I felt enlightened after watching this video. Learned total probability theorem as well! Best video for Baye's Theorem on UA-cam hands down. Thank you Sir 🙏❤
I have given up my statistics major and almost shifted to arts and commerce but after I saw your 8 minute and 31 seconds of pure awesomeness, I have decided to stick to my major and be a failing student.
I know it's been 5 years, but I hope you passed statistics. It's very hard I know but keep pushing. Do let us know how it went. Wish you best in life 🙏❤
ur such an amazing teacher what i failed to understand in my class after listening to an hour of lecture you thought the same thing in a better way and that too in few minutes . thank you sir
just love your videos, understood why-to-do instead of the traditional what-to-do, ad I just wanted to keep listening, which is not common for me, thanks
Bayes theorem was a puzzle for me for manymany years, because my education was frequentist, and the stats people were still disputing each other. My bosses boss once called me into the office and asked me about the Monty Hall riddle.... I just happened to guess the right answer! So I got to keep my job! Anyway, 'I like how you said 'having balls..'.. all mathematicians must have balls, including the women because it automatically means you have an infinite number spheres and circles too! Beautiful harmonic shapes, and now I understand something that is fundamental to being a scientist and a mathematician! Thank you!
Bayes theorem - underappreciated - probably a bit too new for many people (published after the Rev Bayes' death, he died in 1761). Nice explanation - thanks
I appreciate the breakdown of the first part, but can’t we just look at the two buckets and go “ok there are 12 balls total and 5 are blue...5/12 is the answer”?
Here you can count them, but in many problems it is the denominator of Bayes' theorem - the "given something else happened" bit - which is the hardest thing to calculate. Here's another example: P(criminal is guilty | evidence) = P(evidence | guilty) x P(guilty) / P(evidence) but what is P(evidence)? It is P(evidence and guilty) + P(evidence and not guilty) - or in other words: P(evidence | guilty) x P(guilty) + P(evidence | not guilty) x P(not guilty) - and that is equivalent to the video: P(A | B1) x P(B1) + P(A | B2) x B2 . The whole point is that counting the balls works only when you can indeed count the balls, whereas the process he showed works in a far broader array of situations :)
in this example yes, but the data types in the separate buckets could be different and may not lend themselves to being lumped together, so calculating the probabilities separately is then useful
The way I did it mentally was sort of compute the blueness of each bucket, i.e. 1/2 for the first and 1/3 for the second. I then added the two bluenesses together to get 5/6. I then computed how much of the blueness was contributed by B1 by dividing the blueness of B1 by the total blueness(1/2 divided by 5/6 = 3/5). This should work for any number of buckets.
Great video, clears up a few things for me :) Just a minor thing you could improve would be to write your equals sign a tad bigger since they now can be mistaken for just a 'times' dot, I guess it's obvious from context but I jumped into one of the videos and wasn't completely sure about the equation first
Knowing there are 2 buckets, one with 3 blue balls and one with 2 blue balls, my initial thought was I don't care how many yellow balls there are. Therefore, of a total of 5 blue balls, 3/5 chance of being in bucket 1. I'm not sure if this is flawed logic, but it made more sense given the initial question. The actual math is important for more complicated questions.
Your logic works if number of balls in each bucket is the same, but breaks down if that's not the case. Consider if instead bucket 2 had 200 blue balls and 400 nonblue balls. Bayes theorem still gives 3/5 as answer. It doesn't give 3/203 as answer.
Didn't get the P(A) = P(A∩B1) + P(A∩B2) bit. Probability of A (getting a blue ball) is the probability of A and B1 (blue ball from bucket 1) plus probability of A and B2 (blue ball from bucket2). Why the "intersect" ∩ ?
P(bucket 1 | blue) = p(blue from bucket 1)/p(blue) = (1/2 * 1/2)/(1/2 * 1/2 + 1/2 * 1/3) = (1/4)/(5/22) = 3/5. Im not doing any bayes theorem set up here, but it looks like im implictely using the formula.
Someone correct me but seeing the final answer my mind immediately jumped to the fact that there are only 5 blue balls total. Given that the ball is the blue, our new sample space should be 5, so when we ask what the odds are of it being in bucket one, we just look at how many blue balls are in bucket one, still giving 3/5ths.
Your reasoning only works if number of balls in each bucket is the same, as is the case in the video example. If bucket 2 had 20 blue and 40 nonblue, the answer would still be 3/5, if you do the Bayes theorem, not 3/23 as would be suggested by your reasoning. This is assuming probability of picking from each bucket is still 1/2.
