This guy just explained an entire chapter on one of my modules in 15mins , compared to my useless prof and teaching assistant who couldn't do it with 3 lectures and 120 pages of notes . Hands down . Best . Tutor . Ever . Hats off to you Sir .
Very good video, just like the others, on the playlist! This line made me chuckle though: (5:20) "There are 26 successes in the population, the 26 females and 32 failures, the 32 men."
Education needs to be totally revamped. We don't need so many professors giving lectures, one for each subject in every schools in the world. We just need a handful of great teachers creating online content per subject. Instead, we need lots of great coaches and mentors to help individual learners. Learners will watch online lectures at their own pace. Coaches and mentors will work with individual learners to guide them, introduce good learning resources, push them to achieve and most important of all, motivate them. That's the new model of education.
Actually that could be a way to progress, however while dealing with larger masses who are willing to have a tutor or professor(i bet 3/4 univeristy undergrads are willing to have) it would still be needed to create some kind of classes, and for classes you need halls, and that cluster of halls would be called universitt ahaim. So basically we would go on the same cycle, of course with small (compared to scale) changes. But the system is actually worth thinking.
I've been having so much trouble with these concepts, especially because I am taking summer session and my professor teaches incredibly fast and is barely available for office hours. These videos have been helping me more than my professor in the last couple weeks! Thanks so much!
0:15 Binomial Distribution 4:32 Hypergeometric Distribution 7:27 Poisson Distribution 9:31 Geometric Distribution 13:02 Does not fit into any of the distribution
what a lecture. Just superb. whenever feel like revising probability i just watch these lectures and concepts get refreshed. Thanx is just a small word. Hats off professor.
Thanks for the help, sir, really helpful video as i started from 1st video , i understood everything you taught. after 9 years of you uploaded this video it is still the best video on statistic on youtube.
Dear JBStatistics can you put more applied examples like these , it's always good to see real world application of statistics than what we read in books like the ball , coin toss , choosing cards from a deck etc in book. These examples are amazing and the more students get to see these explained and solved by you. The more kids will start loving math, probability and statistics. Hope you keep posting these awesome videos
Thanks for the feedback and kind words. I try my best to incorporate real-world examples in my problems, and I'll continue to do that when I make new videos in the near future.
Been searching youtube looking to help me study for a probability test, pretty glad I've found jb statistics! Thankyou, sir, I like the way you teach, best I've found so far. Anybody want to recommend some other youtube channel for maths/stats tutorials? You can never have too much learnin'! (Must be up to this standard or close)
Hands down, one of the best channels for Probability & Statistics on RUclips! Finished my entire semester's worth of course content in only a few hours thanks to you.
sir when you started with the hyper geometric distributions i laughed so much as i remembered the problem about counting females independently and that would cost you a job :D awesome videos really.. thank you!
BIG massive thanks to you JB, I nearly learn my whole first year Stat from your video, please go on and post more The way you explain and teach is way better than my professors I reckon Once again Cheers mate
On the first example you would have a success rate for one mission, but the real numbers to consider when designing a system would be the neg binom for failure in order to ensure its reliability over time. Given the size of our military, if there were 10k systems used once on 100 missions, that would be a rather high failure rate over the 1m uses, and 7 or 8 components would be much more suitable when it comes to reliability. Practicality must be assessed however, so 6 is probably as good as you'll get
when Tom goes crying to his boss that he ruined the first part, his boss should tell him “ oh , sorry Tom , forgot to tell you …that’s exactly how we wanted that one…good job!”
this is a little late but just wanna say for 12:56, summing many terms isn't a lot of writing cause the geometric sum is commonly known, so we can use a_1=p and (common ratio)=q.
I spent 2 days creating a tool to calculate I think...binomial I forget which one specifically. My online college course rushed us through 30 day classes I learned way more from the internet - and videos like this. Thank you for keeping it simple ;)
@jbstatistics I am curious about the answer at @12:59. i believe that the P(X>6) = 1 - 0.0183 (the value that was computed at the end). If we have an infinite amount of trials wouldn't we eventually get one red slot? We just won't get it in the first 6 trials but certainly red will come up at some point.
Yes, and that's partly what the probability I calculate is based on. The question asks for the probability that the **first** time the ball lands in a red slot comes after the 6th spin. That's definitely not a certainty. The probability the first red happens at some point is 1, yes.
