be much less likely than it is but it comes to about 1 in 833 of a pure chance, perfect game. I honestly figured it was well into the thousands. The big wheel is also very interesting. The dollar value that is received on the first spin is also the chance of going over a dollar on the second spin. For example, if .75 is spun on the first try, there is (theoretically) a 75% chance that the next spin will add up to more than a dollar.
This helped me a lot. Thank you for making such helpful video. I am in seventh grade, and I immediately think of this video whenever I do probability. Plus, I also like your voice.
kolumath No, I really was asking, how do they predict the odds? I think it has something to do with the ratios of how many bets are placed on each horse. But I'm not sure how that workds mathematically. And if that's true, doesn't that place the odds on the expectations of the crowd rather than the skill of the jockey and horse?
Hi! This is Compound like to take several trials like two coins or a coin and a head or taking two numbere from your set. I think. In your case just 1/128 × 1/127 × 1/126 in 3 trials without replacement. I watched Probability crash course ,- much better. Here is a link th-cam.com/video/ZKdvO8ynuGI/w-d-xo.html Very nice and lean explained
Here's a question... Say you've got a 6 sided dice, and you roll it twice (or more), what's the chance that you'll roll a 6 either time? ... Also, is there a formula I could use for that.
its 1/6x1/6 because like he said no matter how many times u roll it , the number 6 wont dissapear from the dice so its 1/36. but its a different story if there are 2 6 sided dice and they are rolled at the same times then it will be 2/12 since there are 2 dice with 1,2,3,4,5,6 on both of them and they are both rolled at the same time there will be 2 6s on each of the dice and there are a total of 12 numbers since its 6 on each dice so its 2/12 or 1/6 chance of rolling 6 if 2 dice are rolled at the same time
Something's wrong with that math... If one flips a coin twice, then there are four possible outcomes: Tails/Tails, Heads/Heads, and two different versions of Heads/Tails. Which if we're looking for all results with any Tails at all, then that makes a 3/4 chance of Tails showing up either time. With the dice, 1/36 can't be right, since its lower than even the original chance. 2/12 can't be right either, as its identical. I think logically, one has to take the fractional chance of 5/6 miss chance, and times it by the chance of the second die, before adding that chance back onto the original die. As... Say: 1/6 + (5/6x1/6)... I'm not sure how to work that out more acurately, but doing it on a calculator, it works out to be about 30.5% (with a 5/9ths at the end)
I think you're overthinking it. You need to know how many times you're rolling the dice exactly. If it's two times - then do it like this: There is replacement so you do 1/6x1/6=1/36. And it has to be lower than the original chance. It's less likely to get two of the SAME number consecutively....rather than just get it once. Watch the video.
hi, i understand all of these types of probability, but i have found a question that i don't understand. The questing states that a guy has 6 coins; one 5 cent coin, two 10 cent coins, and three 20 cent coins. and it then asked me to calculate the probability of taking two coins, the second coin being a higher value than the first one. how could i do this???
Izzy Kindon I'd solve this with a probability tree. So you'd have the first coin selected, then the second possibility. Put the number of times the second coin is bigger than the first (Ie, the first 5c coin, each one picked after would be larger, so that's 5 possibilities where the second coin is larger than the first) and place it over the total number of outcomes. If you need more help, sign up for a half hour of tutoring. Good luck!
kolumath I think the '6-coin-problem' is more complicated than that - the first coin must be either 5 cents or 10 cents (if it is 20 cents, the next coin cannot be 'higher'), so the probability is 3/6 (which is 1/2). If the first coin is 5 cents, the second coin MUST be higher, but if it is 10 cents, the probability of the second coin being higher is 3/5. You surely are required to combine these probabilities in some way to answer the question?
