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
Hi, I am a junior college student at the UoPeople enrolled in Statistics. Thank you for sharing so valuable material. Looking forward to learn from you!😇😇🤣😂🙂
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 🙏😁
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.
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'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! 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
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!
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!
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
Can you help me understand why "how many lights can you expect to hit before making it through one?" is the mean? :( I didnt understand that at all. :(
Out of the 20 videos I’ve watched on geometric distribution this is the first to explain it in a way I understand. Thank you so much
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!
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!
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!
Hi, I am a junior college student at the UoPeople enrolled in Statistics. Thank you for sharing so valuable material. Looking forward to learn from you!😇😇🤣😂🙂
Hey, me too...
One of the most clear tutorial channels on UA-cam, underrated
You guys are single-handedly raising my stats grades🙌 thank you SO much
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!
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.
Omg, found an amazing resource material for my Statistics class!
Thank you so much for your positive feedback and kind words! :)
Short and comprehensive good job man
Thank you for your feedback and for watching!
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
ohh my you are amazingggggggggggggg the best so far on youtube God BLESS YOU
No you are amazing! Thank you for watching!
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!
Best way of teaching
Thank you so much for your kind words!
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.
breathtakingly easy breezy !!!!! I am so relieved.
you made it so easy thank you.. god bless you
You are sooo welcome! Thanks for watching!
Thanks for the video. I would like to point out the "p-value" which might be confusing for first time learners.
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 liked the way of teaching. Please keep it up.. ♥︎
Thank you so much! I will! :)
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
I finally understand geometric distribution!
amazing explanation now i don't need to cram formula
it is so informative. Thank u so much
Awesome explanation
Thanks for the video,, it has been of use to me
You are very welcome! Thanks for watching!
perfectly explained 👍
Thank you for your kind words, Bentley!
Great work!
Great explanation!
it's amazing thank you so match good luck
You are welcome! :)
Well explained, provide the one for poisson distribution.
The videos are soo helpful thanks alot 😊
But please can you try doing more than 1 examples
Well presented
Thank you for your kind words!
Thanks its really amazing
I'm so glad you think so! Thanks for watching!
great video !!!
Amazing man
Thank you so much
Thank you very much
You are very welcome! 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".
cool tut!!!
Thank you again!
very helpful
Thanks again Jonita! I appreciate your kind words! :)
thank you sir❤
You are so very welcome!!
Thanks a lot sir
Thank you Ryan for watching.
Very clear
Thank you! I'm happy you think so!
Good video, but I have the problem to find out the k, when the propability is given - how to do this??
I like it!
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!
Aye great explanation! :)
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
can you do a video on Poisson distri
great video
Thank you! 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) ?
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
Very good
Thank you!
Can you tell prob. Of 3rd light being the 4th one that is green
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!
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
Where does square root q come from 🤔
hi js wanted to ask how did you get the q=0.8? how?
1-0.2=0.8
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
Can you help me understand why "how many lights can you expect to hit before making it through one?" is the mean? :( I didnt understand that at all. :(
🙏✨
:)
i did not get it, Where did the 8 come from?
1-0.2=0.8
♥️
But ig mean is qover p
Hi, I'm not exactly sure what you are referring to. Can you provide some more information?
Volume is so low barely can hear you
❤
great video
Thank you Richard!