The conditions of a hypergeometric are not satisfied there. For a random variable to have a hypergeometric distribution, we need to be drawing items without replacement from a source that contains a certain number of successes and a certain number of failures. That is not happening in the last example. Don't get trapped into the false line of thinking that if the trials are dependent, then it must be hypergeometric; it simply doesn't work that way. There are many different ways for dependence to arise.
I love your videos, they helped me so much! But, at 10:57, I thought chance of failure should be 19/37 since the green slot is an additional slot of failure added to the 18 black slots...correct me if I am wrong!
In that example, the probability of failure is 19/37, as you state, but that is the probability I use in the video. I write the probability of failure as 1-18/37, which is 19/37.
You are awesome......really helped me so much....Why don't you make videos on joint random variable or conditional distribution........I needed those badly ....love from bangladesh :)
Thanks! I'm glad to be of help. I frequently get requests for videos on joint probability distributions, and they are high on the priority list, but I'm struggling to find time to make them. One of these days.
thanks for the reply... :) you know, you make videos so much up to the point....i have never seen such a wonderful combination of explanation along with a very relevant example.....i got very good grade in mid semester only just watching those videos bt sadly my semester final syllabus werent covered in your videos which include joint random variable......i recommended all of my friends about your videos before probability exam...now after watching ur videos they call you "JB BOSS" :) keep making videos keep saving lives :p
In 10:18, why is that formula used? Why isn't it "The permutations P(18, 1) times P(19, 4) divided by P(37, 5), all divided by 5"? Then we multiply "the number of ways we could choose 1 red spot" with "the number of ways that we could choose 4 other ones". Divide that by all possible cases (P(37, 5)) and we get the possibility of getting one red spot out of five. Divide by five to get the possibility of not having the red spot until the last round. Why am I getting a different answer? What is wrong with my formula, and why is the one used in the video the correct one?
Independence happens in only one way. Dependency happens in an infinite number of ways. The hypergeometric distribution is appropriate for one specific type of dependency, where we are choosing x items without replacement from a source containing a certain number of successes and a certain number of failures. That situation is not the case in that example in this video.
This video is Great. 😍😍😍 But for the very first example (military equipment problem) , why it is not a geometric random variable? I am sorry if I asked a stupid question, but can you please help me?
No, what I state in the video is correct. Yes, there are 19 non-reds out of the 37 possibilities, and (1-18/37) gives the probability of one of those 19 non-reds occurring.
After the P of just 1 left working was calc to 0.0137 ..as the P of a single failure is 0.32 why isn't the P of all 6 failing [ie the last one after the others ] >> 0.32 x 0.0137
There are a number of examples in this video, and all involve different scenarios and different methods of finding the required probability. Which example are you referring to?
The trials are not independent there. As people are selected, the probability of getting a man (or woman) changes, depending on what has occurred previously. For example, if the first person selected is male then the probability of getting on a male on the second trial decreases. One of the requirements of a binomial rv is that the trials are independent. I discuss this difference in more detail in my intro to the hypergeometric distribution video.
Hi can you please solve this and explain why please I need help:The probability that a new drug prevents infection by a certain flu strain is 40%. What is the probability that the drug will be effective in one out of 5 exposed persons?
+Anooshiravan ensafian From the last question, the second paragraph is basically saying that Coz Tom is a bit Shaky, his probability to destroy an equipment after he destroyed one will be increased. So since the events aren't independent (coz destroying the first eq meaning he is going to have a higher chance to destroy the second eq, so the first "destroy event is related to the second destroy event", therefore, it is not independent events), The basic Stat distribution we can use is Hypergeometric, but obviously, the conditions for Tom's situation don't quite fit the Hypergeo distribution. So we need to think of an alternative distribution to deal with it. Since the question didn't give us further information about how will the probability change after Tom destroy the first eq, this question is simply can't answer unless further information provided.
+Anooshiravan ensafian To notice a trial is it an independent trial is more like a sense of statistics knowledge rather than a Mathematical skill, although you can provide it is independent or dependent Mathematically, but generally just think of the "events" in the trial, do they affect and related to each other, if they don't, then it is an independent trial.
I wonder how does that logic work with a geometric distribution so beautifully... if p(x>6) then x=7,8,9... In that case, x = 8 is also possible..so how can we limit ourselves to only 1st 6 trials to be failures when the 7th can also be a failure. That makes sense to me if p(x=7). Yes I have seen the mathematical proof but my logic isn't satisfied...
