This is quite possibly the best lecture, and certainly the best explanation of how the formula for combinations is derived from the formula for permutations, that I've ever seen.
Simple most, step by step, taking all kinds of examples with not a single extra word ! So, everybody praises MIT . Thanks MIT for the OCW for the whole world .♥️🙏
This is indeed a very good lecture, despite its speed. But pausing the video and solving the problems on your own makes it very understandable. I was actually impressed at how easy the formula for the total number of possible subsets, which is 2 to the power of N, is easily and clearly derived. Thank you to MIT.
I struggled for the longest time with the discussion around 40:39 about "how many cases" there are. I believe what confused me was the instructor (who is excellent) started an approach with conditional probabilities, and then changed back to an approach with counting. ("If you're told that the first two are heads" sounds like a description of a conditional probability on HH in the first two spots, when we're really conditioned on "3 H total.") What I _think_ he means is that he's counting every instance in which HH are the first two options. The "the only uncertainty" phrase leads you , again, to think he's going to talk about probabilities. However, he seems to mean something more along the lines of "there are 8 cases where the first two of the three Hs are at the start, namely one case for placing the remaining H in each of the eight remaining positions in the list." Hope this helps someone!
Well, the number of outcomes or combinations of having 3H in 10 tosses is given by (10 3), eg. one possible outcome is (HTHTHTTTTT), you can see that those outcomes are the subset B, once you have 3H in 10 tosses is certainly that you "live" in that subset, now that your working in subset B there are certain outcomes that begins with HH (our event A) and the third H is somewhere between those 8 remaining tosses and therefore there is 8 possible combinations(subset A inside B)...but to do that assumption or inference (sorry for my loose use of language) you have to know that your working in B that is that event B "happened".
I also struggled at 40:39. This is conditional probability problem solved with counting. For events A and B, the conditional probability is P(A|B) = P(A∩B)/P(B). The probability for the denominator (with counting) is 10 choose 3. The confusing part is that since A already occurred, our new universe (omega) has 8 slots available out of the original 10; therefore the probability for the numerator (again with counting) for the last head, is 8 choose 1 = 8.
I read a book in probability which I found to be well written and I spent 8 hours going over the different counting methods and realizing that only the outcome nodes were counted and the outcomes were lists and not sets. I did this for all of the basic methods except for the possible number of subsets. I hadn't figured out why that one was true yet. In the first ten minutes of this lecture the professor shows clearly how each method is derived with perfectly clear examples. Amazing work!
The way to substitute all the information for values and take it to the sample space is the key to everything... the probabilities come out on their own (if you use common sense of course!).
40:39 Assumptions: 1) independent tosses 2) P(H) = p event B: 3 out of 10 tosses were heads event A: first 2 tosses were heads Given that B occured, the conditional probability that the first 2 tosses were heads is P(A|B) = P(A ∩ B) / P(B). P(A ∩ B) = P(H1 ∩ H2 ∩ 3 out of 10 tosses are heads) = P(H1) * P(H2 | H1) * P(1 out of 8 tosses is heads | H1 ∩ H2) = P(H1) * P(H2) * P(1 out of 8 tosses is heads) (independence) = p^2 * 8C1 * p * (1-p)^7 P(B) = P(3 out of 10 tosses were heads) = 10C3 * p^3 * (1-p)^7 P(A ∩ B) / P(B) = 8C1 / 10C3 = 8/120
For those who did not understand the problem at 40:00 about having 3 heads appear in a toss and the first 2 are heads. Step back from conditional probability and let’s walk through this problem step by step. We are given that 3 heads occurred, and therefore 7 tails(I think it is self explanatory). So the first step is to calculate how many different permutations of 3 Heads are 7 Tails are there?. Well let’s think about one of the possible outcomes ( this is a totally random one possible outcome) : HTHHTTTTTT. Now in order to solve this imagine just for a moment that each head is distinct and each tail is distinct from each other( of course they are indistinguishable but lets assume for now that they are, you will see why in a moment). So let’s label the the three heads as H1 H2 H3, and 7 tails as T1 T2 T3… and so on until T7. Hope that you are with me so far. Now, draw 10 lines, separated by space, horizontally on a piece of paper, like so : __. ___. ____ ….. 