There is something confusing here. When you write P(A) = P(A intersect B1) + P( A intersect B2), you are implying that A, B1 and B2 are subsets of some universal set, say U. What is U?
I totally agree with you. I tend to have mind map when probability topics are discussed, and would try to visualize the entire sample space vs the individual subspaces, usually via tree diagram, or venn diagram, some sort of diagram that shows event and sample space in one view.
for finding P(A), why can't we directly count the total number of balls and the total number of blue balls in both the buckets that comes to be 5/12 so we can directly put values in the formula?
@@liiiinder I just explained the logic of doing this to a guy in the comments, his name was hehehahaha sth. Basically you'd be merging the two buckets, and only keeping the blues and then changing the blues in bucket 1 to be different from blues in bucket 2 and then just taking a simple dist. what is the chance of bucket 1 given a blue ball is now what is the chance of say red given we have drawn a ball. (imagining blue balls of bucket 1 are now red, and blue balls of bucket 2 are like black)
I think what remains is how well your logic generalises. Bayes is good because it generalises well regardless of number of variables and complexity of the question.
hey, your video was really helpful! If there were three different variables to consider like what if we have three different coloured balls, will the same apply?
Impossible to express this particular problem without audio, visual, and pictorial format. Impossible to complete any amount of psych research without those 3 variables
I'm enamored with Bayes theorem. I watch the videos and understand them, but I don't find myself thinking intuitively using Bayes or conditional probability. So I want to know if thats ever a thing. Do the people who study this a lot, or teach it, ever reach a point where they see it all around them? I enjoy it either way but I have wondered whether I'm just not wired see and use Bayes creatively in life.
Isn't it easier to compute the 5/12 probability by just observing that one has a total of 12 balls with 5 of them being blue? Given that the process of extracting one ball is overall random (as if all the balls were in one single bucket for this process).
It was nice sir....you solved the problem at the point where finding P(A) and differentiating it with P(A|B) was concerned.... Thank you sir..... But please tell me what do you call that stylus thing using which you write this ??
Question: when moving P(B) by multiplication... how does that change the relationship from an intersect to a given that? Example: P(A|B) = P(A n B)/ P(B) Multiply the denominator by P(B) = 1 Then the numerator by P(B) = it becomes a given that | rather than remain and n. OR P(A|B)*P(B).... How/why does it go from an intersect to a given when the denominator is multiplied to the numberator?
Eureka. I see where my thinking broke. Forgotten algebraic rule. You must do the act on both sides of the =. Given that P(A|B) = P(A n B)/ P(B) AND I multiply up P(B) THEN I must do the multiplication on both sides of the equation... Resulting in: P(A|B) *P(B) = P(A n B) Right??
Hi, I randomly tried to solve the question of "B1 given blue balls". 1)Given blue balls, hence I eliminate the yellow balls in both bucket, 2) that leave B1 with 3 blue balls and B2 with 2 Blue balls. 3) So the chances of selecting B1 is 3/5. May i know is this concept correct? Don't want to learn the wrong concept. lol
Analytically yes. It checks out. I tried it with a different example. Let's check the logic. I think you're just not doing as many steps as bayes but logically you should be doing it. We want to know if we drew a blue ball, what's the chance it's from bucket one. so by eliminating yellow you're getting P(A|B1) and P(A|B2). You're actually neglecting P(B1) = P(B2) and I can't fully fathom where you've brought that into computation, I think at this point you simply merged the two and changed the nature of the elements, like now the blues in bucket 1 are different from blues in bucket two. so then you just gave a simple probability dist based on the bucket 1 blues and the total pool. I think your concept may be correct, not sure how well it'll generalize to all the other concepts that bayes covers but I think your logic is correct. If you're not sure what you did, you've basically merged the two buckets, removed yellow to just have the blues and to keep the fraction standing you then changed the blues of each bucket, like now you're solving for what's the chance to draw a red from a bucket with 4 reds and 3 blacks, 4 reds being the blues in bucket 1 and 3 blacks being the blues in bucket 2. "What is the chance of getting bucket 1 given drawing a blue has been converted to what is the chance of drawing a red given we have drawn a ball."
The matter rn is how well your logic generalises, bayes is good because it generalises well regardless of the number of variables and complexity of the question.
I felt enlightened after watching this video.
Learned total probability theorem as well! Best video for Baye's Theorem on UA-cam hands down. Thank you Sir 🙏❤
You taught me this for free much quicker than what my college tuition-paid professor ever could.
same tbh
amen lmao
I have given up my statistics major and almost shifted to arts and commerce but after I saw your 8 minute and 31 seconds of pure awesomeness, I have decided to stick to my major and be a failing student.
