Bayes' Theorem - Example: A disjoint union

in my introduction to Bayes theorem I gave you this formula that the probability of a given B could be written by the conditional probabilities the other way around that is the probability of B given a but you had to multiply by the probability of a and divided by the probability of B if we wanted to can take that formula rearrange and I figure out what it was the probability for be given a in terms of the probability of a given B as well now in this video we're gonna upgrade this for me a little bit we're going to investigate what happens if we have some sort of division in our sample space that we can separate it into different quote-unquote buckets so the example I want to use to motivate this is one run quite literally going and drawing two different buckets and I'm having balls of different color inside of these two buckets you'll notice that the one on the left the one I'm calling be one well it's gonna have three blue balls and three yellow balls so the probability of a blue ball being drawn from that bucket would be 50/50 and then B 2 is it bit different it's got two balls but it's got four yellow balls and so the probability of drawing a blue ball from the second bucket would be one-third so now I'm saying well look what if I draw from a bucket but I don't know which it is like I randomly choose a bucket and then I randomly choose a ball from within that bucket well now if I know that I've drawn a blue ball what what's the probability that I came from that first bucket or alternatively I could ask well if I have a blue ball what's the probability that came from the second bucket this is the problem we're gonna try to investigate and I'm gonna begin by just computing a few different things that I can compute and then we're gonna try to put them together and I sort of upgraded Bayes theorem now the first thing that I know that I'm confident about is certain conditional probabilities probabilities it going in one direction not the direction I'm looking at but the ones that go in the other direction as in let me first just denote by a the event that we're going to select a blue ball and I'll let b1 be the event that I chose her bucket b1 and b2 the event that I chose from about to be - they're not so the thing I was confident about is that the probability of choosing a blue ball what I call a given that I am looking at bucket b1 well I know that this is going to be equal to 1/2 there's three blue balls there's three yellow balls it's a 50% probability I am also pretty confident if I look over here at b2 I can say that the probability of drawing a blue ball given that I'm in the second bucket well two blue balls six over also one-third okay so that's not so bad I we've done some computations here what else do I know well other things I might be able to compute is there's two different buckets here there's a b1 and the b2 and equally likely to be drawn from so I can say that the probability of drawing from the b1 or the probability of drawing from the b2 this is just a 50/50 so another 1/2 that I'm asserting a sort of part of the setup of my problem that my two buckets are equally likely to be drawn from now I'm gonna do something a little harder I mean I'm trying to investigate what is the probability of a what's the probability of drawing this this blue ball where where I first have to randomly pick between these two buckets and then when I get a bucket I randomly draw a ball from that that could be blue that could be not who knows no the first thing I want to note is that there's a disjoint union here the the b1 and the b2 are completely separate there's no overlaps between them but by design this is a disjoint union of our sample space so what we can do is we can investigate what the probability of a given that I'm inside of b1 is and then I can add to that the probability of being a inside of my beaching now the reason why this works is is first of all that the the sample space of any can be dividing these two things the A's that are the b1 and the A's are in the b2 which I formally write down as a probability of a intersect B wanted a intersect B do and then the second thing I'm using here is knowing that four disjoint unions that the probability of a disjoint union is the sum of those individual probabilities indeed you were doing a counting problem you said what is the number in a disjoint you need to figure out the number in a and you add to it the number and beats the Zack same idea is what we've seen before okay so I have these two different intersections but what I know is I now know how to deal with intersections in terms of conditional probabilities so let's remind ourselves how conditional probability worked we had this this particular property that the the probability of a given B was by definition just the intersection of a intersect B all divided by the probability of B so so that was my broader formula and I can I can use that formula I can put in here so how does it work well okay this is a intersect B one so any intersect B one is the probability of a given B one and then I have to take the denominator multiply it up multiplied by the probability of B and then the probability of a intersect B is the probability of a given now it's the b2 and multiplied by the probability of my b2 and because I was planning ahead I've already computed all four of those numbers so the probability of a given B one was a half the probability of just be one here I'm gonna add in the subscript B 1 was equal to another half the positi is a given B two was a third and the probability of B 2 was another half equally likely to choose B 1 is b2 all right so this is going to be 1/4 plus 1/6 and then this is going to be 3 12 plus 2 12 which is going to be a total of 512 and that's my final answer all right now I've gone and computed some some some values here but well why is that even asking in this problem let's go back and try to remember I was given these two different buckets and I was saying if I draw a blue bottle then what is the chance that I was in the bucket one that's what I'm asking I'm asking this conditional probability was the probability of being in bucket 1 given that I've drawn a blue ball so I can apply Bayes theorem and I've written it down for you already it is going to be that the probability of B bucket 1 given a that's what I have here is going to be the probability of a given B 1 times the probability to be 1 all divided by the probability of a this is base all right well let's try to compute this we've done the probability of a given B 1 already that was just 1/2 remember there were three blue balls and four yellow balls look at the extent probability the probability of being B 1 we did that already there was the two buckets they were equally likely it was another value of 1/2 and then we just went through the sort of somewhat arduous task and figuring out the probability of a so that was the 5 twelfths we did an anticipation of this and now I can get there so multiplying by the 12 up that's gonna leave me with the final value of 3/5 now we should verify does this make sense I think so if you remember in the 2 different buckets that we had that be one had more blue balls then as a as a probability it was 3 6 then the B 2 would only had two cents so if you've gone in a blue ball it is somewhat more likely that you're in the bucket 1 than you are in the bucket 2 and so we get this value 3/5 that's a little bit bigger than the 50% probability of randomly choosing between the two buckets so this makes sense at least at a broad picture for us so this is an example of how we can apply Bayes theorem in a scenario where there's two different buckets it's basically exactly the same thing if you have three or four and different buckets you just have to divide it up into more cases your P of a is gonna have n different subdivisions inside of it but other than that it's the same idea

Channel: Dr. Trefor Bazett

Views: 167,003

Rating: 4.9441929 out of 5

Keywords: Calculus, Education, Math, Solution, Class, College, Bayes Theorem, Disjoint Union, probability, conditional, two buckets

Id: k6Dw0on6NtM

Length: 8min 32sec (512 seconds)

Published: Sat Nov 18 2017