Suppose you intend to toss a coin until you get heads once. If the first six tosses are tails, then we know that heads comes at some point after that. (On trial number 7, 8, 9, ...) So P(First 6 trials are failures) = P(X > 6) = (1-p)^6. This implies P(X
I undstand the latter and the former but not the between. I quote you "we know that heads come at some point after the 6 tails" This is my question. So we can be getting 7 tails, 8 tails, 9 tails....too and then heads as there is no upper limit for failure (getting a tail) In other words how can we get a 'fixed' constant number for it. P(x>6)=P(x=6) X>6 has an increasing number of failures more than 6 But x=6 we limit ourselves to calculating probability for STRICTLY 6 failures and then a success. I understood the complement and the rest. I guess I'll have to rote learn this concept as I fail to see the flaw in my logic. :(
Ahmad Shumayal (1-p)^6 gives the probability that the *first 6 trials are all failures*. Since we are assuming independence between trials, then P(first 6 trials are failures) = P(First trial is a failure)* P(Second trial is a failure)*...*P(Sixth trial is a failure) = (1-p)(1-p)(1-p)(1-p)(1-p)(1-p)=(1-p)^6 (This isn't based on the geometric probability mass function, it's based on probability basics. I just realized that perhaps this is where the confusion arises.) So we are left with P(First six trials are failures) = (1-p)^6, which implies P(X>6) = (1-p)^6. I hope this helps!
Ahmad Shumayal I know this is late, and this may not satisfy your query: By probabilities the ball will EVENTUALLY land on red. But we know the first 6 are GUARANTEED failures, so we can say that at some eventual point it will land on red, which is 1-p(x
I know very little about the GCSE standards (only what a brief google search tells me), and I haven't tailored the material to any specific international standard. I'm simply trying to teach introductory statistics the way I think is best (pitched at the level of an introductory course for non stats majors, with some of the mathematical underpinnings included).
@@jbstatistics I see what you did there now that you reformulated it I sure doesn't make any sense. It was meant to be a joke anyway. However, tell that to the rectorate in our university who is the head of two other faculties.
![Kodak Black - Versatile 4 [Official Video]](http://i.ytimg.com/vi/VcP_1Ua__vY/mqdefault.jpg)
This guy just explained an entire chapter on one of my modules in 15mins , compared to my useless prof and teaching assistant who couldn't do it with 3 lectures and 120 pages of notes .
Hands down . Best . Tutor . Ever .
Hats off to you Sir .
lol
They have to stretch the material out so they can keep their jobs. Online educations would destroy Universities.
What an oversimplified and ignorant worldview.
@@MSloCvideos Sad thing is he's right tho, for some classes (Like this one)
Very good video, just like the others, on the playlist! This line made me chuckle though:
(5:20) "There are 26 successes in the population, the 26 females and 32 failures, the 32 men."
Education needs to be totally revamped. We don't need so many professors giving lectures, one for each subject in every schools in the world. We just need a handful of great teachers creating online content per subject. Instead, we need lots of great coaches and mentors to help individual learners. Learners will watch online lectures at their own pace. Coaches and mentors will work with individual learners to guide them, introduce good learning resources, push them to achieve and most important of all, motivate them. That's the new model of education.
Actually that could be a way to progress, however while dealing with larger masses who are willing to have a tutor or professor(i bet 3/4 univeristy undergrads are willing to have) it would still be needed to create some kind of classes, and for classes you need halls, and that cluster of halls would be called universitt ahaim. So basically we would go on the same cycle, of course with small (compared to scale) changes. But the system is actually worth thinking.
I've been having so much trouble with these concepts, especially because I am taking summer session and my professor teaches incredibly fast and is barely available for office hours. These videos have been helping me more than my professor in the last couple weeks! Thanks so much!
You are very welcome!
0:15 Binomial Distribution
4:32 Hypergeometric Distribution
7:27 Poisson Distribution
9:31 Geometric Distribution
13:02 Does not fit into any of the distribution
When you have 14:51 minutes before an exam
After watching all the previous videos in this series, I got all these answers right. Thanks for great instruction!
Really great videos, you get straight to the point without losing the viewer, well done
Thanks Francois.
what a lecture. Just superb. whenever feel like revising probability i just watch these lectures and concepts get refreshed. Thanx is just a small word. Hats off professor.
I'm glad to be of help. Thanks for the kind words!
everything he speaks is just valuable
so simple so clear
thank you😊
Thanks for the help, sir, really helpful video as i started from 1st video , i understood everything you taught. after 9 years of you uploaded this video it is still the best video on statistic on youtube.
Thanks for the kind words! I'm glad to be of help!
