That was a true example of connecting provable formulas to intuition. Nice job man. keep it going
3 года назад+5
This was ridicolously helpful compared to classic way of "teaching" calculus where you are Just exposed to formulas that are completely out of the blue. Thank you so much
Man you are awesome!! That was a great derivation of poisson distribution without any introduction of binomial distribution. You basically derived exponential from minimum set of assumptions and that's always beautiful to watch. Thanks
for the generalization , how do we prove that : N x (N-1) x ...... (N-k+1) / (n-k) approaches--> mean ^ k as N --> infinity . anyways love the take on this proof.
I have a question about your first example. Since the berries are indistinguishable, surely there would be less than 12^N combinations, since swapping two berries would count as the same combination? I thought you would need to use the stars and bars method so I got (N+11)C(11) different combinations. Am I mistaken that distributions are the same as combinations?
@@Zero-tg4dc imagine that we have 2 berries and 2 jams, and there are exactly 4 = 2^2 choices to put our berries to our jams, simulate to 1 berry and 3 jams (3^1 = 3 choices), 3 berries and 1 jam (1^3 = 1 choice) (^),...and finally x berries and y jams we have y^x choices, that how i understand it.
That was a true example of connecting provable formulas to intuition.
Nice job man. keep it going
This was ridicolously helpful compared to classic way of "teaching" calculus where you are Just exposed to formulas that are completely out of the blue. Thank you so much
Man you are awesome!! That was a great derivation of poisson distribution without any introduction of binomial distribution. You basically derived exponential from minimum set of assumptions and that's always beautiful to watch. Thanks
Thank you ! I liked this derivation so much that I decided to share it with others
for the generalization , how do we prove that : N x (N-1) x ...... (N-k+1) / (n-k) approaches--> mean ^ k as N --> infinity . anyways love the take on this proof.
without thinking about it, wouldn't you have (n-k)^k or smith at the bottom of this fraction ?
thank you a lot for the exquisite explanation
Very good, but can you carry it out for k berries, I tried, but not getting it.
this was really good!
I have a question about your first example. Since the berries are indistinguishable, surely there would be less than 12^N combinations, since swapping two berries would count as the same combination? I thought you would need to use the stars and bars method so I got (N+11)C(11) different combinations. Am I mistaken that distributions are the same as combinations?
Jars are not indistinguishable for us in this case. So each berry has N choices, n^N in total
@@MetaMaths But is each berry indistinguishable?
@@Zero-tg4dc imagine that we have 2 berries and 2 jams, and there are exactly 4 = 2^2 choices to put our berries to our jams, simulate to 1 berry and 3 jams (3^1 = 3 choices), 3 berries and 1 jam (1^3 = 1 choice) (^),...and finally x berries and y jams we have y^x choices, that how i understand it.
Gold
Thanks ! Loved your bridge game video !
@@MetaMaths thanks!!!