Literally the most wholesome professor I met at Stanford. Chris is one of the most genuinely good people that I have ever met that used ML for the good of this world. I miss taking CS109.
Agree ! What a way to teach and what a pleasurable way to learn! I am so glad I can at least experience this online - thank you prof Chris for posting your videos online
@@ovge5696 Course materials are already there in the description. You are the one creating drama for no reason. I will compliment the person who are good at their job. You are no one to stop me.
00:07 Introduction to CS109 and Professor Chris 02:14 Passion for teaching and research in education technology 05:46 CS109 course relies heavily on material from math 51 or CME 100. 07:32 Understanding the unit choice and workload for CS109 11:10 Encouraging lecture attendance through incentive system 12:46 Multiple avenues for getting help and resources are available for CS109 students. 16:40 Underestimation of complexity in language translation. 18:09 Advancements in AI and technology have led to solving complex problems like speech transcription and protein folding. 21:32 Introduction to the complexity of problem-solving and understanding images with distributions 23:15 Understanding images through pixel values and the concept of visual cortex 26:28 Neurons and layers of hidden neurons are combined to create a bigger probability machine. 28:13 Intelligence in model comes from well-set weights 31:40 Using probability theory to solve real-world problems. 33:22 Fundamental probability theory in CS109 drives understanding of natural phenomena and application. 36:36 Probability enhances understanding and enables cool applications 38:04 Importance of using CS109 algorithm for accurate, shorter exams 41:09 Probability theory helps update beliefs based on evidence. 42:46 Probability theory foundation starts with counting 46:00 Understanding counting outcomes with dice rolls. 47:38 Understanding the step rule of counting in probability 51:09 Understanding the steps in determining unique images using probability 52:50 There are 5.8 times 10 to the power of 77 unique images from 12 pixels. 56:37 Finite resources but infinite combinations 58:27 Counting outcomes from mutually exclusive sets using addition. 1:02:12 Using counting and or rule to break down large problems into smaller pieces. 1:03:58 Understanding binary counting and possibilities 1:07:29 Counting with overlapping sets 1:09:24 Counting with steps or counting with or leads to specific formulas for counting outcomes. 1:12:50 Using step rule of counting to determine total ways of organizing letters
This professor has the nicest most hypercurious most friendly most exploratory most lovely etc. energy ever, he's amazing, I love him, he's super contagious 😄 💗 🥰 💖
What a beautiful way of teaching. First make them understand the WHY of it then the HOW of it. The WHY is so important because this is where students might fall in love with the subject and maybe learn on their own. Its all about igniting that curiosity. No one class can teach them everything. But if you make them curious and the foundations strong, they will go on and teach themselves. Because they are in love with it. These Teachers are a gift to the world.
A nice way to visualize the last problem. let us first assume all the letters are distinct in the word 'BOBA' (or BOB'A). then there would naturally be 4! (24) ways to arrange it. Now, we will manage the B that was repeated. Notice how how in every permutation we have counted, the B was permuted as it was distinct but now all those permutations must be counted only once. To approach this, we would ask "What fraction of these 4! permutations are actually distinct?". To answer this, in any given random permutation lets say OB'BA, if we only look at the two Bs we can arrange them in 2! ways which is OB'BA and OBB'A, therefore the probability of any random permutation being unique 1/2!. Since this is true for each one of the 24 permutations, the total distinct permutation of the word BOBA is 4!*(1/2!) = 12. The illustrated explainantion was just to extend the idea up to multiple repeating letters. Therefore in case of the word MISSISSIPPI. the total number of distinct permutations are 11!*(1/4!)(1/4!)(1/2!).
Finally, I landed on this series and completed the first lecture. What an incredible lecture! I am determined and hopeful for the next lectures. Probability isn't as scary as it was before this lecture. And yeah!! after graduation, I'll be one of those who would love to build further upon the foundation of probability.
