Independent Events (Basics of Probability: Independence of Two Events)
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- čas přidán 29. 08. 2024
- An introduction to the concept of independent events, pitched at a level appropriate for the probability section of a typical introductory statistics course. I give the definition of independence, work through some simple examples, and attempt to illustrate the meaning of independence in various ways. (Note: I use the phrase "not independent" rather than "dependent" almost exclusively. There is nothing wrong with calling events dependent when they are not independent, but I prefer to use "not independent" for a couple of reasons.)
(I'm on a bit of a probability run, but looking forward to getting back to statistics videos in the near future.)
This one turned out to be long, as I had a number of points I wanted to discuss. Here's the breakdown:
0:20. The definition of independence, showing what that means in terms of conditional probability, and some hand-waving discussion of what independence means.
3:48. Very simple examples (P(A) = x, P(B) = y, etc.).
6:20. Die rolling examples.
10:03. Discussion of the fact that if A and B are independent, so are (A and Bc), (Ac and B), and (Ac and Bc), including a hand-waving justification. The previous example involved a lead-in to this. (The hand-waving is legit happening behind the scenes.)
11:15. Visual illustration of independence.
14:39. Playing card examples.
18:19. Discussion of how we sometimes assume events are independent (e.g. heads on first toss of a fair coin, heads on second toss), and how this is an *assumption*, and not something that can be proven mathematically (despite what you might see elsewhere).
19:24. Discussion of how the term "independent" can have a different meaning in everyday English compared to its usage in probability, and how that is sometimes a cause for confusion.
Simply incredible lesson. Exactly what I was looking for, with rigorous and practical explanations. Thank you.
Thanks for the kind words. I'm glad to be of help!
Glad to see you back. I have watched your videos on inference and hypothesis testing over and over again. I have used your videos to educate non-english speaking students in Greece. Hope you carry on with the good work. You and Khan Academy and 3blue1brown are amazing!!!! Your contribution to education and therefore to science is huge!
Thanks for the kind words! I'm very glad to be of help to people around the world!
Sir, don't forget our society is growing better because educational CZcamsrs like you exist. I even learn programming by engaging online and I need to learn statistics a bit for my Python programming. Huge thanks from South Korea!
10 months ago, but just noticing now. Thanks for the very kind words! I'm very glad I can be of a little help to people around the world.
Thank you for this post,
Realizing independence(in the probability context) means equally proportional distribution is a great realization for me.
Glad to be of help!
So easy to understand. The voiceover is silky smooth too. Great combination for learning.
Thanks. This video helps me greatly. The meaning of Independent in probability is different from normal day-to-day English.
it's crystal clear now what INDEPENDENCE means Thank you
I'm glad to be of help!
The content here is exceptional ! Great job Sir !
you saved me on my econometrics, thank you so much
thank you, this really helped me get a thorough gist of independence!
I'm glad to be of help!
The last example wakes me up and reminds me to think out of the box. Thanks!
I’m currently in 7th grade and I have a quiz on this. Thank you for making this video it really helped
Nice stuff. Found your channel through some random comment on a 3Blue1Brown video
Thanks! I'd have to work on my animation skills for quite some time to get up to 3Blue1Brown level :)
You are awesome. Don't deny it.
I've never denied it :)
I don´t understand how knowing a die came up odd doesn´t change if our event one or two can take place. only one can take place now so shouldn´t the two events be dependant? cheers adn great vid
What's the probability the die comes up with an odd number? 1/2. If we know only that the die came up with a 1 or a 2, what's the probability the number is odd? 1/2. What's the probability of rolling a 1 or a 2? 1/3. If we know only that the die came up with an odd number, what is the probability it is a 1 or a 2? 1/3. Knowing the die is a 1 or a 2 doesn't change the probability of odd, and vice-versa. Sure, if the number is odd then 2 is no longer a possibility, but 1 is twice as likely as it was without that knowledge.
Damn, you're the best.
this is really helpful explained simply and in a really easy way to understand. thank you so much for this.