10 lines like this. In the first slot we have one of the 10 different outcomes( either one of the 3 of the heads, (h1 h2,…) or a one of the 7 tails( t1, t2,,, and so on until t7)). Now, the basic principle of counting states that the number of ways of rearranging these 3 heads and 7 tails into these 10 slots, assuming that each of the 10 elements is distinct from each other is 10! Of course. Now, we need to drop the assumption that the heads are distinct from each other, and that tails are distinct from each other. In order to do this let’s look at a simple example. Imagine that one of the outcomes is H1 H2 H3 T1 T2 T3 T4 T5 T6 T7. Just one of the many outcomes. Now, within this outcome, is another permutation, like say H2 H1 H3 T1 T2 T3 T4 T5 T6 T7 different than the -previous one I just mentioned? .Since now heads are the same, these outcomes are indistinguishable, and there are 3! permutations of H1 H2 and H3, so the first step is to divide 10! By 3! To remove all these cases of different permutations of H1 H2 H3. By a similar analogy, we should divide 10! By 7! To remove the cases of different permutations of Tails. Therefore, the total number of outcomes of event B is ( 10! / (3!*7!) ). The second step is the following: we are told that the first two throws are heads. So this means that now if we were to draw 10 seperate slots on the piece of paper, we know that it will look like this: H H and then the remaining 8 slots is any combination that contains 7 tails and 1 head. One example of such an outcome in event A is => H H T T H T T T T. Now since we already know that the first two slots are hard and stone 2 heads, lets cover them up, and look at the remaining 8 slots on the right. How many different ways can you put this remaining 1 head into the 8 slots. Well you have 8 ways of doing this. And there you have it, you have 8 permutations of having the first 2 slots occupied by heads, and the remaining 8 occupied between.1 head and 7 tails. So the final answer is 8 / 120, where 8 is the number of combinations of 2 heads in the first slot and 1 head in either one of the remaining 8 slots, and 120 is (10! / 7! * 3!) which we calculated above. And here is the explanation. I hope that it makes sense to anyone who did not understand professor’s explanation
what a nice proof at 35.27! we had to proof it algebraically in analysis 1 for EECS, took me and my friends about 5 hours haha. this way its a lot more intuitive - what a great lecturer he is. Thanks!
Since I always hated it when sth is defined "just so that it works" or seemingly arbitrary.. The way I explain 0! = 1: 3! divided by 3 = 2! 2! divided by 2 = 1! 1! divided by 1 = 0! Same goes for x to the power of 0 Anyways the prof in this video explains things 100* better than my profs.. They just confuse me all the time and their slides look very complicated. Even these easy counting rules, that have been explained at high school pretty well..Luckily I found these videos on youtube
In 38:30 , I didn't understand how elements inside B are equally likely if elements inside sample space are not. Even if B occurred, how elements of B become equally likely?
it is assumed that the die isn't fair so for example the probability of obtaining a sequence with say 9 heads and 1 tail could be much much higher than 9 tails and 1 heads, that wouldn't be the case if the coin was fair though. However, in B it is a FACT that there is exactly 3 heads and 7 tails so all that is left is to arrange the the 3 heads and 7 tails in different ways together and since coin tosses are individually independent of each other all possible arrangement of the sequence have an equal chance of occurring.
Thanks for the lectures! Just a comment there's posibility of putting subs for the students questions? Some times de the camera man dosen't have the time to change the audio setup. Thanks!
He got me confused with the ordering. I thought he said fixed order earlier but then used "n choose and order k" formula. I guessed he meant different ways to order at: 18:31
The text for this course is: Bertsekas, Dimitri, and John Tsitsiklis. Introduction to Probability. 2nd ed. Athena Scientific, 2008. ISBN: 9781886529236. For more info and materials see the course on MIT OpenCourseWare at: ocw.mit.edu/6-041F10. Best wishes on your studies!
Rommel,we are fixing the positions of two heads in the table of outcomes(in fact this can be fixed anywhere in the slots,but here it is the first two slots),so we have no control over the third one,it can be put anywhere in the slots.so there are 8c1=8 choices for this particular outcome.
1. Are students permitted to arrive late for lectures thereby disrupting the attention of other, more courteous students? 2. Must every sentence start with "so?"
