Incredibly helpfull ! As you have explained both the binomial and geometric distributions you could sum up the Poisson and Non-binomial dists each in another video.The Poisson dist. will be easy to explain as a generalization of the binomial dist. while the non binomial dist will be slightly more challenging. Anyway i love your videos keep up the great work !!!
Hello mate. Your videos are much better to understand the consent of each distribution. If you wanna add a book recommendation with enough examples (for exercises) it would be helpful Best regards.
That is so great to hear! There are many great books out there, but unfortunately, we do not have a good resource that we know off-hand for this material.
hello thanks for the video! it was great~ There is sth that i wonder, can we find the right answers with cross solutions? For example, suppose there are 20 people, 8 of whom are girls, in a group. Let's form a team of 5 people and ask to have 3 girls. I'm trying to do cross solutions. When the answer to this question is solved with binomial, 0.2304 is found. I also want to solve it with geometric distribution. Statistics is a very rich content, is it not possible for us to do this? i couldn't solve it with geometric one, if i can i'll try to find the solutions with bernoulli and then possion etc
Great question! I'm not completely sure whether there is a solution to your problem using the geometric distribution. If you find a way to do it, I'd be very interested in seeing the solution!
i lost hope with probability , final test is one week later. I used chatgpt to teach me some stuff then lost . so I asked it if it can give me the lesson keyword . I wrote it an found You. A great hero indeed. Thanks bro for your hard working
I've never heard a more clear explanation of geometric distributions. The demonstration of multiplying 4 fails in a row followed by 1 success at 2:59 really helped me conceptualize this topic. Incredible video.
Great question! This video just provides them at a high level. If you would like to look more into the derivations and proofs for these formulas, take a look at a couple resources like this: www.cs.cornell.edu/courses/cs280/2008sp/280wk11_x4.pdf math.stackexchange.com/questions/1299465/proof-variance-of-geometric-distribution
You're saying the probability of the 3rd light being green is 12.8% but isn't this the same probability as long as it is green once and red the two other times? The calculation is 0.8 * 0.8 * 0.2 as if the order does not matter (you can "move around" the 0.8 and still get the same odds e.g. 0.2 * 0.8 * 0.8 = 0.128), but in your question, it does. How can we be sure that we are calculating the correct odds? Or is the more appropriate question "What is the probability of making it through once in three attempts?"
Great question! We calculated the probability in this way because we are only looking for the probability of it specifically being red then red then green in that order. If order did not matter and we were trying to determine the probability of making it through one light out of 3, we would need to add together all the different situations that would satisfy this. That would be (0.2 * 0.8 * 0.8) + (0.8 * 0.2 * 0.8) + (0.8 * 0.8 * 0.2) or 3 * (0.8 * 0.8 * 0.2). I hope this helps!
Great question! They are not quite the same. A good resource explaining them is below. We plan on making a video on this topic in the future. stats.libretexts.org/Courses/Saint_Mary's_College_Notre_Dame/MATH_345__-_Probability_(Kuter)/3%3A_Discrete_Random_Variables/3.4%3A_Hypergeometric_Geometric_and_Negative_Binomial_Distributions
I kept asking my TA on a geometric distribution problem I've been stuck on for days, but he was never clear on the concept of it. Thanks so very much for this video 🙏😁
I'm assuming you are asking how you would calculate the probability of hitting at most 3 red lights before hitting your first green light. If that is true, then you have to calculate the probability to hitting your first green light on your first, second, third and forth lights separately using this geometric distribution formula and add them all up. I hope this was helpful. If I interpreted your question incorrectly, please let me know!
This was very helpful, although I have one question: In the example, the question asks how many lights can we expect to hit before making it through one. So on the fifth light, we hit a green light, but we want to know the how many lights before we hit the green one, so I would choose 5-1=4 lights. I think I am interpreting the question wrong or thinking too far into it, but I'm not sure how to understand what the question is exactly asking.
Hi Greta, that is an awesome observation! You are completely right! We did use the expected value formula correctly, but we interpreted the result incorrectly. The expected value, as you mentioned, gives you the expected trial for which you would expect to get the first success. So we can expect to get through our first light on the 5th one. However, like you said, that would mean we hit 4 lights before this! With that being said, we did correctly interpret our final result despite it not explicitly answering the question. We said "on any given night we can expect to come across our first green light by the 5th one".