Galera, isso é ouro. O mais próximo que nós reles mortes oriundos de um interior brabo de alguma cidade do Brasil podemos chegar de um ensino de excelência. Aproveitem
Number of distinct arrangements for "BOBA": The 2 B's create combinations that can repeat themselves, so can use the Sum Rule of Outcomes, and eliminating the duplicates - Placing B in the first spot, we have 3 possible combinations total when placing the second B (B 1 2 3) - Placing B in the second spot, we have 2 remaining combinations possible ( _ B 1 2) - Placing B in the before last position, we have 1 possible combination ( _ _ B B) So total possible combinations of B without repetition is 3 + 2 + 1 = 6 For placing A and O, since it is an independent step from determining the B combinations, we can use the Step Rule of Counting and multiply with our 6 possible B outcomes. We only have 2 possible arrangements for A and O: either A first, O second, or vice versa (AO/OA) So we multiply the sum of possible B combinations, to the possible permutations of AO/OA, which is 2: 6 x 2 = 12 Therefore, there are 12 different ways to arrange the letters in "BOBA".
he said one sentence if you want to study on your own probability topic, you can't maybe you think Calculus Algebra Statistics are difficult to learn but that's not the true really the hardest subject I found in recent day is probability we studied probability in graduation and high school you found probability is easiest topic to solve it but if you want to use probability in real world application like AI it is harder to do because it confuse a lot due to randomness and uncertainty things
Can anyone confirm that the final answer for the BOBA question = 12? 4*3*2*1 then divided by two because the B's have been doubled. The problem set app only available for Stanford students and I'm not. Thanks.
I made the code for it and yes I can confirm that the answer is 12. ['B', 'A', 'O', 'B'] ['B', 'B', 'O', 'A'] ['B', 'A', 'B', 'O'] ['O', 'B', 'A', 'B'] ['B', 'B', 'A', 'O'] ['O', 'A', 'B', 'B'] ['O', 'B', 'B', 'A'] ['A', 'B', 'O', 'B'] ['A', 'O', 'B', 'B'] ['B', 'O', 'A', 'B'] ['A', 'B', 'B', 'O'] ['B', 'O', 'B', 'A'] But I am not sure about the calculation of expected outcome, 4*3*2*1 divided by 2. Why its 4 ? because as my understanding, we only have 3 choice [B, O, A].
@@alif_ni It's a common problem for students of Permutation and Combinations. If there are total n objects, of which m are alike of the same kind, p are alike of 2nd kind, then the total no. of unique combinations can be given by - n!/(m!*p!). For the case of BOBA, Total 4 places out of which 2 are alike (B in this case). Therefore the total no. of combinations become 4!/2! = 12. Similarly for MISSISSIPPI - total unique permutations can be 11!/(4!*4!*2!) = 51975. But here the instructor's intention is not to get the answer but to figure out if you can figure out the cases in counting upto these numbers.
I wonder if they used ML models to automatically blur some frames and bleep the audio... It at least have the models identify them. (Example: 1:00:02) Edit: Lol ... Maybe an ML model to find spelling issues (slide title at 1:01:05) Edit 2: Instructor's excitement is contagious!
No they didn't, as a video editor I can tell u this much. Even if it is hypothetically possible it's just so much easier to go in post prod and do it yourself. for the blurred frames specifically, it was likely made to avoid potential copyright issues (right now it won't but potentially in the future if copyright laws change it might) u can't train a model to detect that lol, as all copyrighted materials are vastly different from each other, can be a photo, can be a real object like a ball with a logo, can be a background song. Models learn to distinguish between similar objects.
Why do computer scientists use Mac instead of Windows? I use Windows but I am interested in Mac a little because so many computer scientists use Mac. I don't know why.
Hi there, thanks for your question! You can find the prerequisites on this page:online.stanford.edu/courses/cs109-introduction-probability-computer-scientists
You are considering 3 spots here. But there are 4 Spots. So more permutations are possible. For Example: ------------------------ with three spots and B, O, A in this order we get one permutation only: BOA but with another B available: we get BBOA and BOAB with the same order.
