I've seen this explained many times by different instructors, and by far this (along with the proceeding videos) is the clearest explanation I've seen on how to derive the general formula.
dude, I'm repeating intro to probability after i failed miserably last semester because of horrible professor. You're short tutorials are helping me countless times more than a semester in my university. Thank you so much!
This is a general rant: I understand this. What I cannot do is read the different questions and work them out. my H.W questions goes "This guy has these probabilities of so and so. if he chooses to do this and that but he is unable to that and that; what is the possibility of he does this but not that and that but not this." and I spend first 10 minutes trying to understand the back and forth and negation and positives and negatives and then trying to identify the usable data, and finally time went by and I have to move on because they gave 2 minutes to solve a paragraph. See that you explain this in 14 minutes. It is impossible to do one of these questions in 2 minutes, at least not for me. Hense my frustration. Much less when the professor and the book choose to work and explain the very easy questions and leave for the students the very hard ones without help. thank you for listening.
Hey Sal, as I'm watching your videos on combinatorics and how they fit into probability, I was having a hard time grasping the calculations to find the numerator (# of outcomes that satisfy the condition) until I thought about it in the opposite way you explained. You name a list of (k) heads and fit them into (n) flip slots (and frankly, I loose my intuition at this point) but I found it much more natural to rather name the flips (1 through n) and fit them into a group of size k. Then the expression (kCn / k Choose n) has a lot of meaning: "how many ways can I Choose exactly n heads from k trials. Also, thank you for all your hard work that you put into Kahn Academy. You really are doing something great!
@Khan Academy Confused on how last term is (n-(k-1)) as it always doesn’t work for example 10 flips with probability of getting four heads. Isn’t it (n-(n-k))
Hi I have a doubt with these expression. what is the probability of K heads in N flips of the fair coin. The expression continues as N*(N-1) … (n-(k-1) …My doubt is that from where does k-1 comes in numerator part. Please assist?
To think I got into this with otherwise zero connection, motive, or use for this math other than to find out how absurd it was that I got 0 copies of the limited 6* Op out of 7 non-limited 6*s. Other words, what's the probability of 0 successes out of 7 tries, when the limited 6* has a 35% chance of appearing when drawing any 6*. Turns out it was ~4.9%. 7! / (0!(7-0)!) * .35⁰ * (1-.35)⁷-⁰ 4.9% doesn't seem too absurdly improbable in hindsight, but it sure felt like it during the agonizing pulls.
Hey Sal, First; thank you for all your work. Next; perhaps you can shed some light on a little oddity..... If I try to determine the number of different ways 2 coin flips could result in 3 heads, and also the number of ways 2 coin flips could result in 4 heads, using the suggested formula, the results are counter intuitive. The result for 3 heads from 2 flips seems somewhat logical, as the result is negative...but 4 from 2 yields a positive (but fractional) result. 2*1/(4*3*2*1)(-2*-1) = 1/24
(ran out of space) . How can there be 1/24th of a way that 2 coin flips can result in 4 heads? .. ....and why does there seem to be more of a way (?) to end up with 4 heads from 2 flips than 3 heads from two flips. These both seem clearly impossible, but 4 heads seems even more impossible than 3 heads as a result of 2 coin flips.
your choosing k things from n buckets. The kth spot is the last spot you have to fill and you have already filled (k - 1) spots of the head k when you reach this kth spot. So there are n - (k-1) possible ways to fill this kth spot. Just like how when you have filled 0 spots of k there are n ways to fill the first spot. Once you have filled 1 spot here are (n - 1) ways to fill the second spot etc etc
I know he is far better than any math teacher anyone has had, but his teaching style has issues. The way he writes things down is really messy, he talks too fast, and he makes a lot of mistakes, which he then realizes and erases, starting over. It doesn't even fulfill half of UA-cam's standards.
I've seen this explained many times by different instructors, and by far this (along with the proceeding videos) is the clearest explanation I've seen on how to derive the general formula.
It's interesting how a UA-camr can explain something in 11 minutes better than a college professor in 90 minutes.
This is better and free (you can donate if you want to), college is overpriced and worse.
It's a strange world we live in.
GOLD. I need to remember to donate when I get paid.
Let's get some real estate over here!! - best KhanAcademy line
Well, you can't blame him. It is, after all, free real estate. Insert meme face.
WOWW, NOW I UNDERSTAND FACTORIAL!
Thanks Khan Academy!
dude, I'm repeating intro to probability after i failed miserably last semester because of horrible professor.
You're short tutorials are helping me countless times more than a semester in my university.