+Burntsider depends on the prof, sometimes students have stupid study plans so they can't make it in time. And if someones getting distracted merely by someone entering the room, then he probably isn't paying attention anyways lol. Fillwords like "so" helps some people to keep the flow instead of stuttering. People have their flaws.
He is a genius! I have read many different versions and understood only after this lecture. Highly recommended!!
me too
Have u even finished his book noob?
Agree!
This is quite possibly the best lecture, and certainly the best explanation of how the formula for combinations is derived from the formula for permutations, that I've ever seen.
and the relationship between the two
For this semester, I am counting on you professor.
Pun indented?
Did you pass?
Lecturers with dry humor are the best
Simple most, step by step, taking all kinds of examples with not a single extra word ! So, everybody praises MIT . Thanks MIT for the OCW for the whole world .♥️🙏
This is indeed a very good lecture, despite its speed. But pausing the video and solving the problems on your own makes it very understandable. I was actually impressed at how easy the formula for the total number of possible subsets, which is 2 to the power of N, is easily and clearly derived. Thank you to MIT.
I struggled for the longest time with the discussion around 40:39 about "how many cases" there are. I believe what confused me was the instructor (who is excellent) started an approach with conditional probabilities, and then changed back to an approach with counting. ("If you're told that the first two are heads" sounds like a description of a conditional probability on HH in the first two spots, when we're really conditioned on "3 H total.") What I _think_ he means is that he's counting every instance in which HH are the first two options. The "the only uncertainty" phrase leads you , again, to think he's going to talk about probabilities. However, he seems to mean something more along the lines of "there are 8 cases where the first two of the three Hs are at the start, namely one case for placing the remaining H in each of the eight remaining positions in the list." Hope this helps someone!
Well, the number of outcomes or combinations of having 3H in 10 tosses is given by (10 3), eg. one possible outcome is (HTHTHTTTTT), you can see that those outcomes are the subset B, once you have 3H in 10 tosses is certainly that you "live" in that subset, now that your working in subset B there are certain outcomes that begins with HH (our event A) and the third H is somewhere between those 8 remaining tosses and therefore there is 8 possible combinations(subset A inside B)...but to do that assumption or inference (sorry for my loose use of language) you have to know that your working in B that is that event B "happened".
I also struggled at 40:39. This is conditional probability problem solved with counting. For events A and B, the conditional probability is P(A|B) = P(A∩B)/P(B). The probability for the denominator (with counting) is 10 choose 3. The confusing part is that since A already occurred, our new universe (omega) has 8 slots available out of the original 10; therefore the probability for the numerator (again with counting) for the last head, is 8 choose 1 = 8.
Thank you all, that's the answer I am looking for.
Amazing. This man is a real genius. Thanks MIT.
His lecture is not fast, quite helpful for non-native speakers.
Thanks prof. John Tsitsiklis and mit to share the knowledge to the whole world
Sorry about my language, not native speaker.
Why did I not have a professor like this? It could’ve made my life so easier
I read a book in probability which I found to be well written and I spent 8 hours going over the different counting methods and realizing that only the outcome nodes were counted and the outcomes were lists and not sets. I did this for all of the basic methods except for the possible number of subsets. I hadn't figured out why that one was true yet. In the first ten minutes of this lecture the professor shows clearly how each method is derived with perfectly clear examples. Amazing work!
Name of the probability book that u read
Hey how can I study pigeonhole principal
@amaniaridja2005 I don't know of a good source. There might be Wikipedia sources on it and you could also search for books in your local library
@amaniaridja2005 I don't know of a good source. There might be Wikipedia sources on it and you could also search for books in your local library
@@thomasbates9189 thanks but to be honest it doesn't help
This is an ingenious way of determining number combinations
the way he teaches counting is a real contribution to humanity
The way to substitute all the information for values and take it to the sample space is the key to everything... the probabilities come out on their own (if you use common sense of course!).
the part of n choose k is a brilliant explaination.
40:39 Assumptions: 1) independent tosses 2) P(H) = p
event B: 3 out of 10 tosses were heads
event A: first 2 tosses were heads
Given that B occured, the conditional probability that the first 2 tosses were heads is
P(A|B) = P(A ∩ B) / P(B).
