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Numerade Educator

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Problem 59 Hard Difficulty

In Exercise $15, A_{i}=\{$ awarded project $i\},$ for $i=1,2,3 .$ Use the probabilities given there to compute the following probabilities, and explain in words the meaning of each one.
(a) $P\left(A_{2} A_{1}\right)$
(b) $P\left(A_{2} \cap A_{3} | A_{1}\right)$
(c) $P\left(A_{2} \cup A_{3} | A_{1}\right)$
(d) $P\left(A_{1} \cap A_{3} \cap A_{3} | A_{1} \cup A_{2} \cup A_{3}\right)$

Answer

(a) .50 $\\$ (b) .0455$\\$ (c) .682$\\$ (d) .0189

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Video Transcript

all right. In a previous problem, we were given a bunch of probabilities that a certain firm will get awarded certain projects and we're supposed to now find some conditional probabilities as well as interpret them in the context of the problem. So let's start with party. The probability of a two given a one. And for this we're gonna use our definition of conditional probability. So that's gonna be probability of a two intersect a one all over the probability of a one were given both of these values. The probability of a two intersect A one is 0.11 The probability of a one is there a 10.22? This reduces down to 1/2 or 0.5. What this means in the context of the problem is that there is a 50% chance that, given the firm is awarded Project number one, it will also be awarded Project number two. All right, next problem. We have the probability of a two intersect a three given a one. Once the debt, once again, we're gonna use our definition of conditional probability eight to intercept a three intersect a one all over the probability of a one. All right. Since Intersection is both communities and associative, this is the same thing as the probability of a one intersect h, you intersect a three, which turns out to be 0.1 over 0.22 for a probability of a one. And this is 0.0 455 Now, what this means is that, given the given that the firm is awarded project number one, there is a 4.55% chance that the same form will be awarded both projects too. And three, Remember that intersection generally means. And when we're interpreting a problem like this, now we have probability of 18 Union A three given a one. Once again, we can use our definition of conditional probability. All right, now, if we look at this, we look at this top and we see, huh? Intersection distributes so we can expand that out. Union A three, a one. There we go. All right. Now we can use our descent rule for probability to expand the numerator two probability of a to intersect a one plus probability of a three intersect. A one minus probability of would be the intersection of both of these. But that's just a one. Intersect H two intersect a three because the A one is redundant all over the probability of a one. So that's gonna be 0.11 plus 0.5 minus 0.1 all over 0.22 just substituting it. And that ends up being 0.682 In the context of this problem, what this means is that given that the firm is awarded project number one, there is a 68.2% chance that'll either be awarded Project One or project number three. Sorry, Project number two or project number three. Keep in mind a union generally means or all right, Finally, we have the business. The probability of a one intersect A to intercept a three, given a one union A to union A three once again using our definition of conditional probability, it's a one intersect A to intercept a three intersect a one union, a choo union, a three all over probability of a one union A to union a three. Now, in order, some fly this top. I'm going to use a Venn diagram is a visual, so let's call a one the green. A two Sorry, a one Laurent H of the green in a three, the blue. And if we make a Venn diagram like so where that's these colors correspond to each of these ays. Then we know that a one intersect A. To intercept a three is the center section here and that a one union A to union A three is the entirety of what's inside these circles. Now, in order to find the intersection, we need to figure out what's common between the two. Well, that's just a one intersect A to intercept a three. So this top, it's just the probability of a one intersect a two, and it's like a three. Now we have to use our addition rule for three variables on the bottom Here, probability of a one intersect. Sorry. Plus the probability of a two lost the probability of a three minus. The probability of a one intersect a two minus The probability of a one intersect a three minus The probability of a two intersect a three plus the probability of a one intersect h you intersect a three by the way. Yes. I am aware this was a require part of the problem where these probabilities come from. But that's just a jog. Your memory, in case you don't remember that we did that earlier or if he just wanted to computer by hand again. All right. This top part is 0.1 and this bottom part ends up being 0.22 plus 0.25 plus 0.28 minus 0.11 minus 0.5 minus zero points. Said then that doesn't work, right? It should be 07 shouldn't it? All right. Yeah, that should be 07 My fault. Just looking back at the problem, I just realized that looked way too big. Plus the probability of zero Windsor one that ends up being 0.53 If you divide that out, that become 0.189 Now, what does this mean in the context of the problem? Well, that means that given that this project this firm got awarded either projects 12 or three, that there is a 1.89% chance that it got awarded all three projects projects one and two and three. And there you have it