You guys are single-handedly raising my stats grades🙌 thank you SO much
You're doing an AMAZING job teaching this material ; may God help you the same way you help us.
Thank you so much for your positive feedback! I appreciate your support!
Thanks for the video,, it has been of use to me
You are very welcome! Thanks for watching!
I don't know why you have so low views. Your tutorials are so good.
Your support and kind words helps us continue to grow, and we really appreciate it! :)
@@AceTutors1 I understand it is difficult, but please do post more videos. I would love to learn more.
@@souravdey1227 We are definitely trying to put out more content soon!
Incredibly helpfull ! As you have explained both the binomial and geometric distributions you could sum up the Poisson and Non-binomial dists each in another video.The Poisson dist. will be easy to explain as a generalization of the binomial dist. while the non binomial dist will be slightly more challenging. Anyway i love your videos keep up the great work !!!
Thanks so much for your feedback! Those are great points and definitely things we want to cover soon!
Omg, found an amazing resource material for my Statistics class!
Thank you so much for your positive feedback and kind words! :)
Thanks for the clear and slow explanation, it solved my doubts and helped me learn the concept.
Thank you so much for your positive feedback and support! :)
@@AceTutors1 Very well said, clear AND slow.
Short and sweet.....1h lecture made me mess in this bt u make me understand in 7min.....wow thanks XD
Thank you so much for your kind words. I'm glad we were able to help
Hello mate.
Your videos are much better to understand the consent of each distribution.
If you wanna add a book recommendation with enough examples (for exercises) it would be helpful
Best regards.
That is so great to hear! There are many great books out there, but unfortunately, we do not have a good resource that we know off-hand for this material.
hello thanks for the video! it was great~
There is sth that i wonder, can we find the right answers with cross solutions?
For example, suppose there are 20 people, 8 of whom are girls, in a group. Let's form a team of 5 people and ask to have 3 girls. I'm trying to do cross solutions. When the answer to this question is solved with binomial, 0.2304 is found. I also want to solve it with geometric distribution. Statistics is a very rich content, is it not possible for us to do this? i couldn't solve it with geometric one, if i can i'll try to find the solutions with bernoulli and then possion etc
Great question! I'm not completely sure whether there is a solution to your problem using the geometric distribution. If you find a way to do it, I'd be very interested in seeing the solution!
i lost hope with probability , final test is one week later. I used chatgpt to teach me some stuff then lost . so I asked it if it can give me the lesson keyword . I wrote it an found You. A great hero indeed. Thanks bro for your hard working
That makes me so happy! I'm glad we were able to help!
Thank you so much! I really find this helpful. It is my first time to learn this.
You are welcome so much! I'm glad we were able to help!
I've never heard a more clear explanation of geometric distributions. The demonstration of multiplying 4 fails in a row followed by 1 success at 2:59 really helped me conceptualize this topic. Incredible video.
where do the formulae for the mean and sd come from?
Great question! This video just provides them at a high level. If you would like to look more into the derivations and proofs for these formulas, take a look at a couple resources like this:
www.cs.cornell.edu/courses/cs280/2008sp/280wk11_x4.pdf
math.stackexchange.com/questions/1299465/proof-variance-of-geometric-distribution
Good video, but I have the problem to find out the k, when the propability is given - how to do this??
You're saying the probability of the 3rd light being green is 12.8% but isn't this the same probability as long as it is green once and red the two other times? The calculation is 0.8 * 0.8 * 0.2 as if the order does not matter (you can "move around" the 0.8 and still get the same odds e.g. 0.2 * 0.8 * 0.8 = 0.128), but in your question, it does. How can we be sure that we are calculating the correct odds?
Or is the more appropriate question "What is the probability of making it through once in three attempts?"
Great question! We calculated the probability in this way because we are only looking for the probability of it specifically being red then red then green in that order. If order did not matter and we were trying to determine the probability of making it through one light out of 3, we would need to add together all the different situations that would satisfy this. That would be (0.2 * 0.8 * 0.8) + (0.8 * 0.2 * 0.8) + (0.8 * 0.8 * 0.2) or 3 * (0.8 * 0.8 * 0.2). I hope this helps!
how is it possible that the teacher expects us to memorize this and 20 other statistic topics in our head with no sheets allowed
Statistics courses can definitely be pretty intense! I hope we were able to help a bit!