P(A ∩ B) = P(H1 ∩ H2 ∩ 3 out of 10 tosses are heads)
= P(H1) * P(H2 | H1) * P(1 out of 8 tosses is heads | H1 ∩ H2)
= P(H1) * P(H2) * P(1 out of 8 tosses is heads) (independence)
= p^2 * 8C1 * p * (1-p)^7
P(B) = P(3 out of 10 tosses were heads)
= 10C3 * p^3 * (1-p)^7
P(A ∩ B) / P(B) = 8C1 / 10C3 = 8/120
Omg you know how to teach,wow👏👏👏. You make life with stats so so easy for students.
For those who did not understand the problem at 40:00 about having 3 heads appear in a toss and the first 2 are heads. Step back from conditional probability and let’s walk through this problem step by step. We are given that 3 heads occurred, and therefore 7 tails(I think it is self explanatory). So the first step is to calculate how many different permutations of 3 Heads are 7 Tails are there?. Well let’s think about one of the possible outcomes ( this is a totally random one possible outcome) : HTHHTTTTTT. Now in order to solve this imagine just for a moment that each head is distinct and each tail is distinct from each other( of course they are indistinguishable but lets assume for now that they are, you will see why in a moment). So let’s label the the three heads as H1 H2 H3, and 7 tails as T1 T2 T3… and so on until T7. Hope that you are with me so far. Now, draw 10 lines, separated by space, horizontally on a piece of paper, like so : __. ___. ____ ….. 10 lines like this. In the first slot we have one of the 10 different outcomes( either one of the 3 of the heads, (h1 h2,…) or a one of the 7 tails( t1, t2,,, and so on until t7)). Now, the basic principle of counting states that the number of ways of rearranging these 3 heads and 7 tails into these 10 slots, assuming that each of the 10 elements is distinct from each other is 10! Of course. Now, we need to drop the assumption that the heads are distinct from each other, and that tails are distinct from each other. In order to do this let’s look at a simple example. Imagine that one of the outcomes is H1 H2 H3 T1 T2 T3 T4 T5 T6 T7. Just one of the many outcomes. Now, within this outcome, is another permutation, like say H2 H1 H3 T1 T2 T3 T4 T5 T6 T7 different than the -previous one I just mentioned? .Since now heads are the same, these outcomes are indistinguishable, and there are 3! permutations of H1 H2 and H3, so the first step is to divide 10! By 3! To remove all these cases of different permutations of H1 H2 H3. By a similar analogy, we should divide 10! By 7! To remove the cases of different permutations of Tails. Therefore, the total number of outcomes of event B is ( 10! / (3!*7!) ). The second step is the following: we are told that the first two throws are heads. So this means that now if we were to draw 10 seperate slots on the piece of paper, we know that it will look like this: H H and then the remaining 8 slots is any combination that contains 7 tails and 1 head. One example of such an outcome in event A is => H H T T H T T T T. Now since we already know that the first two slots are hard and stone 2 heads, lets cover them up, and look at the remaining 8 slots on the right. How many different ways can you put this remaining 1 head into the 8 slots. Well you have 8 ways of doing this. And there you have it, you have 8 permutations of having the first 2 slots occupied by heads, and the remaining 8 occupied between.1 head and 7 tails. So the final answer is 8 / 120, where 8 is the number of combinations of 2 heads in the first slot and 1 head in either one of the remaining 8 slots, and 120 is (10! / 7! * 3!) which we calculated above. And here is the explanation. I hope that it makes sense to anyone who did not understand professor’s explanation
26:30 start doing algebra furiously :D
what a nice proof at 35.27! we had to proof it algebraically in analysis 1 for EECS, took me and my friends about 5 hours haha. this way its a lot more intuitive - what a great lecturer he is. Thanks!
MIT always does the best💗
I LOVE MIT LECTURES
I go to CSULB yet I pay more attention to these lectures and learn more. I should just pay you guys.
The choice you can make at a stage multiplied by each other.
I like how the prof says factorial.
fucktorial
.' .░░░💕*so much thankful to M.I.T and you all kindness; beautiful courses.'
Thank you for this class. The teacher is very good.
15:46 or you could do: (1)(5/6)(4/6)(3/6)(2/6)(1/6)
the total number of choices/the total number of possibilities
The latter implies no order
elegantly explained, Prof! respect
Exceptionally well explained, thank you
I'm yet to complete this video, commenting in the middle of it. What an explanation, amazing!
Thank you so much. I have finally understood this subject.