I'm surprised how Stanford is a top Uni, but their pedagogy is excellent such that any student irrespective of their academic level would understand what's is taught in class. Very different from the pedagogy approach in MIT where they have a expect the student to be excellent so they have a more strict/pedantic approach to teaching.
I'm certainly grateful for having free content but I'd like to point out that it would be fitting to be a bit more precise numerically. Rounding is fine but getting a something times 10^77 instead of 10^86 is to be off by several orders of magnitude. I don't think that's just me being nit-picky...
Literally the most wholesome professor I met at Stanford. Chris is one of the most genuinely good people that I have ever met that used ML for the good of this world. I miss taking CS109.
What an energtic professor!! Very amazed by his energy and the way of his teaching!!
Agree ! What a way to teach and what a pleasurable way to learn! I am so glad I can at least experience this online - thank you prof Chris for posting your videos online
Just give me the course material, no need for theatrics.
@@ovge5696 if you dont like, dont comment. Simple.
That defies the entire purpose of the comment section.
@@ovge5696 Course materials are already there in the description. You are the one creating drama for no reason. I will compliment the person who are good at their job. You are no one to stop me.
With the energetic voice and great patience of the professor, looking forward to the next section!
00:07 Introduction to CS109 and Professor Chris
02:14 Passion for teaching and research in education technology
05:46 CS109 course relies heavily on material from math 51 or CME 100.
07:32 Understanding the unit choice and workload for CS109
11:10 Encouraging lecture attendance through incentive system
12:46 Multiple avenues for getting help and resources are available for CS109 students.
16:40 Underestimation of complexity in language translation.
18:09 Advancements in AI and technology have led to solving complex problems like speech transcription and protein folding.
21:32 Introduction to the complexity of problem-solving and understanding images with distributions
23:15 Understanding images through pixel values and the concept of visual cortex
26:28 Neurons and layers of hidden neurons are combined to create a bigger probability machine.
28:13 Intelligence in model comes from well-set weights
31:40 Using probability theory to solve real-world problems.
33:22 Fundamental probability theory in CS109 drives understanding of natural phenomena and application.
36:36 Probability enhances understanding and enables cool applications
38:04 Importance of using CS109 algorithm for accurate, shorter exams
41:09 Probability theory helps update beliefs based on evidence.
42:46 Probability theory foundation starts with counting
46:00 Understanding counting outcomes with dice rolls.
47:38 Understanding the step rule of counting in probability
51:09 Understanding the steps in determining unique images using probability
52:50 There are 5.8 times 10 to the power of 77 unique images from 12 pixels.
56:37 Finite resources but infinite combinations
58:27 Counting outcomes from mutually exclusive sets using addition.
1:02:12 Using counting and or rule to break down large problems into smaller pieces.
1:03:58 Understanding binary counting and possibilities
1:07:29 Counting with overlapping sets
1:09:24 Counting with steps or counting with or leads to specific formulas for counting outcomes.
1:12:50 Using step rule of counting to determine total ways of organizing letters
Great energy from Professor! Give this man a thumbs up.
I didn't think that machine learning lectures could be this fun, thank you very much.
He is so wonderful and handsome and an awesome teacher. Chris, thank you so much for everything you do. The world needs more good people like you.
Oh! I just love the energy of the professor. So engaging.
This professor has the nicest most hypercurious most friendly most exploratory most lovely etc. energy ever, he's amazing, I love him, he's super contagious 😄 💗 🥰 💖
love him! thankyou from the small town in India!
What a beautiful way of teaching. First make them understand the WHY of it then the HOW of it. The WHY is so important because this is where students might fall in love with the subject and maybe learn on their own. Its all about igniting that curiosity. No one class can teach them everything. But if you make them curious and the foundations strong, they will go on and teach themselves. Because they are in love with it. These Teachers are a gift to the world.
Wow, this guy is great and amazing and everything else which is related to great professor!