@@AceTutors1 fuck that teacher and his exam
Can you tell prob. Of 3rd light being the 4th one that is green
Short and comprehensive good job man
Thank you for your feedback and for watching!
Volume is so low barely can hear you
But ig mean is qover p
Hi, I'm not exactly sure what you are referring to. Can you provide some more information?
Where does square root q come from 🤔
Is hypogeometric and geometric probability distribution the same?
Great question! They are not quite the same. A good resource explaining them is below. We plan on making a video on this topic in the future.
stats.libretexts.org/Courses/Saint_Mary's_College_Notre_Dame/MATH_345__-_Probability_(Kuter)/3%3A_Discrete_Random_Variables/3.4%3A_Hypergeometric_Geometric_and_Negative_Binomial_Distributions
ohh my you are amazingggggggggggggg the best so far on youtube God BLESS YOU
No you are amazing! Thank you for watching!
I kept asking my TA on a geometric distribution problem I've been stuck on for days, but he was never clear on the concept of it. Thanks so very much for this video 🙏😁
You are so welcome! Thanks for watching!
I have to study probability for a quant trading interview and I have never studied geometric distribution. This helped a lot cheers
That makes me so happy! Thanks for watching!
Can we use
Mean= q/p
Great question! Unfortunately no. That would be something else. The mean is 1/p for this distribution. I hope this helps
perfectly explained 👍
Thank you for your kind words, Bentley!
Awesome explanation
I liked the way of teaching. Please keep it up.. ♥︎
Thank you so much! I will! :)
Well presented
Thank you for your kind words!
it's amazing thank you so match good luck
You are welcome! :)
The videos are soo helpful thanks alot 😊
But please can you try doing more than 1 examples
hi js wanted to ask how did you get the q=0.8? how?
1-0.2=0.8
I liked the video it was short and up to the point
That's definitely what we go for! I'm glad we could help
Amazing man
Thank you so much
Great explanation!
very helpful
Thanks again Jonita! I appreciate your kind words! :)
you made it so easy thank you.. god bless you
You are sooo welcome! Thanks for watching!
If the question ask for the at most three traffic how should we solve it?
I'm assuming you are asking how you would calculate the probability of hitting at most 3 red lights before hitting your first green light. If that is true, then you have to calculate the probability to hitting your first green light on your first, second, third and forth lights separately using this geometric distribution formula and add them all up. I hope this was helpful. If I interpreted your question incorrectly, please let me know!
@@AceTutors1 is it like this:
P(x≤3)=1-p(x=0)+p(x=1)+p(x=2) ?
Well explained, provide the one for poisson distribution.
great video !!!
One of the most clear tutorial channels on RUclips, underrated
🙏✨
:)
breathtakingly easy breezy !!!!! I am so relieved.
I like it!
Thanks its really amazing
I'm so glad you think so! Thanks for watching!
♥️
cool tut!!!
Thank you again!
Best way of teaching
Thank you so much for your kind words!
it is so informative. Thank u so much
i did not get it, Where did the 8 come from?
1-0.2=0.8
Thank you very much
You are very welcome! Thanks for watching!
thank you sir❤
You are so very welcome!!
Thanks a lot sir
Thank you Ryan for watching.
great video
Thank you Richard!
Very clear
Thank you! I'm happy you think so!
great video
Thank you! Thanks for watching!
This was very helpful, although I have one question: In the example, the question asks how many lights can we expect to hit before making it through one. So on the fifth light, we hit a green light, but we want to know the how many lights before we hit the green one, so I would choose 5-1=4 lights. I think I am interpreting the question wrong or thinking too far into it, but I'm not sure how to understand what the question is exactly asking.
Hi Greta, that is an awesome observation! You are completely right! We did use the expected value formula correctly, but we interpreted the result incorrectly. The expected value, as you mentioned, gives you the expected trial for which you would expect to get the first success. So we can expect to get through our first light on the 5th one. However, like you said, that would mean we hit 4 lights before this! With that being said, we did correctly interpret our final result despite it not explicitly answering the question. We said "on any given night we can expect to come across our first green light by the 5th one".
Very good
Thank you!
Great work!
MAN this vdo is really really best.....
Thank you so much for your kind words! I'm glad we were able to help out!
❤