Izzy Kindon I replied to kolumath but I think I have worked out your answer - you only have to work out how many combinations of two there are in a total of six. Call the coins A (5 cents), B (10 cents) and C (20 cents) so the six coins are ABBCCC. There are fifteen (15) different pairs (eg AB, BB, BC, etc) but these can be drawn out in either order (AB could also be BA etc). This makes thirty (30) possible sequences of two coins. The total of which the first is 'lower' in the alphabet than the second (eg AB, BC, etc but not BA, CB, etc) is eleven (11), so the answer to your question is a probability of 11/30 which is roughly 37 percent.
ok so, i get this very well ( thank you!) but then in my math book it isn't one sack of beans, its's 3 cups with different color balls in it and i just don't get that anymore... if there would be more then one sack of beans with different colors and different amount of beans, then how do you know the odds?
Kim Wilson I can't be sure as i haven't read the problem... but if they DON'T tell you which one is in each cup, you can treat it like the one bag.... However, if they DO tell you which is each cup (cup A has 2 white, 3 blue... cup B has 4 white and 1 blue...), then you calculate the probability for each cup, then multiply your results. Hope that helps. If not, contact me for a half hour of tutoring.
Does it also apply to unknowns? For example if we did not know the skill of the chess players... Surely probability would have a role to play. For example, if a chess tournament is taking place in the room next door, and you are shown the layout of a room and where everyone is sitting (3 people on the left hand side and 12 on the right side), could you not use probability to answer the question: "What are the chances that the winner of the chess tournament is sitting on the left hand side of the room?"
01:36 It _does_ make sense, as long as you put the bean back into the bag after drawing it in the first trial ;) 03:00 Let me play Joe Know this time: This is not true. If we assumed that the trials were successful, we wouldn't have to calculate the probability, because it already happened, so this particular outcome is now _certain_ (100% probable). What we assume as successful is just the _first_ outcome in a row (that is, the _condition_). The _total_ outcome of the _entire_ experiment (that is, whether we drew two green beans or not) is _still uncertain_. We want to know what are the chances of a particular outcome (drawing two green beans in a row) to actually happen, provided that we succeeded to draw a green bean the first time. 05:18 It would be nice to explain _why_ should we multiply and not something else. We multiply, because we take 1/2 of that 1/2 of possibilities where the first coin was heads. From all possible outcomes (4 in total), only 2 have heads as the result of the first flip, which gives 2 out of 4, or 2/4, or 1/2. But only 1/2 of these two outcomes (of that 1/2) is what we want: two heads in a row. So we take 1/2 of that 1/2. And 1/2 of 1/2 is 1/4, or (1/2)·(1/2)=1/4. 10:10 Haha this shows a nice "bug" in your spinner example: there is a possibility that the arrow will stop precisely at the boundary between two wedges ;) This means that there are events (very unlikely, but still) which you didn't include in your sample space ;) How would it influence the probability of getting each of the wedges? :> OK is there a part 3?
but the question is what the probability of taking 2 green beans from the bag 2 times in a ROW this question was structured assuming u are not stupid, because lets say u have a 2 in 6 chances to get hurt 2 tries to save yourself 4 of the bags are empty and 2 bags got a bear trap so whats the probability of u hitting the 2 bear traps in a row so lets say the first time u hit the bear trap and it destroyed your right hand to no fixing u dont put the bear trap back in the bag again to try it the 2nd time
bruh..... thats not the question .-. the question is what the chance of getting 2 greens in 2 tries its we dont know the outcome but we just assume and find the possiblity of getting 2 greens in a row dont go all technical here... no one cares
I don't see why real teachers can't explain stuff as simple as people do on TH-cam. Thanks man!
This is by far the VERY BEST math instruction on probability that I have ever seen. ABSOLUTELY the very best. I KNOW understand probability. THANK YOU! THANK YOU!
Joe Know frightened the shit outta me.
Yeah he is SOOOO FRIKIN LOUD
excuse me bitch um how tf do u know u got a green one
Both of these probability videos were WONDERFUL!! I had NO clue how to explain either of these skills to a 9 y/o (esp. a visual learner) these will be perfect. Clear, simplified and easy to pay attentions to. We really appreciate it, THANK YOU!!!
yes
why tf are you teaching a 9 yr old probability
@@incognitobrowser7537 bruh he 18 now
Thanks, I really enjoy how structured this video is, it's clearly thought out and everything is explained right down to its core. (What I've been browsing through TH-cam looking for for ages!!)