A nice way to visualize the last problem.
let us first assume all the letters are distinct in the word 'BOBA' (or BOB'A).
then there would naturally be 4! (24) ways to arrange it. Now, we will manage the B that was repeated.
Notice how how in every permutation we have counted, the B was permuted as it was distinct but now all those permutations must be counted only once. To approach this, we would ask "What fraction of these 4! permutations are actually distinct?".
To answer this, in any given random permutation lets say OB'BA, if we only look at the two Bs we can arrange them in 2! ways which is OB'BA and OBB'A, therefore the probability of any random permutation being unique 1/2!.
Since this is true for each one of the 24 permutations, the total distinct permutation of the word BOBA is 4!*(1/2!) = 12.
The illustrated explainantion was just to extend the idea up to multiple repeating letters.
Therefore in case of the word MISSISSIPPI. the total number of distinct permutations are 11!*(1/4!)(1/4!)(1/2!).
Thank you bro
Finally, I landed on this series and completed the first lecture. What an incredible lecture! I am determined and hopeful for the next lectures. Probability isn't as scary as it was before this lecture. And yeah!! after graduation, I'll be one of those who would love to build further upon the foundation of probability.
15:20 the lecture starts
48:20 step rule of counting
58:00 sum rule of counting
1:01:00 6 bit problem and inclusion exclusion rule
Sir,you are the only one teacher who make me more energetic after attending lecture....👌👌
Looking forward to other course taught by this teacher, he is so energetic and enthusiastic! What a great lecture
How good are you! This lecture is the best one I have ever seen regarding to introduction of probability theory.
The spiritual art of counting it all joy. Having faith that no matter what input 😂) expect the best outcome
Checked to see if I had 1.25x speed enabled at the beginning. Lol. Thank you for this highly valuable course!
Galera, isso é ouro. O mais próximo que nós reles mortes oriundos de um interior brabo de alguma cidade do Brasil podemos chegar de um ensino de excelência. Aproveitem
Amazing Professor!!! Thank you Chris
Thank you for this great course, having access to the Problem sets would be amazing! they are a very important part of the learning experience 😢
Seconded!
Number of distinct arrangements for "BOBA":
The 2 B's create combinations that can repeat themselves, so can use the Sum Rule of Outcomes, and eliminating the duplicates
- Placing B in the first spot, we have 3 possible combinations total when placing the second B (B 1 2 3)
- Placing B in the second spot, we have 2 remaining combinations possible ( _ B 1 2)
- Placing B in the before last position, we have 1 possible combination ( _ _ B B)
So total possible combinations of B without repetition is 3 + 2 + 1 = 6
For placing A and O, since it is an independent step from determining the B combinations, we can use the Step Rule of Counting and multiply with our 6 possible B outcomes.
We only have 2 possible arrangements for A and O: either A first, O second, or vice versa (AO/OA)
So we multiply the sum of possible B combinations, to the possible permutations of AO/OA, which is 2:
6 x 2 = 12
Therefore, there are 12 different ways to arrange the letters in "BOBA".
he said one sentence if you want to study on your own probability topic, you can't maybe you think Calculus Algebra Statistics are difficult to learn but that's not the true really the hardest subject I found in recent day is probability we studied probability in graduation and high school you found probability is easiest topic to solve it but if you want to use probability in real world application like AI it is harder to do because it confuse a lot due to randomness and uncertainty things
its wowww giving that perspective make me love the "Chris"
Now I see why Stanford and many other American unis are regarded so highly
Thank you, Chris! Thank you so much
33:34 Kannada 💛❤ which is the National Language of Indian Subcontinent
Can anyone confirm that the final answer for the BOBA question = 12?
4*3*2*1 then divided by two because the B's have been doubled.
The problem set app only available for Stanford students and I'm not. Thanks.
Answer is 12 👍
I made the code for it and yes I can confirm that the answer is 12.