Thank you so much!
i LOVE ur videos
that really helped me on my science test, Thank you😊
Its free real estate
This is a general rant: I understand this. What I cannot do is read the different questions and work them out. my H.W questions goes "This guy has these probabilities of so and so. if he chooses to do this and that but he is unable to that and that; what is the possibility of he does this but not that and that but not this." and I spend first 10 minutes trying to understand the back and forth and negation and positives and negatives and then trying to identify the usable data, and finally time went by and I have to move on because they gave 2 minutes to solve a paragraph. See that you explain this in 14 minutes. It is impossible to do one of these questions in 2 minutes, at least not for me. Hense my frustration. Much less when the professor and the book choose to work and explain the very easy questions and leave for the students the very hard ones without help. thank you for listening.
very nice as always Mr. Khan
I am studying for the FE exam, and I never took statistics in school, thanks so much!
What's FE exam?
Please I need to know what program you're using.
Thanks!
Thank you.
Hey Sal, as I'm watching your videos on combinatorics and how they fit into probability, I was having a hard time grasping the calculations to find the numerator (# of outcomes that satisfy the condition) until I thought about it in the opposite way you explained. You name a list of (k) heads and fit them into (n) flip slots (and frankly, I loose my intuition at this point) but I found it much more natural to rather name the flips (1 through n) and fit them into a group of size k. Then the expression (kCn / k Choose n) has a lot of meaning: "how many ways can I Choose exactly n heads from k trials.
Also, thank you for all your hard work that you put into Kahn Academy. You really are doing something great!
The phrase is nChoosek.
I luv Khan Academy
Beautiful summary :)
Very helpful, thank you!
ɷ Heey Friendss I Have F0unddddd Workingg Online Hacck visitt : - t.co/LGoGHuEMnA
i am saved now from my test!
@Khan Academy Confused on how last term is (n-(k-1)) as it always doesn’t work for example 10 flips with probability of getting four heads.
Isn’t it (n-(n-k))
youre videos are very helpful
Hi I have a doubt with these expression. what is the probability of K heads in N flips of the fair coin. The expression continues as N*(N-1) … (n-(k-1) …My doubt is that from where does k-1 comes in numerator part. Please assist?
same
thank u
Hi Sal. How is the site maintenance going? I haven't been able to do any exercises for a while now...
i think it's working now
To think I got into this with otherwise zero connection, motive, or use for this math other than to find out how absurd it was that I got 0 copies of the limited 6* Op out of 7 non-limited 6*s.
Other words, what's the probability of 0 successes out of 7 tries, when the limited 6* has a 35% chance of appearing when drawing any 6*.
Turns out it was ~4.9%.
7! / (0!(7-0)!) * .35⁰ * (1-.35)⁷-⁰
4.9% doesn't seem too absurdly improbable in hindsight, but it sure felt like it during the agonizing pulls.
Can you do the Birthday problem for us? ( k or is that n is 23, say)?
tanks
Hey Sal,
First; thank you for all your work.
Next; perhaps you can shed some light on a little oddity.....
If I try to determine the number of different ways 2 coin flips could result in 3 heads, and also the number of ways 2 coin flips could result in 4 heads, using the suggested formula, the results are counter intuitive.
The result for 3 heads from 2 flips seems somewhat logical, as the result is negative...but 4 from 2 yields a positive (but fractional) result. 2*1/(4*3*2*1)(-2*-1)
= 1/24
beng Factorials work for positive integers only I believe. It's in the definition.
@vpletap thanks for the info chief. I was getting the shakes from withdrawal XD
I was almost hoping to myself that he'd be introducing more Dr Seuss when he started with thing 1 and thing 2...
To think I got here from calculating differentials.
This must be a breeze
Let's get some real estate over here
(ran out of space)
.
How can there be 1/24th of a way that 2 coin flips can result in 4 heads?
..
....and why does there seem to be more of a way (?) to end up with 4 heads from 2 flips than 3 heads from two flips.
These both seem clearly impossible, but 4 heads seems even more impossible than 3 heads as a result of 2 coin flips.
Is God a counciousness or he's just the natural order of things?
both
how the last term was (n -(k-1))
n=5 k=3 so 5-(3-1)= 5-(2) = 3
Isn’t it (n-(n-k) ?
your choosing k things from n buckets. The kth spot is the last spot you have to fill and you have already filled (k - 1) spots of the head k when you reach this kth spot. So there are n - (k-1) possible ways to fill this kth spot. Just like how when you have filled 0 spots of k there are n ways to fill the first spot. Once you have filled 1 spot here are (n - 1) ways to fill the second spot etc etc
I know he is far better than any math teacher anyone has had, but his teaching style has issues. The way he writes things down is really messy, he talks too fast, and he makes a lot of mistakes, which he then realizes and erases, starting over. It doesn't even fulfill half of UA-cam's standards.
cope