I'm glad to be of help!
Crystal clear. Thank you
You are very welcome!
Thank you so much for your response. A subject ignored in texts but important nevertheless less.As usual you have not disappointed..
Vrey nice sir thank you very much understood very clearly
I'm glad to be of help!
19:25 This bit seems especially important
You just saved my stats grade. thank you!
excellent videos
Jb, you are very clear in your teaching, makes it easy to grasp the subtle points that cause most beginners to come to the wrong answer. Can you please make videos when the unconditional probabilities are not given but depend on each other? For example, what is the probability of a 3rd girl if there are two girls in a family already OR what is the probability of the second child being a girl if the oldest is a girl? I have difficulty when the sample space becomes the second even , do I make sense? Thank you so very much for making this world more tolerable. Some post pitiful hate contents or violent videos and some are like you...
9:50
Since they are independent then why A entersection C don't equal pA*pC
P(A n C) = 1/6, P(A) = 1/3, P(C) = 1/2. So P(A n C) = P(A)P(C).
It is very cool and understandable thank you
Your videos are so nice.Please upload more videos on statistics. Your videos are helpful.
Thanks for the kind words. I'll be uploading more videos soon.
You are super awesome Professor! Thank you kindly for your help! This is a good brush up for Quant Trading assessments. :D
For the last example, would it be correct to say that events A and B are conditionally independent given the weather, but are not unconditionally independent? If W is the event of good weather, then P(B | A,W) = P(B | W), P(A | B, W) = P(A | W).
They wouldn't be unconditionally independent since knowledge of B would increase the probability of A (if Tom goes fishing, there's a good chance the weather's decent, so Pete will probably go golfing. P(A|B) > P(A)).
I'm still a little confused on how you would formally explain the effect of good weather (W) in the unconditional case.
What happens when A and B are mutually exclusive? P(A intersection B)=0 but A and B are independent.
" but A and B are independent." Nope. If P(A n B) = 0, then A and B are independent if and only if P(A) = 0 or P(B) = 0. I have a full video discussing that very topic here: czcams.com/video/UOfsad9WWwk/video.html&ab_channel=jbstatistics
I don't understand the main purpose of part 13:44. Can you summarize and explain briefly the big concept here?
Anw, thankyou for another great video
It's a different visual illustration of what independence is, and how if P(B | A) > P(B) , say, then P(B|A^c) < P(B). I don't think any summary I give here will be of use compared to what's in the video.
Very nice video
What is the difference between independent and mutually exclusive events, if independent events are those which don't change the probability of the other events?
Hello,
that means we would need to always prove they are independent events because to me it would look like the example on 8:15 show that A and C are dependent since they overlap on the {2}, but in reality they are not. Is this ok to conclude of the need to test before concluding?
thank you
I'm not sure what you are asking. Events A and B are independent if and only if P(A n B) = P(A)P(B), so yes, you can't say A and B are independent unless you know that condition is true.
Thanks for help sir
Very helpful thanks a lot!
Thanks, subscribed
Informative, Well done: )
You are very good! Thank you very much!
Really very helpfull Lectures
but cannot the coin example be mathematically proven with an ordered pair? like for 2 coins : {(T,H),(H,T),(T,T),(H,H)} the fact that we got some result on the first element doesnt affect what we could get in the second element.
If we have a situation where say P(A) = 0.45 and P(B) = 0.75, what could we say about P(A n B) with only this information? And thank you for these quality videos, sir!
kormkor there’s not enough information to solve p( A n B)
refer the video at 4:09
In this scenario, we aren't told if A and B are independent or not. If they are independent, there is no way to work out the intersection. If A and B aren't independent, then the intersection is P(A)*P(B).
Superb video
Nice explanation
Great video !
Could you please set up a Patreon account and publish a little more often? :) You're awesome!
Well, that's certainly a nice compliment! I'm not sure about the Patreon aspect, but I will soon get back to video production!
thank u very much u really helped me a lot
Damn, this is wonderful! Setup a pateron page!
