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(Spts) Use the Law of Total Probability to prove the following:If P(AIB) P(A[BC ) , then A and B are independent: IE P(AJc) > P(BIC) and P(Alcc) > P(BICc) , t...

Question

(Spts) Use the Law of Total Probability to prove the following:If P(AIB) P(A[BC ) , then A and B are independent: IE P(AJc) > P(BIC) and P(Alcc) > P(BICc) , then P(A) > P(B):

(Spts) Use the Law of Total Probability to prove the following: If P(AIB) P(A[BC ) , then A and B are independent: IE P(AJc) > P(BIC) and P(Alcc) > P(BICc) , then P(A) > P(B):



Answers

Use Theorem 2.8 , the law of total probability, to prove the following: a. If $P(A | B)=P(A | \bar{B}),$ then $A$ and $B$ are independent. b. If $P(A | C) > P(B | C)$ and $P(A | \bar{C}) > P(B | \bar{C}),$ then $P(A) > P(B)$

Yeah. This problem, we were told that A and B are independent that the probability that neither occurs is a. So in other words the probability of a compliment and yeah be complement. This is where neither of them occur. This is your forte and the probability of B is B. So the probability of B. We want to find the probability of it. Now. Working off of this first statement here, this means that the probability of a union with the probability of B of compliment is able to A. That's just using the morgans law. Sure. Now from here this means that the probability of a union with the probability of B. I'm sorry, I should write this differently. Let me write this As the probability of a union be with him inside of there. Actually like this. Now if the content was equal to a then this is equal to one -A. Yeah start complement ruler. But we do know the probability of a union be is equal to the probability of A plus. The probability of B minus the probability of A and B. What this tells us is that the probability of A plus, the probability of me minus the probability of A times the probability of B because they are independent, There's one -A. So we have the probability of A plus B minus. The probability of A Times Me is one -A. So the probability of a Times 1 -7 Is April one -A -7. And so the probability of A does one minus a minus B Over 1 -7, which is what we wanted to show.

And this problem, we're going to determine whether or not mutually exclusive events. E. And we can be independent. Now, since A. And B are mutually exclusive. That means that a intersection B will be close to fight, since the events cannot happen together. And that's the probability of a intersection B must be equal to zero because intersection base an impossible event. So the probability will be close to zero. Now, if A. And B are independent then P. R. E. Intersection. He should be equals two B. Of A times PlB. And if this is equal to zero, then that would imply that the O. P. Zero or P. R. B. C. Now he and we are not impossible events. So neither preservation P zero nor piano piece should be zero. So this line is a contradiction. So that means that our assumption that A. And B. Are independent must be wrong. And because of that, we should not be able to write this line. So that means that the events A. And B are not independent. So that means that A. And B. Are too dependent events.

In here asks us to find the probability of a given that we're taking the probably of subset be. So In order to solve this, we have to take the general formula of probably of A and B over the probability of finding beat. So this particular case, we know that a is your 0.342% and B is going to be your 0.279%. So to find a probably of A and B together, we need a multiplied by each other. And we divided to spy the probably be, which is your 0.279 So, aggregate, we're gonna get 0.342 um, for the probably of a given that be in this particular sample space on the second part of this question here asked us to find the probability of not be so. We know that the probably of B is your 0.279 So to find probably have not be is to take the compliment of it. So it's one minus 0.279 here, and lastly, it wants us to find the probably of A and B, which is what we found earlier, where probably of A and B together is 0.34 to multiple adviser 0.279 and this is going to the answers his question here.

And this problem, we're going to determine whether or not to events A and B are independent. If A is a subset of people. Now, if A and B are independent events, then P of a given B should be equal to be of aid. Now, in this case, if we calculate P O E U and me. Yeah, we can use the definition of conditional probability to say that the of A even be is equals two P of a intersection be divided by P O. P. Now A is said to be a subset of B because there is a subset of be so intersection B is equals to eat. So this will be P O. A. And we have this S. P O. P. So P L E given B is equals two p.m. E divided by PlB and this is not equals two P of A. Because PlB is said to be greater than zero, but it might not be equals to one. So if P f B is any number other than one, any positive number other than one, Pf, given B is not equals two P o P. Because of that, based on the given information, we can say that p of a given B is not equal to p L. A, and hence the events A and B are not independent, they are dependent events.


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