Even though I already know all this (watching for tutoring ideas cause my friend's struggling in gr. 7) I find it easy to follow along, as a beginner or a person with some previous knowledge.
I'm definitely subscribing and using your videos as my future studying materials.
Just thought you'd like to know. Keep up the good work!!! :D
Thanks for making this video! My math teacher doesn't even teach the concepts to us and he just expects us to know everything!
plz keep making videos, ppl like me need someone like you to learn easy
Exactly what I need. I’m stuck with homework. Thanks.
Great video. I've been using this to calculate the probability of winning games on the price is right. I just watched a woman play a perfect game on "Three Strikes" which is essentially 4 green jelly beans (numbers) and 3 orange beans (strikes) Then, which each number drawn, the contestant has to decide which of the four spaces the number belongs in. I actually thought it would be much less likely than it is but it comes to about 1 in 833 of a pure chance, perfect game. I honestly figured it was well into the thousands.
The big wheel is also very interesting. The dollar value that is received on the first spin is also the chance of going over a dollar on the second spin. For example, if .75 is spun on the first try, there is (theoretically) a 75% chance that the next spin will add up to more than a dollar.
Best lecture ever i have got because of creative presentation, dynamic speaking style.......waiting for more......please upload.......
This helped me a lot. Thank you for making such helpful video.
I am in seventh grade, and I immediately think of this video whenever I do probability.
Plus, I also like your voice.
:P
Me still being confused cause I am dumb :👁👄👁
Me too sweetie, me too.
ha
Your not alone
Me to ☠️💀
the way of explaining is incredible!!
This video will be great for my Math 2 students! I will share this link! :)
Hello again, do you have more videos on probability?
One of the great explanation i ever listen
ITS REALLY A NICE EXPLANATION.AND YOUR VOICE IS VERY NICE
THANKS great videos
thanks a lot sir! Great vids. You need to do more of these. Probability density function?
Please don't make loud sudden noises in your videos besides that great video:)
+rszara Yup, R.I.P headphone users :P
kmsl
thanks mr kolu ur the best
@@bonbonpony i use headphones >:V no hate doh.
it just give me chills down my spine when i heard the buzz.
This helped so much with college math!
Thank you so much for the video. I have a test tomorrow.
Thank you for both part of probability its really helpful.....
finally found a good video
sir u r excellent - u r from which country ?
Plz add more ur help is easy to understand
What is the difference between compound probability and conditional probability?
Umm... Can you do more videos like this please!!!!!
Helped alot!!!! But can you do more tree diagram videos because I need to understand that for my upcoming test.?!
This helped a lot, Thank you!!!
BTW is there a part 3
Thank you very much.you are a good teacher..........
Good job, looking for more!
If probability only works for games of random chance, how do they predict the odds in a horse race?
kolumath No, I really was asking, how do they predict the odds? I think it has something to do with the ratios of how many bets are placed on each horse. But I'm not sure how that workds mathematically. And if that's true, doesn't that place the odds on the expectations of the crowd rather than the skill of the jockey and horse?
hey can you show me the math on having 1/128 chance, happen 3x in a row?
Hi! This is Compound like to take several trials like two coins or a coin and a head or taking two numbere from your set. I think. In your case just 1/128 × 1/127 × 1/126 in 3 trials without replacement. I watched Probability crash course ,- much better. Here is a link th-cam.com/video/ZKdvO8ynuGI/w-d-xo.html Very nice and lean explained
Wow it's a great video thx for doing so many problems
Missy Sj I jfjri2jx
You are the best!
so sad that don't have any more vidoes
a really nice video
helped a lot thank you
Here's a question... Say you've got a 6 sided dice, and you roll it twice (or more), what's the chance that you'll roll a 6 either time?