Thanks for the compliment!
Regarding problem of independence of E: The card is 8 , H: The card is Hearts, mathematically they seem to be independent. This is due to proportion of number of 8's in sample space is equal to proportion of 8 in Hearts? .. Is my argument correct?
Yes, your argument is correct. But the "proportion" argument that you use only works if the sample points are equally likely (as they are in the example).
Hi. In part "20:00" of video, you are saying A and B are dependent events based on weather etc. However, you said that given just two probabilities A and B we can never tell if its independent unless we know the intersection. How is A dependent on B and how can we show that? Thanks.
I said that we cannot determine the probability of the intersection of A and B from only the individual probabilities A and B, which is different from what you are stating I said.
The point of the weather example was to try to give an intuitive understanding of the meaning of independence, and not just crank numbers. I have cranking numbers examples in this video. Then I move on from there. We can have an intuitive understanding that events are dependent from the situation under discussion, without knowing the probability of either event or the probability of the intersection. There's some things I know without looking at numbers:
For a randomly selected person:
A: They kill somebody in a car accident today.
B: They drive drunk today.
These are clearly not independent. I don't know P(A). I don't know P(B). I don't know P(A n B). I do know that P(A | B) > P(A).
Suppose you're getting married two Saturdays hence.
A: It rains on the Friday.
B: It rains on the Saturday.
I don't know P(A). I don't know P(B). I don't know P(A n B). But if you were getting married in two weeks on Saturday, and the only info you were privy to was Friday's weather, would you prefer that it rained on Friday, or be bright sunshine?
@@jbstatistics Thanks for clarification. I thought 'weather' was some third unknown factor because when events are dependent then event 'A' depends on 'B'. But there I couldn't see how "Pete goes golfing next Saturday" was dependent on event "Tom goes fishing next Saturday." But now I am beginning to see it because the example with car accident and "being drunk" is more clear! Great videos!
@@djskyski I had a different example of mine in my head when I wrote that last reply, but it does still apply. The point of the Pete/Tom example is that it's a dangerous road to just assume independence simply because two events seem somewhat disconnected. Weather was indeed a hidden factor, connected to both events, and the likely the main (but not only) link between those two events.
the video will be better if @jbstatistics can show the calculation for chart about the independence of A complement and B complement. I have been confused on that part because i only saw the chart with calculated data but don't know how to get there. Any charts should show watchers how to derive data so they won't get confused during watching. Thanks
I don't know what you mean by "chart" here, and I don't know what you mean by "calculated data" or "how to derive data" either. I really have no idea what your feedback means. Can you give a specific example of what you're talking about? Cheers.
@@jbstatistics I think they're referencing the bar charts at the 12:07 and 13:55 mark. Personally, I understood the bar representing P(A). However, in the case of the bar representing P(A Complement), I wasn't sure how you were able to determine the proportions of P(B) and P(complement of B) within the bar.
Edit: Never mind. I understand now. :) The blue has to add up to 0.7 (i.e. P(B)) so that's how you were able to determine the proportion in the P(Ac) bar.
cant it be the case the probability values (the product and the intersection) are equal due to chance?
If I know a number is even then I know it cannot be a 1 so how can you say it doesn't change the probability that it is a 1 or a 2.
For the reasons I outlined in the video. P(A) = P(1 or a 2) = 2/6 = 1/3. P(A|C) = P(1 or a 2 given the number is even) = 1/3. These two probabilities are the same. Sure the chance of getting a 1 has been eliminated, but the chance of getting a 2 has doubled. These cancel each other out perfectly, and P(A|C) = P(A).
I think there is a mistake in 7:15 mins. P(A) should be equal to 1/2 instead of 1/3 .
I could be wrong, so if I am then can you plz explain how did you get 1/3
That's not an error. A is the event that we roll a 1 or a 2 when we're rolling a balanced six-sided die. There are 6 equally likely outcomes, and two of them result in A, so P(A) = 2/6.
@@jbstatistics Thank you that was helpful.
