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)

You are very welcome!

Thanks Francois.

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.

I'm glad to be of help!

You are very welcome!

I'm glad to be of help. Thanks for the kind words!

You are very welcome. Thanks for the compliments!

Thanks for the kind words! I'm glad to be of help!

I'm glad to be of help! I'm always happy to help my friends in Montreal!

I'm glad I could be of help.

Well, I hope I can get to the point where you like it a lot :)

You are very welcome! I'm glad to be of help.

+Cadeem Musgrove Thanks! I'm glad I could help!

Great! I'm glad to be of help!

Thanks!

You are very welcome!

+Aakash Jaiswal Thanks Aakash!

You are very welcome!

I designed them to be everlasting :)

+Warren Wilson Thanks! I like that one too!

+MMA king Thanks!

+Maanvendra Kholiya You are very welcome!

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 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.

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

Thanks!

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.

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!

Thanks!

you are most welcome...

Thanks!

Thanks!

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?

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

it is combination not permuration

+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.

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.

Which is why I work through the question to find P(X=1). I then move on to another question.

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

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.

@@atalaynalldere1827 Mine was a joke too :) I understand what you're saying.

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).