Counting Probabilities with Combinatorics and the Factorial

[@MrMaxusy]

You have a fantastic talent for teaching. Your videos on control theory helped me through my studies and now that I'm finished I still come back just to learn more about all sorts of topics, refreshers, whatever. Keep it up!

[@JoseAguilar-fz9mb]

Great teacher, great lessons. The only word I can say is: Thanks. 🙏

[@charlieawesome5986]

Thank you these videos are absolutely golden ☺.

[@panizzutti]

I love this subject so much

[@oigxam1]

Excelent Video Doc! Love your work! From Mexico!

[@RolandoLopezNieto]

Great videos, thank you very much

[@ScienceMasterHK]

Waiting for a complete course Steve! Love it so far!

[@Carrymejane]

My man of the history

[@eliegakuba]

Thanks for these fantastic videos, Steve! Are there any accompanying exercises to help us grasp the concepts even better?

[@alarissacamila]

up up up!! i know we can probably just google it, but it would be nice to have some recommended list or any resources links

[@panizzutti]

This is so great thank you holy

[@PrntScrpt]

Thanks professor

[@magtazeum4071]

Lovely ❤

[@CaravaggioRoma]

thank you

[@alarissacamila]

I was so focused I got sad when the video ended. Is there any schedule for the next lessons? Im here for it either way, great explanation!!!

[@AmirHX]

17:49 What is the probability that I don't have a repeated letter?---> 7.85% ((26)!/(26-4)!)+10^2 / 26^4+10^2

[@samsung6980]

I think it should not be added, but multiplied instead.

[@Khazloo]

It should be multiplied. ((26)!/(26-4)!)*10^2 / (26^4)*10^2

[@Arickjd]

@@Khazloo I thought the same

[@arashamirian212]

awesome

[@studybuddy1015]

good video

[@dostdarali3999]

❤

[@andrezabona3518]

Professor, I have one question... lets say that order doesn't matter in your example of flip coins, i.e, for me ht==th (maybe I am trowing it the same time). Should it change the probability? Maybe now we should have P(A)=2/3 instead of 3/4? I did this experiment on python and it tells me that the probability continues to be 3/4 but I am not getting why!

[@TomaszBruxelles]

Should be 2/3. Imagine you have 3 sticks. One with ends h-h, 2nd t-h and last one t-t. Now you see you were right. Something is wrong in Python code.

[@andrezabona3518]

@@TomaszBruxelles My code looks like that: import random # Gera um número aleatório: 0 ou 1 a=0 for i in range(10000): r1 = random.randint(0, 1) r2 = random.randint(0, 1) if r1+r2==1 or r1+r2==2: a+=1 print(a*100/10000) I can't see the problem 😢

[@anthonybernstein1626]

It won't be 2/3 because your events don't have the same probability. Think of it this way: let's say we both observe the same coin flips, but I can tell the coins apart while you can't. From my perspective, every event (hh, ht, th, tt) has the same probability (1/4). You will observe hh and tt exactly when I do, so you'll agree that they have the probability 1/4. Then, since the probabilities of all possible events sum up to 1, the remaining event that you observe (ht or th) must occur with the probability 1/2.

[@chrlemes]

@@andrezabona3518 Andreza, você não modelou corretamente o experimento. No seu código, r1 tem 50% de probabilidade de ser 0 e 50% de ser 1, assim como r2. Então r1+r2=1 quando r1=0 e r2=1 ou r1=1 e r2=0. Como r1 e r2 são independentes, P(r1∩r2)=P(r1)*P(r2). Daí, P(r1=0∩r2=1)=P(r1=1∩r2=0)=0.5*0.5=0.25. Assim, você terá 50% de probabilidade de ter r1+r2=1. r1+r2=2 ocorrerá quando ambos forem 1, ou seja, P(r1=1∩r2=1)=0.5*.05=0.25. Dessa forma, no seu modelo, P(r1+r2=1 ∪ r1+r2=2) = 0.75.

[@nehalahane457]

How many five cards Run can I deal off top of deck.. What's the meaning of this sentence?

[@igorg4129]

But what is the formula for a case "with replacment" when order "does NOT matter"?

[@chrlemes]

The k-combination of n elements with replacement equals the k-combination of n+k-1 elements without replacement. It would be (n+k-1)!/k!

[@Aminasajeel]

n^r / r!

[@phillipisrael4343]

Need practice questions to assimilate this cognitive treasure.

[@nightskymusic5163]

Sir upload next videos

[@benstallone6784]

n choices, r samples, without replacement, order matters: n! / (n-r)! n choices, r samples, without replacement, order doesn't matter: n! / ((n-r)!*r!) n choices, r samples, with replacement, order matters: n^r n choices, r samples, with replacement, order doesn't matter: ???????

[@benstallone6784]

never mind. found it n choices, r samples, with replacement, order doesn't matter: (n+r-1)! / (r!*(n-1)!) not obvious at all! you have to translate the problem to r objects being separated by n-1 separators

[@benstallone6784]

n! / ((n-r)!*r!) is always an integer?!? 🤨

[@anthonybernstein1626]

Yes, think of Pascal's triangle. The the k-th number in the n-th row is nCk. Also every number is the sum of the two numbers above it, and the sides are when k=n or k=0, in both cases nCk=1, so every element is an integer.