I ve failed statistics three times but through ur tutorials I ve regained my hopes in passing thi module. thank you
You got this girl 💪❤️
Musa Ncube I hope you passed bro
I know it's been 5 years, but I hope you passed statistics. It's very hard I know but keep pushing. Do let us know how it went. Wish you best in life 🙏❤
Crazy how he explains it the way I want it to be explained. That’s an ideal teacher right there.
ur such an amazing teacher what i failed to understand in my class after listening to an hour of lecture you thought the same thing in a better way and that too in few minutes . thank you sir
I don't think you know what a daddy you are. You help us uni guys so much, thank you so much!
I tried understanding this for ages from my textbook unsuccessfully, this helped immensely
This was incredible. Your excitement for the problem helped make the theory exceptionally clear and concise. Thank you!
Thank you sir for your clear and simple explanation of Bayes theorem.
just love your videos, understood why-to-do instead of the traditional what-to-do, ad I just wanted to keep listening, which is not common for me, thanks
Bayes theorem was a puzzle for me for manymany years, because my education was frequentist, and the stats people were still disputing each other. My bosses boss once called me into the office and asked me about the Monty Hall riddle.... I just happened to guess the right answer! So I got to keep my job! Anyway, 'I like how you said 'having balls..'.. all mathematicians must have balls, including the women because it automatically means you have an infinite number spheres and circles too! Beautiful harmonic shapes, and now I understand something that is fundamental to being a scientist and a mathematician! Thank you!
Years later, thank you sir. This helped me so much. Very detailed!
It has been 6 years, but ur video still helps me solve math prob, thanks!
I am so glad I've found this video. The Bayes' theorem is made so easy, thanks very much, great job!
I also calculated 5/12 for P(A) by simply counting up all the balls (12) and all the blue balls (5) and voilá 5/12. It's nice to know they match.
Great work sir the video helped me so much. your idea is a bit different and simple from the commonly taught expression of Baye's theorem.thanks
I literally watched just this one video and now I can solve soooo many problems!! Thanks a lot 😄
Glad it helped!
Thanks for this simple yet clear explanation.
I really enjoy ypur teaching. Makes maths so simple to understand. Glad i found you 🤩🤩
I have learned more in 1 day than in a whole semester
The people who downvoted this got here accidently while searching for Beyonce's latest video
Michael Bauers hahahah
I was looking for Beyoncé’s theorem too.
True Story
Thank you Sir for the beautiful explanation.
Bayes theorem - underappreciated - probably a bit too new for many people (published after the Rev Bayes' death, he died in 1761).
Nice explanation - thanks
You deserve more recognition 🙇
i love you, i have literally never understood anything in life better
Brilliant example!
Its so hard method even the balls thing is really so hard but you put alot of effort in to it so thank you ❤️❤️
Thank you so much atleast am gonna pass college finally
Thanks sir wish i had a math sur like u
I appreciate the breakdown of the first part, but can’t we just look at the two buckets and go “ok there are 12 balls total and 5 are blue...5/12 is the answer”?
exactly what im thinking
Here you can count them, but in many problems it is the denominator of Bayes' theorem - the "given something else happened" bit - which is the hardest thing to calculate. Here's another example:
P(criminal is guilty | evidence) = P(evidence | guilty) x P(guilty) / P(evidence)
but what is P(evidence)?
It is P(evidence and guilty) + P(evidence and not guilty)
- or in other words:
P(evidence | guilty) x P(guilty) + P(evidence | not guilty) x P(not guilty)
- and that is equivalent to the video: P(A | B1) x P(B1) + P(A | B2) x B2 .
The whole point is that counting the balls works only when you can indeed count the balls, whereas the process he showed works in a far broader array of situations :)
in this example yes, but the data types in the separate buckets could be different and may not lend themselves to being lumped together, so calculating the probabilities separately is then useful
The way I did it mentally was sort of compute the blueness of each bucket, i.e. 1/2 for the first and 1/3 for the second. I then added the two bluenesses together to get 5/6. I then computed how much of the blueness was contributed by B1 by dividing the blueness of B1 by the total blueness(1/2 divided by 5/6 = 3/5). This should work for any number of buckets.
Amazing pedagogy!
You taught me this in 10min ,while my teacher couldn’t in 5 months
It would be bit easy for me to understand with subtitles. But you explain good😊
Beautifully explained.
very well explained ....Man i wish I had your logic!
Great job sir, may God bless you!