['B', 'A', 'O', 'B']
['B', 'B', 'O', 'A']
['B', 'A', 'B', 'O']
['O', 'B', 'A', 'B']
['B', 'B', 'A', 'O']
['O', 'A', 'B', 'B']
['O', 'B', 'B', 'A']
['A', 'B', 'O', 'B']
['A', 'O', 'B', 'B']
['B', 'O', 'A', 'B']
['A', 'B', 'B', 'O']
['B', 'O', 'B', 'A']
But I am not sure about the calculation of expected outcome, 4*3*2*1 divided by 2.
Why its 4 ? because as my understanding, we only have 3 choice [B, O, A].
@@alif_ni It's a common problem for students of Permutation and Combinations. If there are total n objects, of which m are alike of the same kind, p are alike of 2nd kind, then the total no. of unique combinations can be given by - n!/(m!*p!). For the case of BOBA, Total 4 places out of which 2 are alike (B in this case). Therefore the total no. of combinations become 4!/2! = 12. Similarly for MISSISSIPPI - total unique permutations can be 11!/(4!*4!*2!) = 51975. But here the instructor's intention is not to get the answer but to figure out if you can figure out the cases in counting upto these numbers.
super amazing thank you stanford
damn this is so intuitive thanks for posting
Thank you!
Can you please share slides and problem sets?
Nah, they're all his. You get none
I wonder if they used ML models to automatically blur some frames and bleep the audio... It at least have the models identify them. (Example: 1:00:02)
Edit: Lol ... Maybe an ML model to find spelling issues (slide title at 1:01:05)
Edit 2: Instructor's excitement is contagious!
No they didn't, as a video editor I can tell u this much. Even if it is hypothetically possible it's just so much easier to go in post prod and do it yourself.
for the blurred frames specifically, it was likely made to avoid potential copyright issues (right now it won't but potentially in the future if copyright laws change it might)
u can't train a model to detect that lol, as all copyrighted materials are vastly different from each other, can be a photo, can be a real object like a ball with a logo, can be a background song. Models learn to distinguish between similar objects.
Why do computer scientists use Mac instead of Windows? I use Windows but I am interested in Mac a little because so many computer scientists use Mac. I don't know why.
How do you get access to the program sets? :(
pow(17e6, 12)=5.8e86
33:48 that is ಕನ್ನಡ
Amazing professor
What are prerequisites? Calculus?
Hi there, thanks for your question! You can find the prerequisites on this page:online.stanford.edu/courses/cs109-introduction-probability-computer-scientists
I think the answer is 6. Since B and A is repeated so we are left with 3 unique choices B O A = 3 * 2 * 1 = 6
You are considering 3 spots here. But there are 4 Spots. So more permutations are possible.
For Example:
------------------------
with three spots and B, O, A in this order we get one permutation only: BOA but with another B available: we get BBOA and BOAB with the same order.
Is there any way for students outside Stanford could solve the problem?
Love him!!!
Thanky you for sharing
Good vibes, salam dari Indonesia!
what a great video man i love it
Thanks for watching and for your comment!
They put an "I've learned more in this video than in a whole semester of university!" kinda youtuber in the classroom lmao
Can we have pdf lecture?
I'm surprised how Stanford is a top Uni, but their pedagogy is excellent such that any student irrespective of their academic level would understand what's is taught in class. Very different from the pedagogy approach in MIT where they have a expect the student to be excellent so they have a more strict/pedantic approach to teaching.
good content
this is why it's called Stanford.
Like him or not, he definitely has the Ch-ris-ma 😉
Came here from the Wired video. Who signs off on my degree when I’m done?
I'm certainly grateful for having free content but I'd like to point out that it would be fitting to be a bit more precise numerically.
Rounding is fine but getting a something times 10^77 instead of 10^86 is to be off by several orders of magnitude.
I don't think that's just me being nit-picky...
I am sorry, I don't try to be rude but, whats the point of this vids if there is no any homework/assignment available on the actual website ?
If u check cs109 2022 the problem sets are avaliable on the website
Come back to Kenya 😂
,💻🔨🚀
learning probability for 12th grade from standford lol
Try the hw
Math does bloom😂