As you can tell I'm not the sharpest tool in the shed when it comes to probability so sorry if I caused any trouble
8:17 "knowing the probability that it is an even number doesn't change the probability that it is a 1 or a 2". I don't follow... using the argument from the previous example, it eliminates 1 as a possibility, so it should change the probability of 1 to zero.
What am I missing?
Knowing the number is even doesn't change the probability of A = {1, 2}. The probability of rolling either a 1 or a 2 is 1/3, whether we know the number is even, we know the number is odd, or we have no idea of what the number is. Knowing the number is even changes the probability of 1 from 1/6 to 0, as you state, but it also changes the probability of 2 from 1/6 to 1/3.
Thanks. That's helpful. Could I have some help with another question I'm struggling with for a while now. If in the same example at 8:15, I add another element to the sample space, let's say 9. This will change the probabilities so that the events are not independent anymore as P(A and C) is not equal to P(A)*P(B). Mathematically, I get it, but I fail to grasp it intuitively. Why should another element in the sample space change the events A and C from being independent to becoming dependent, especially since there is no change to the events themselves?
@@rickytyagi Well I think I could help you with that. When we say that events are independent we go forward to having that P(A/B)= P(A) beacuse the sample space of the event B on which sample space of event A is being conditioned upon tends to balance the (we could say some sort of loss or) decrease in probablity (because of all elements of event A not occuring when event B has occured) by the reduced sample space of event B . Therefore if we go literally then we get to know that the changes in probability is balanced by the reduced sample space S of random experiment which is the sample space of event B i.e S' s subset.
At last, every event conditioned upon changes the probability of the other event it's just that the probability tends to balance the changes . It's a sort of number magic performed
GOAT
i got it thank you so much !!!
Can you do a video on the Laplace distribution? I know it's not as useful for introductory statistics but I find it interesting.
Thanks for the suggestion. I may get to that at some point.
Hi had a doubt
1) How independent evnnt is different than mutually exclusive event?
2) How independent event is different than unconditiona ll event
1) The concepts of independence and mutual exclusivity are simply different. I discuss mutually exclusive events here: czcams.com/video/B1v9OeCTlu0/video.html, and I have a video titled "Are Mutually Exclusive Events Independent?" here: czcams.com/video/UOfsad9WWwk/video.html.
2) First, no single event can be labelled called independent. Independence speaks to the relationship between events. I don't know what precisely you mean by "unconditional event". It's possible that some people would use the term "unconditional events" to mean "independent events" but I would never use that wording.
8:16 How does the knowledge that the number is even not change the probability that the number is either a 1 or a 2? That makes zero sense. Is the number is even, then the probability of it being a one is 0?
"Is the number is even, then the probability of it being a one is 0?"
Sure, and the probability of it being a two is 1/3. So given it's even the probability of it being either a 1 or a 2 is 1/3, which is what the original probability was.
Hi, your videos are amazing, thank you very much, I'm from Colombia and some videos in Spanish are not as complete as yours, 😁
You are very welcome Elian! Thanks for the compliment!
yeah im still confused
thanks !!
I love U
Fact: Some guy who commented on the original fruit ninja ad put this in a playlist.
Do you know you are amazing??
Hello thanks for your video it is really helpful but It might be there a mistake in 9:50 minute that p(A\C') = 1/2 not 1/3 because here {p(A)*p(c')}/p(c') = {(1/2)*(1/3)} / 1/3 = 1/2 !
.
so it should be P(A) = 1/2
and P(C)= 1/3
No, that isn't a mistake in my video, that's a classic mistake on your part. You have calculated the probability of the intersection by multiplying the individual probabilities together, but that is only correct if the events are independent. So, you've implicitly assumed independence, then calculated the probability based on that assumption, when the assumption is incorrect. My values are correct in that example (and the rest of the video), and I explain the rationale in detail in the video. Cheers.
oooooh I am really sorry I got it, thanks a lot
I thought this was khan academy based on the thumbnail
Nice explanation