... Also, is there a formula I could use for that.
its 1/6x1/6 because like he said
no matter how many times u roll it , the number 6 wont dissapear from the dice
so its 1/36.
but its a different story if there are 2 6 sided dice and they are rolled at the same times
then it will be 2/12 since there are 2 dice with 1,2,3,4,5,6 on both of them
and they are both rolled at the same time
there will be 2 6s on each of the dice and there are a total of 12 numbers since its 6 on each dice
so its 2/12 or 1/6 chance of rolling 6 if 2 dice are rolled at the same time
Something's wrong with that math... If one flips a coin twice, then there are four possible outcomes: Tails/Tails, Heads/Heads, and two different versions of Heads/Tails. Which if we're looking for all results with any Tails at all, then that makes a 3/4 chance of Tails showing up either time.
With the dice, 1/36 can't be right, since its lower than even the original chance. 2/12 can't be right either, as its identical.
I think logically, one has to take the fractional chance of 5/6 miss chance, and times it by the chance of the second die, before adding that chance back onto the original die. As... Say: 1/6 + (5/6x1/6)... I'm not sure how to work that out more acurately, but doing it on a calculator, it works out to be about 30.5% (with a 5/9ths at the end)
I think you're overthinking it. You need to know how many times you're rolling the dice exactly. If it's two times - then do it like this:
There is replacement so you do 1/6x1/6=1/36. And it has to be lower than the original chance. It's less likely to get two of the SAME number consecutively....rather than just get it once. Watch the video.
Thanks this really helped me!
would it be dependent and independent
Thank you.
hello
i-
I like the quality of the video itself. What tools did you use to create the video?
Thanks. The basic presentation was completed in PowerPoint, a little special editing in sony vegas.
Great videos but not color blind friendly. Maybe consider using shapes or differentiating color of jelly beans with another visual difference as well.
joe know WHAT THE FUCK I HAD MY VOLUME UP
There is a sort of chance in chess if your opponent makes a blunder.
really help
Wow ! Thanks 😊
hi, i understand all of these types of probability, but i have found a question that i don't understand. The questing states that a guy has 6 coins; one 5 cent coin, two 10 cent coins, and three 20 cent coins. and it then asked me to calculate the probability of taking two coins, the second coin being a higher value than the first one. how could i do this???
Izzy Kindon I'd solve this with a probability tree. So you'd have the first coin selected, then the second possibility. Put the number of times the second coin is bigger than the first (Ie, the first 5c coin, each one picked after would be larger, so that's 5 possibilities where the second coin is larger than the first) and place it over the total number of outcomes. If you need more help, sign up for a half hour of tutoring. Good luck!
kolumath I think the '6-coin-problem' is more complicated than that - the first coin must be either 5 cents or 10 cents (if it is 20 cents, the next coin cannot be 'higher'), so the probability is 3/6 (which is 1/2). If the first coin is 5 cents, the second coin MUST be higher, but if it is 10 cents, the probability of the second coin being higher is 3/5. You surely are required to combine these probabilities in some way to answer the question?
Izzy Kindon I replied to kolumath but I think I have worked out your answer - you only have to work out how many combinations of two there are in a total of six. Call the coins A (5 cents), B (10 cents) and C (20 cents) so the six coins are ABBCCC. There are fifteen (15) different pairs (eg AB, BB, BC, etc) but these can be drawn out in either order (AB could also be BA etc). This makes thirty (30) possible sequences of two coins. The total of which the first is 'lower' in the alphabet than the second (eg AB, BC, etc but not BA, CB, etc) is eleven (11), so the answer to your question is a probability of 11/30 which is roughly 37 percent.
this is amazing
THANK YOU
u are the best ♥
great help thank you soooo much !!
Joe is awesome
Y u gay
@@subject_84 Y u gay
And if you close it, when it goes into the corner, it reopens it XDXDXD.
ok so, i get this very well ( thank you!) but then in my math book it isn't one sack of beans, its's 3 cups with different color balls in it and i just don't get that anymore... if there would be more then one sack of beans with different colors and different amount of beans, then how do you know the odds?