Great video. Must view video to understand Counting.. Thank you guys
Since I always hated it when sth is defined "just so that it works" or seemingly arbitrary..
The way I explain 0! = 1:
3! divided by 3 = 2!
2! divided by 2 = 1!
1! divided by 1 = 0!
Same goes for x to the power of 0
Anyways the prof in this video explains things 100* better than my profs.. They just confuse me all the time and their slides look very complicated. Even these easy counting rules, that have been explained at high school pretty well..Luckily I found these videos on youtube
sounds tough, how you doing Henry? hope you ain't stressing bud.
i am confused at 42:45
so if all the xi are dependent why does the formula for xi not change?
Really simplified! Thank you!
can we call him "Mr. strang" of probability?
Strange is meme
the logicality is unreal
extremely
thanks
Thanks Mit
Amazing material.
From 9:10 was professor saying Putin in the subset????
Great video, finally I understand! :) thanks, MIT!!!
TY
Thank you Sir.
explanations of counting is great by a geneous prop...
Simply genius
Nice hair cut ;) And great lecture of course!
''HHHrwhhhaat is the sum.'' Great lecture.
In 38:30 , I didn't understand how elements inside B are equally likely if elements inside sample space are not.
Even if B occurred, how elements of B become equally likely?
it is assumed that the die isn't fair so for example the probability of obtaining a sequence with say 9 heads and 1 tail could be much much higher than 9 tails and 1 heads, that wouldn't be the case if the coin was fair though. However, in B it is a FACT that there is exactly 3 heads and 7 tails so all that is left is to arrange the the 3 heads and 7 tails in different ways together and since coin tosses are individually independent of each other all possible arrangement of the sequence have an equal chance of occurring.
Thanks for the lectures! Just a comment there's posibility of putting subs for the students questions? Some times de the camera man dosen't have the time to change the audio setup. Thanks!
respect!!!
Best prob lect!
Has anyone managed to complete the 3 heads problem using conditional probabilities?
at 47:53 omg left bottom corner: hooded "A" taking a class at MIT! that explains everything!
shouldn't it be 8/(10 choose 3)?????????? wtf?? Does he correct it later in the video?
he corrected it right after he finished writing that
Hocam sıhatler olsun.
Bizim okullarda anlatılan derslere bak bide bunlara bak, sonra ülkeler arasındaki farklara bak; sebebi belli oluyor.
20:08 dang Prof flipped us off 😂
He got me confused with the ordering. I thought he said fixed order earlier but then used "n choose and order k" formula. I guessed he meant different ways to order at: 18:31
legend!!!
What is the name of the textbook for this class?
The text for this course is: Bertsekas, Dimitri, and John Tsitsiklis. Introduction to Probability. 2nd ed. Athena Scientific, 2008. ISBN: 9781886529236. For more info and materials see the course on MIT OpenCourseWare at: ocw.mit.edu/6-041F10. Best wishes on your studies!
Thanks for sharing.
Hello, where does 8 come from?
Rommel,we are fixing the positions of two heads in the table of outcomes(in fact this can be fixed anywhere in the slots,but here it is the first two slots),so we have no control over the third one,it can be put anywhere in the slots.so there are 8c1=8 choices for this particular outcome.
This is so fun !! I love math I take precal and I'm in 10 grade
I got Turing award at age 5.
watching it with speed - 2X option is highly recommended! :D
sounds funny, 1.5x is pretty sufficient lol
how AnB is 8??
Who is watching 👀 his video in 2x speed ? 😃
Nice haircut professor
4:42 Terry?
What a lecture
Counting means adding.
32:46
1. Are students permitted to arrive late for lectures thereby disrupting the attention of other, more courteous students? 2. Must every sentence start with "so?"
+Burntsider depends on the prof, sometimes students have stupid study plans so they can't make it in time. And if someones getting distracted merely by someone entering the room, then he probably isn't paying attention anyways lol.
Fillwords like "so" helps some people to keep the flow instead of stuttering. People have their flaws.
No, usually doors are welded shut as the lecture starts.
@@MrCmon113 LOL
@@MrCmon113 ahahahaha!
Joe Pesci..
8:28 so n fuck-torial
עכשיו אני בוגר הנדסת אלקטרוניקה. אָמֵן
who is here in 2024? :)
Me
Me
26:50