Great video, clears up a few things for me :)
Just a minor thing you could improve would be to write your equals sign a tad bigger since they now can be mistaken for just a 'times' dot, I guess it's obvious from context but I jumped into one of the videos and wasn't completely sure about the equation first
Simply explained. Thanks
Very helpful, thank you
Thank you this was so helpful!!!!!!!!!
Knowing there are 2 buckets, one with 3 blue balls and one with 2 blue balls, my initial thought was I don't care how many yellow balls there are. Therefore, of a total of 5 blue balls, 3/5 chance of being in bucket 1. I'm not sure if this is flawed logic, but it made more sense given the initial question.
The actual math is important for more complicated questions.
Your logic works if number of balls in each bucket is the same, but breaks down if that's not the case. Consider if instead bucket 2 had 200 blue balls and 400 nonblue balls. Bayes theorem still gives 3/5 as answer. It doesn't give 3/203 as answer.
5 greens balls in total
3 greens in B1 and 2 greens in B2
P of B1 given that the ball is green is = 3/5
Similarly, since both buckets are equally likely, P(A) is total blue balls out of total balls which is 5/12
Didn't get the P(A) = P(A∩B1) + P(A∩B2) bit.
Probability of A (getting a blue ball) is the probability of A and B1 (blue ball from bucket 1) plus probability of A and B2 (blue ball from bucket2).
Why the "intersect" ∩ ?
Very Nice way of explanation
Thank you!
Thank You so much for helping out with this.
Awesome teaching - thanks for the knowledge
Thanks for having the courage to talk about your blue balls!
Thank you ,this is very useful to me.
Very helpful👏
Could u please show with an example how can we use Bayes theorem to predict who is more likely to win the presidential bid?
Thanks
Thank you Infinity times....
Fenómeno Trefor, besos de Riquelme!
Você explica muito bem, parabéns.
You're great! You just need a better mic 🎤 and that's it 😉
Nice explanation thank you
Thanks so much!
P(bucket 1 | blue) = p(blue from bucket 1)/p(blue) = (1/2 * 1/2)/(1/2 * 1/2 + 1/2 * 1/3) = (1/4)/(5/22) = 3/5.
Im not doing any bayes theorem set up here, but it looks like im implictely using the formula.
God bless you. Thats all i have to say.
Amazing explanation thanks .
Someone correct me but seeing the final answer my mind immediately jumped to the fact that there are only 5 blue balls total. Given that the ball is the blue, our new sample space should be 5, so when we ask what the odds are of it being in bucket one, we just look at how many blue balls are in bucket one, still giving 3/5ths.
Your reasoning only works if number of balls in each bucket is the same, as is the case in the video example. If bucket 2 had 20 blue and 40 nonblue, the answer would still be 3/5, if you do the Bayes theorem, not 3/23 as would be suggested by your reasoning. This is assuming probability of picking from each bucket is still 1/2.
Suhail's Theorem to get P(A):
Count Blue = 5
Count Yellow = 7
Sum = 12
P(A) = 5/12
Nobel Prize here I come
bruh
Awesome explanation!
There is something confusing here. When you write P(A) = P(A intersect B1) + P( A intersect B2), you are implying that A, B1 and B2 are subsets of some universal set, say U. What is U?
I totally agree with you.
I tend to have mind map when probability topics are discussed, and would try to visualize the entire sample space vs the individual subspaces, usually via tree diagram, or venn diagram, some sort of diagram that shows event and sample space in one view.
Best explanation ❣️
Thanks, u are a great teacher, however could u repeat the dame examples using table & tree diagram?
@@DrTrefor Thanks for the reply. Watching your UA-cam videos on Statistics is my favorite pastime. I love them. Thanks once again
you're a hero man!!!
thank you so much
that's pretty awesome thank you very much
Can u make more videos this is really helpful
for finding P(A), why can't we directly count the total number of balls and the total number of blue balls in both the buckets that comes to be 5/12 so we can directly put values in the formula?
or just count the blue balls 5 , and 3 in b1 to get 3/5 ...
@@liiiinder I just explained the logic of doing this to a guy in the comments, his name was hehehahaha sth. Basically you'd be merging the two buckets, and only keeping the blues and then changing the blues in bucket 1 to be different from blues in bucket 2 and then just taking a simple dist. what is the chance of bucket 1 given a blue ball is now what is the chance of say red given we have drawn a ball. (imagining blue balls of bucket 1 are now red, and blue balls of bucket 2 are like black)
I think what remains is how well your logic generalises. Bayes is good because it generalises well regardless of number of variables and complexity of the question.