Dear JBStatistics can you put more applied examples like these , it's always good to see real world application of statistics than what we read in books like the ball , coin toss , choosing cards from a deck etc in book. These examples are amazing and the more students get to see these explained and solved by you. The more kids will start loving math, probability and statistics. Hope you keep posting these awesome videos
Thanks for the feedback and kind words. I try my best to incorporate real-world examples in my problems, and I'll continue to do that when I make new videos in the near future.
Been searching youtube looking to help me study for a probability test, pretty glad I've found jb statistics! Thankyou, sir, I like the way you teach, best I've found so far.
Anybody want to recommend some other youtube channel for maths/stats tutorials? You can never have too much learnin'! (Must be up to this standard or close)
Hands down, one of the best channels for Probability & Statistics on RUclips! Finished my entire semester's worth of course content in only a few hours thanks to you.
I'm glad to be of help!
Really hopeful,the best professor I met in my life!
good example, good approach, simple realistic example, good learning tool. Thank you.
You are very welcome. Thanks for the compliments!
man, you're like Khan Academy for statistics. thank you so much for your work. very very helpful
sir when you started with the hyper geometric distributions i laughed so much as i remembered the problem about counting females independently and that would cost you a job :D awesome videos really.. thank you!
Organic chemistry tutor ain't got shit on this man
may the odds forever be in your favor...you saved my university life with this playlist
+Cadeem Musgrove Thanks! I'm glad I could help!
Fantastic video! It's hard to find nice and understandable videos like this for stats. Thanks!
You are very welcome!
When you explain a subject the way you do, I begin to like statistics (bit only a little bit) :).Thank you.
Well, I hope I can get to the point where you like it a lot :)
BIG massive thanks to you JB, I nearly learn my whole first year Stat from your video, please go on and post more
The way you explain and teach is way better than my professors I reckon
Once again Cheers mate
excellent presentation of formula concepts with real-world examples.
Thanks!
GG, explaining not just one condition thus make us understand the whole picture
I am really grateful for finding your channel! thank you so much for your work.
On the first example you would have a success rate for one mission, but the real numbers to consider when designing a system would be the neg binom for failure in order to ensure its reliability over time.
Given the size of our military, if there were 10k systems used once on 100 missions, that would be a rather high failure rate over the 1m uses, and 7 or 8 components would be much more suitable when it comes to reliability.
Practicality must be assessed however, so 6 is probably as good as you'll get
you're the best! you helped me understand everything perfectly in a short time span. I can't thank you enough! 😁
Great! I'm glad to be of help!
thank you, you saved me and my degree
Why can’t you solve the last one with hypergeometrics?
5 years later and still saving our test scores
I designed them to be everlasting :)
when Tom goes crying to his boss that he ruined the first part, his boss should tell him “ oh , sorry Tom , forgot to tell you …that’s exactly how we wanted that one…good job!”
this is a little late but just wanna say for 12:56, summing many terms isn't a lot of writing cause the geometric sum is commonly known, so we can use a_1=p and (common ratio)=q.
you sir are a life savior!!!! thank you so much !!
You are very welcome!
This Video was awesome, Thanks so much. exactly what I needed, nothing more, nothing less. This was perfect!
Appreciate the help. It supplemented exactly what I need to know for my final.
I'm glad I could be of help.
Thank you so much, Professor!! Your videos have made my summer class a way easier time!:)
Greetings from McGill!
I'm glad to be of help! I'm always happy to help my friends in Montreal!
Wow, this video cleared up so much; thank you!!!
Your videos are excellent.. To the point!
+Aakash Jaiswal Thanks Aakash!
I spent 2 days creating a tool to calculate I think...binomial I forget which one specifically. My online college course rushed us through 30 day classes I learned way more from the internet - and videos like this. Thank you for keeping it simple ;)
You made my life easy, Thanks a million.
You are very welcome!
@jbstatistics I am curious about the answer at @12:59. i believe that the P(X>6) = 1 - 0.0183 (the value that was computed at the end). If we have an infinite amount of trials wouldn't we eventually get one red slot? We just won't get it in the first 6 trials but certainly red will come up at some point.
Yes, and that's partly what the probability I calculate is based on. The question asks for the probability that the **first** time the ball lands in a red slot comes after the 6th spin. That's definitely not a certainty. The probability the first red happens at some point is 1, yes.
Wow, it help me a lot to recall the distribution. Thanks! :D
You are very welcome! I'm glad to be of help.
Thank you for the explanations. They bring a lot of clarity.
