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(2.5 pts: _ Give examples of A,B € Mz to show thatP(A+ B) > P(A) + P(B) is possible P(AB) P(A)p(B) > 0 is possible...

Question

(2.5 pts: _ Give examples of A,B € Mz to show thatP(A+ B) > P(A) + P(B) is possible P(AB) P(A)p(B) > 0 is possible

(2.5 pts: _ Give examples of A,B € Mz to show that P(A+ B) > P(A) + P(B) is possible P(AB) P(A)p(B) > 0 is possible



Answers

Let $A, B,$ and $C$ be events such that $P(A)>P(B)$ and $P(C)>0 .$ Construct an example to demonstrate that it is possible that $P(A | C)<P(B | C)$.

Yeah. In this problem we are going to use the addition law of probability. And the first part of the question is given that P of E is equal to 0.7 and P L B is equal to 0.4. Now the question is whether or not it is possible for A and B to be very exclusive. Now, if A and B are usually exclusive, then by the addition law, the probability of E or B will be equal to be a way must be Obi, P M E is equal to 0.7, and P L V is equal to 014 so 0.7014 is equal to 1.1. And this is greater than one. So that implies that P O. E or B is greater than one. And this is not possible because the probability of any event must be less than or equal to one. So that means that our assumption is wrong and that means that A N. B cannot be mutually exclusive, so no A and B are not mutually exclusive. Not in the 2nd part of the question. The value of P E and E is given and we need to find the value of P A or B. Now by the addition rule of probability this is equal to P. Of A plus P, R v minus P of A and B. The value of P L. E is equal to 0.7. Value of P f B is equal to 0.4 and B. Of A. And B is given to be 0.2, So 0.7 0.4 is 1.1 -0.2 and 1.1 -0.2 is equal to 0.9. So the required probability of P.E or B is equal to 0.9.

So here in this kitchen, we are given that A and B, uh, evens off the same samples piece. Consider the sample space B s on. We are given that probability off. Even a occurring is 0.76 on probability. Off even be occurring is 0.5 now, Since we know that probability off sample space s is always one, this is our, uh, basic rule of probability that probability of whole sample space should be close to one now, in part if we are asked if and we are mutually exclusive events, so no end. We are not mutually exclusive since that some off mutually exclusive event should be one since be off a plus probability of being probability off A. That some of these two probabilities are 0.76 plus 0.58 which becomes 1.34 and naughty close to one. Therefore, these two events are not mutually exclusive on as we can see from the when diagram here, this is the whole sample space on. This is even a This is even to be, since he can see that intersection off those these two events is non empty therefore, probability off even a intersection probability of even be is greater than zero. Therefore, these two events are not mutually exclusive. Nine Part B. We are again as if a compliment and even be complement are mutually exclusive. So again, no, no. These two events are also not mutually exclusive since again drawing a Venn diagram, we can see that this is whole off sample space on this is even a This is even be no, we can see that a compliment is this part I be compliment Is this part now as we can see that intersection off a compliment and be compliment is not fight So intersection off a compliment and be compliment is non empty there for probability off a compliment in the section B compliment is also greater than zero. Therefore, a compliment and be compliment uh, not mutually exclusive. Now, in part C probability off a union be should be less than equals to one. This is ah are formula. This is our fact, since reliability off any even occurring should be less than or equal to one and probability off a union be should be Ah, and we know the formal other tablet off a union bees probability off a less probability of be and even start not mutually exclusive. So minus probability, off a intersection B Here we know that probability off a is zero point is zero point 76 and probability will be is 0.5 it minus probability off intersection B. Therefore, we can say that probability off a union B should be greater than or equal to zero point. Uh, we can add these stones and get probability off a union be should be great with our equals. Two 1.34 minus probability off a intersection B So this is our relation or the range off probability off a union be

Here we are asked to explain why the probability of a occurring when B has occurred equals zero if events A and B are mutually exclusive. So here's a Venn diagram that has events A and B that are mutually exclusive. Remember, if these events we're not mutually exclusive, there would have to be some overlap between the event circles something like this. But since these are mutually exclusive, there is no intersection. So if event A occurs, we cannot have it. That event be also occurred. There is no outcome or no set of elements in the sample space that satisfies both Event A and event be simultaneously. So if we're asking the question, what is the probability of event A. Given that D has occurred, that's how we would write it. So we're basically saying Event B has occurred now. What is the probability of event A? But since we know that event be occurred and events A and B are mutually exclusive, we know that event a can't occur. So, in other words, the probability of event a occurring is there for zero. So it's events A and B are mutually exclusive. The probability of event a occurring given that B has occurred is zero

Well this problem we have a theorem that we would like to prove. And so in order to prove it when you start with that left hand side and show that it is equal to the right hands. Yeah, So we have the probability of a union be given C. Now by definition this is equal to the probability of a union. Be nancy all over the probability of C. Mhm. No, this is equal whenever we distribute in that intersection side of the probability of a union. See and be union. See all over the probability of C. Excuse me. Have these signs flipped here? This would be so. Mhm. This in here Inside should be the intersection and that should be the union. Now making use of PM 2.6. That top is the same as the probability of A. N. C. What's the probability of B and C minus? The probability of A. N C and BNC. And each of these are over the probability of C. Just breaking apart of the fraction. By definition. This first part here is the probability of a given C. That's part of the probability of being human. See, And then, lastly, we have the probability of A and B given C. And so we've shown that the theorem holds. Mhm.


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