Why did you move P(B1) and P(B2) from the denominator to the numerator
Good Explanation!
hey, your video was really helpful! If there were three different variables to consider like what if we have three different coloured balls, will the same apply?
Absolutely, the formula generalizes to n variables nicely.
@@DrTrefor Okay thanks!
That was really helpful, thank you very much
Thank yous so much sir
This is so so so good.
Would the probability of finding a blue ball from bucket 1 or bucket 2, be called an independent event or conditionally independent event?
Impossible to express this particular problem without audio, visual, and pictorial format. Impossible to complete any amount of psych research without those 3 variables
you saved me :)
I'm enamored with Bayes theorem. I watch the videos and understand them, but I don't find myself thinking intuitively using Bayes or conditional probability. So I want to know if thats ever a thing. Do the people who study this a lot, or teach it, ever reach a point where they see it all around them? I enjoy it either way but I have wondered whether I'm just not wired see and use Bayes creatively in life.
THANK YOU !!!!
Isn't it easier to compute the 5/12 probability by just observing that one has a total of 12 balls with 5 of them being blue? Given that the process of extracting one ball is overall random (as if all the balls were in one single bucket for this process).
Thank you. This helped
good stuff. i love it
It was nice sir....you solved the problem at the point where finding P(A) and differentiating it with P(A|B) was concerned....
Thank you sir.....
But please tell me what do you call that stylus thing using which you write this ??
thanks. in 7:00 timeline, How P(B1 | A) comes? since A is unknown to us. Shouldn't it be P (A | B1),where B1 is know and A is unknown?
at 3:53 why did we add the intersection? How does (a intersection b) differ from (a|b) in intuitive sense.
@@DrTrefor I did realize it later on. Thanks for clarifying.
Question: when moving P(B) by multiplication... how does that change the relationship from an intersect to a given that?
Example:
P(A|B) = P(A n B)/ P(B)
Multiply the denominator by P(B) = 1
Then the numerator by P(B) = it becomes a given that | rather than remain and n. OR
P(A|B)*P(B)....
How/why does it go from an intersect to a given when the denominator is multiplied to the numberator?
Eureka. I see where my thinking broke. Forgotten algebraic rule. You must do the act on both sides of the =.
Given that
P(A|B) = P(A n B)/ P(B)
AND I multiply up P(B)
THEN I must do the multiplication on both sides of the equation...
Resulting in:
P(A|B) *P(B) = P(A n B)
Right??
Thank you, man! You help me a lot hahaha. Continue the good work!
You're AMAZING !!
Has the equation P(A) = P(A ∪ B1) +(A ∪ B2) been derived from the probability of union of two events rule P(X∪Y) = P(X)+P(Y)?
Dude are you jack whitehall brother??
Hi, I randomly tried to solve the question of "B1 given blue balls".
1)Given blue balls, hence I eliminate the yellow balls in both bucket,
2) that leave B1 with 3 blue balls and B2 with 2 Blue balls.
3) So the chances of selecting B1 is 3/5.
May i know is this concept correct? Don't want to learn the wrong concept. lol
Analytically yes. It checks out. I tried it with a different example. Let's check the logic. I think you're just not doing as many steps as bayes but logically you should be doing it. We want to know if we drew a blue ball, what's the chance it's from bucket one. so by eliminating yellow you're getting P(A|B1) and P(A|B2). You're actually neglecting P(B1) = P(B2) and I can't fully fathom where you've brought that into computation, I think at this point you simply merged the two and changed the nature of the elements, like now the blues in bucket 1 are different from blues in bucket two. so then you just gave a simple probability dist based on the bucket 1 blues and the total pool. I think your concept may be correct, not sure how well it'll generalize to all the other concepts that bayes covers but I think your logic is correct. If you're not sure what you did, you've basically merged the two buckets, removed yellow to just have the blues and to keep the fraction standing you then changed the blues of each bucket, like now you're solving for what's the chance to draw a red from a bucket with 4 reds and 3 blacks, 4 reds being the blues in bucket 1 and 3 blacks being the blues in bucket 2. "What is the chance of getting bucket 1 given drawing a blue has been converted to what is the chance of drawing a red given we have drawn a ball."
The matter rn is how well your logic generalises, bayes is good because it generalises well regardless of the number of variables and complexity of the question.
that helped me. Thanks a lot.
Is it just me or he actually looks like Kevin Peterson?
One question, trefor. If the essence of Bayes theorem is to update the prior belief, what do we update here?
oh okay. so if we repeat the experiment, we might get to a constant value of the probability then?