Just had a question:
Why cant the last example be a hypergeometric distribution?
The conditions of a hypergeometric are not satisfied there. For a random variable to have a hypergeometric distribution, we need to be drawing items without replacement from a source that contains a certain number of successes and a certain number of failures. That is not happening in the last example. Don't get trapped into the false line of thinking that if the trials are dependent, then it must be hypergeometric; it simply doesn't work that way. There are many different ways for dependence to arise.
Your videos are really helpful. Thank you!
What operations used and how did it became 0.0245 before the rounding off?
Haha that last one definitely caught me, i thought it was binomial and i ended up with 1.121E-6. Good question.
+Warren Wilson Thanks! I like that one too!
I love your videos, they helped me so much!
But, at 10:57, I thought chance of failure should be 19/37 since the green slot is an additional slot of failure added to the 18 black slots...correct me if I am wrong!
I think he made a mistake.
In that example, the probability of failure is 19/37, as you state, but that is the probability I use in the video. I write the probability of failure as 1-18/37, which is 19/37.
I don't understand how to use calculator for solving the equation
how do you calculate the hypergeometric probability mass function by hand?
Does Russian roulette is an hypergeometric distribution problem?
You are awesome......really helped me so much....Why don't you make videos on joint random variable or conditional distribution........I needed those badly ....love from bangladesh :)
Thanks! I'm glad to be of help. I frequently get requests for videos on joint probability distributions, and they are high on the priority list, but I'm struggling to find time to make them. One of these days.
thanks for the reply... :) you know, you make videos so much up to the point....i have never seen such a wonderful combination of explanation along with a very relevant example.....i got very good grade in mid semester only just watching those videos bt sadly my semester final syllabus werent covered in your videos which include joint random variable......i recommended all of my friends about your videos before probability exam...now after watching ur videos they call you "JB BOSS" :) keep making videos keep saving lives :p
I think you should also add a timestamp at 4:25
Is there a video on using the binomial tables?
What's the process for the lamda is the poisson one. I understood it, but I can't put it into formula.
Thanks!!
An amazing review thanks!
You are a grade savior :)
In 10:18, why is that formula used? Why isn't it "The permutations P(18, 1) times P(19, 4) divided by P(37, 5), all divided by 5"? Then we multiply "the number of ways we could choose 1 red spot" with "the number of ways that we could choose 4 other ones". Divide that by all possible cases (P(37, 5)) and we get the possibility of getting one red spot out of five. Divide by five to get the possibility of not having the red spot until the last round. Why am I getting a different answer? What is wrong with my formula, and why is the one used in the video the correct one?
it is combination not permuration
why the last question can not be for hypergeometric distribution because of dependency??????????????
Independence happens in only one way. Dependency happens in an infinite number of ways. The hypergeometric distribution is appropriate for one specific type of dependency, where we are choosing x items without replacement from a source containing a certain number of successes and a certain number of failures. That situation is not the case in that example in this video.
@@jbstatistics That was helpful. Thanks
This video is Great. 😍😍😍
But for the very first example (military equipment problem) , why it is not a geometric random variable? I am sorry if I asked a stupid question, but can you please help me?
So roulette is a potential geometric distribution? Now that’s interesting.
you are really helpfull for us.very very thank u .
+Maanvendra Kholiya You are very welcome!
If 5 coins are tossed, what is the probability that all tails come up twice in 5 tosses?
Thanks for the helpful videos! I do have a question:
Shouldn't it be (1-19/37) at 12:38? You have 19 non-reds, right? 18 black + 1 green.
No, what I state in the video is correct. Yes, there are 19 non-reds out of the 37 possibilities, and (1-18/37) gives the probability of one of those 19 non-reds occurring.
Oh, alright, got it! Thanks for answering!
After the P of just 1 left working was calc to 0.0137
..as the P of a single failure is 0.32
why isn't the P of all 6 failing [ie the last one after the others ] >> 0.32 x 0.0137
Best explanation
Thanks!
I am confused why are we using a different approach to the first one which had the equation?
There are a number of examples in this video, and all involve different scenarios and different methods of finding the required probability. Which example are you referring to?
AWESOME VIDEO
Why didn't you use Binomial Distribution in the second example?
The trials are not independent there. As people are selected, the probability of getting a man (or woman) changes, depending on what has occurred previously. For example, if the first person selected is male then the probability of getting on a male on the second trial decreases. One of the requirements of a binomial rv is that the trials are independent. I discuss this difference in more detail in my intro to the hypergeometric distribution video.