Kim Wilson I can't be sure as i haven't read the problem... but if they DON'T tell you which one is in each cup, you can treat it like the one bag.... However, if they DO tell you which is each cup (cup A has 2 white, 3 blue... cup B has 4 white and 1 blue...), then you calculate the probability for each cup, then multiply your results. Hope that helps. If not, contact me for a half hour of tutoring.
I got it! I was confused weather i had to multiply or add it up. Thank you :)
kolumath
Kim Wilson
why is probability incorrect when applied to games of skill or ability ..I don't get it
Does it also apply to unknowns? For example if we did not know the skill of the chess players... Surely probability would have a role to play.
For example, if a chess tournament is taking place in the room next door, and you are shown the layout of a room and where everyone is sitting (3 people on the left hand side and 12 on the right side), could you not use probability to answer the question: "What are the chances that the winner of the chess tournament is sitting on the left hand side of the room?"
helped
Lol instantly puts a link to his channel XD.
nice.
#Manutalk thanks for your help and
interesting, thanks for sharing
Bobby die when he 65 or 64
perfect, thnx
legend :D
I did this in school
Bruh just called them candy beans
I DONT UNDERSDTAND ANYTHING
SOZ
VERY USELESS
git gud ;)
maybe watch the first video if u are that weak at this
cool man
Супер!
those sudden really loud sound effects kmsl
I still don't get it
Same
9:00
1v1 me m8
hey, keep it G, PhreshPhail
Nerddd
01:36 It _does_ make sense, as long as you put the bean back into the bag after drawing it in the first trial ;)
03:00 Let me play Joe Know this time: This is not true. If we assumed that the trials were successful, we wouldn't have to calculate the probability, because it already happened, so this particular outcome is now _certain_ (100% probable). What we assume as successful is just the _first_ outcome in a row (that is, the _condition_). The _total_ outcome of the _entire_ experiment (that is, whether we drew two green beans or not) is _still uncertain_. We want to know what are the chances of a particular outcome (drawing two green beans in a row) to actually happen, provided that we succeeded to draw a green bean the first time.
05:18 It would be nice to explain _why_ should we multiply and not something else. We multiply, because we take 1/2 of that 1/2 of possibilities where the first coin was heads. From all possible outcomes (4 in total), only 2 have heads as the result of the first flip, which gives 2 out of 4, or 2/4, or 1/2. But only 1/2 of these two outcomes (of that 1/2) is what we want: two heads in a row. So we take 1/2 of that 1/2. And 1/2 of 1/2 is 1/4, or (1/2)·(1/2)=1/4.
10:10 Haha this shows a nice "bug" in your spinner example: there is a possibility that the arrow will stop precisely at the boundary between two wedges ;) This means that there are events (very unlikely, but still) which you didn't include in your sample space ;) How would it influence the probability of getting each of the wedges? :>
OK is there a part 3?
but the question is what the probability of taking 2 green beans from the bag 2 times in a ROW
this question was structured assuming u are not stupid, because lets say u have a 2 in 6 chances to get hurt
2 tries to save yourself
4 of the bags are empty and 2 bags got a bear trap
so whats the probability of u hitting the 2 bear traps in a row
so lets say the first time u hit the bear trap and it destroyed your right hand to no fixing
u dont put the bear trap back in the bag again to try it the 2nd time
this is simple probability .-. no one cares about the wedges ;-;
get your ass outta here boii
bruh.....
thats not the question .-.
the question is what the chance of getting 2 greens in 2 tries
its we dont know the outcome but we just assume and find the possiblity of getting 2 greens in a row
dont go all technical here... no one cares
Lmao I love Joe
Why do people who teach math have a SUPER BORING VOICE??!!
i think its not the voice
Yeah but he still taught great you don't have to let your personal bias impact the person's ability.