U R THEEEE BEST!!!!!!
+MMA king Thanks!
Hi can you please solve this and explain why please I need help:The probability that a new drug prevents infection by a certain flu strain is 40%. What is the probability that the drug will be effective in one out of 5 exposed persons?
how can we notice that a trial is independent or not ??? (14:06)
+Anooshiravan ensafian
From the last question, the second paragraph is basically saying that Coz Tom is a bit Shaky, his probability to destroy an equipment after he destroyed one will be increased. So since the events aren't independent
(coz destroying the first eq meaning he is going to have a higher chance to destroy the second eq, so the first "destroy event is related to the second destroy event", therefore, it is not independent events),
The basic Stat distribution we can use is Hypergeometric, but obviously, the conditions for Tom's situation don't quite fit the Hypergeo distribution. So we need to think of an alternative distribution to deal with it.
Since the question didn't give us further information about how will the probability change after Tom destroy the first eq, this question is simply can't answer unless further information provided.
+Anooshiravan ensafian
To notice a trial is it an independent trial is more like a sense of statistics knowledge rather than a Mathematical skill, although you can provide it is independent or dependent Mathematically, but generally just think of the "events" in the trial, do they affect and related to each other, if they don't, then it is an independent trial.
thaaaaaaaaanks alot habebe you are the best
you are awesome.... speechless
Thanks!
you are most welcome...
thank you!!!
Excellent
Thanks!
I wonder how does that logic work with a geometric distribution so beautifully... if p(x>6) then x=7,8,9...
In that case, x = 8 is also possible..so how can we limit ourselves to only 1st 6 trials to be failures when the 7th can also be a failure. That makes sense to me if p(x=7).
Yes I have seen the mathematical proof but my logic isn't satisfied...
Suppose you intend to toss a coin until you get heads once. If the first six tosses are tails, then we know that heads comes at some point after that. (On trial number 7, 8, 9, ...) So P(First 6 trials are failures) = P(X > 6) = (1-p)^6. This implies P(X
I undstand the latter and the former but not the between.
I quote you "we know that heads come at some point after the 6 tails"
This is my question.
So we can be getting 7 tails, 8 tails, 9 tails....too and then heads as there is no upper limit for failure (getting a tail)
In other words how can we get a 'fixed' constant number for it. P(x>6)=P(x=6)
X>6 has an increasing number of failures more than 6
But x=6 we limit ourselves to calculating probability for STRICTLY 6 failures and then a success.
I understood the complement and the rest.
I guess I'll have to rote learn this concept as I fail to see the flaw in my logic. :(
In short,
P(x>6) = q^6 how ?
There can be more failures than six isn't it sir?
Ahmad Shumayal
(1-p)^6 gives the probability that the *first 6 trials are all failures*. Since we are assuming independence between trials, then P(first 6 trials are failures) = P(First trial is a failure)* P(Second trial is a failure)*...*P(Sixth trial is a failure) = (1-p)(1-p)(1-p)(1-p)(1-p)(1-p)=(1-p)^6 (This isn't based on the geometric probability mass function, it's based on probability basics. I just realized that perhaps this is where the confusion arises.)
So we are left with P(First six trials are failures) = (1-p)^6, which implies P(X>6) = (1-p)^6. I hope this helps!
Ahmad Shumayal I know this is late, and this may not satisfy your query:
By probabilities the ball will EVENTUALLY land on red. But we know the first 6 are GUARANTEED failures, so we can say that at some eventual point it will land on red, which is 1-p(x
The first question should be p(x=1) because the problem said EXACTLY one, not at least.
Which is why I work through the question to find P(X=1). I then move on to another question.
You're amazing
Thanks!
Ive watched 5 videos im still confused so much for wanting to pass
HELAL OLSUN ABI
I LOVE YOU
Yo is this GCSE level or not?
I know very little about the GCSE standards (only what a brief google search tells me), and I haven't tailored the material to any specific international standard. I'm simply trying to teach introductory statistics the way I think is best (pitched at the level of an introductory course for non stats majors, with some of the mathematical underpinnings included).
I see that you have never been to Turkey, same people can be in a committee more than once surely if they have multiple roles...
Can a Turkish "committee of 7 people" have fewer than 7 people?
@@jbstatistics I see what you did there now that you reformulated it I sure doesn't make any sense. It was meant to be a joke anyway. However, tell that to the rectorate in our university who is the head of two other faculties.
@@torealityAN Mine was a joke too :) I understand what you're saying.
4:45 ohhh dear :v