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# For customers purchasing a full set of tires at a particular tire store, consider the events$A=\{$ tires purchased were made in the USA $\}$$B=\{ purchaser has tires balanced immediately \}$$C=\{$ purchaser requests front-end alignment $\}$along with $A^{\prime}, B^{\prime},$ and $C^{\prime} .$ Assume the following unconditional and conditional probabilities:(a) Construct a tree diagram consisting of first-, second-, and third-generation branches and place an event label and appropriate probability next to each branch.(b) Compute $P(A \cap B \cap C)$ .(c) Compute $P(B \cap C) .$(d) Compute $P(C)$ .(e) Compute $P(A B \cap C),$ the probability of a purchase of US tires given that both balancing and an alignment were requested.

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Probability Topics

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### Video Transcript

alright for this problem, We are considering customers buying a full set of tires from a particular tire store. Vent A is where the tires were made in the USA Event B is where the purchaser had the tires balanced immediately and events see where the purchaser requests a front end alignment. And then there are a prime b prime and see prime, which are the the knots of each of those events. We also have these given probabilities and conditional probabilities. I'm not gonna bother reading all those out. So for part A, we want to construct a tree diagram for the 1st, 2nd and 3rd generation branches and plays an event label and appropriate probability next to each branch. So we know we're going to have our starting point at our first decision. We're gonna have to our second decision. We are going to have four. They're not for sorry. We are going to have eight. One second. 123 Okay, I'm messing that up. Yeah, for our second decision, we should have total four. And then for our third decision, we should have eight. So I'm gonna zoom in quite a bit here because it's going to get a little nasty. So we start off and then we have event A with probability of a is 0.75 and we have a prime. The probability of a prime is going to be one minus 10.75 or 0.2 to 5, then here for this branch, we have B where the probability of be given a is equal to 0.9 and we have be prime where the probability of the prime given a is equal to 0.1 Up here we have C probability of see, given B and A is equal Thio 0.8. And this event here that event is um yeah, that event is Yeah. No, that's the best way of labeling that actually. Sorry, but my hesitation there we have c prime down here. Probability of C prime given B and a equal to 0.2 than this branch have see ability of see given be prime and a is equal to that 0.6 probability of C prime. Just this branch down there ability of C prime given be prime and a is equal thio 0.4 Then we have offer a prime branch here B and B prime again. Where we have probability of be given a prime is equal 2.8 probability of be prime given a prime is equal Thio point to we have C see Prime and C and C prime Where you gonna zoom? In further probability of C given a prime and be is equal 2.7 the probability of yes ability of see prime given a prime nb equal 2.3 then lastly or the last pair you have a probability of C given a prime and be prime which is that is going to be 0.3. What and we have done here the probability of see prime given a prime and be prime which is equal 2.7 And then these events at the end correspond to so that is see given a or rather not given that is A and B and C A and be and see prime A and be prime and see a and B prime and see prime um, a prime and B and C a prime and be prime and or sorry a prime and be and see prime. Then we have a prime and be prime and see. And we have a prime and be prime and see prime. We can calculate the probabilities of each of those by multiplying down our branches. So I'm going to do that off screen here. Actually, my mistake calculating all of those eyes going to be a bit unnecessary. So we now have the proper branches and event labels for everything. That's all we need for part a part B. Yes. So this is the total tree there, Part B. We want to find the probability of A and B and C, which we confined just by following our tree. That is 0.75 times 0.9 times 0.8, which comes out to 0.54 part C. We want to find the probability of B and C happening, so that means that it's going to be mhm probability of B and C and a plus the probability of B and C and not a which we already have. BNC and a keep in mind that for the ANZ, the order doesn't matter. Forgiven for the conditioning it does, but for and you're ordering doesn't matter. It was 0.54 plus 0.25 Excuse me, M 0.8 time 0.7. That gives us 0.68 for D. We want to find the probability of see occurring. So that is going to be the probability of B and C plus the probability of not be NC. Uh huh. You know that the probability of B and C 0.68 the probability of not being C going to be so that this event can be broken down into actual do it off to the side there. Keep in mind that that's going to be essentially the probability of a and not B and C plus the probability of not a and not be and not be and see not be so probability of a and not B N c his 0.75 times 0.1 time 2.6 and the probability of not a and not B and C is 0.25 times 0.2 times times 0.3. So adding those all together we get 0.74 is the result. Lastly for E, we want to find the probability of a given B and C you can find this using Bayes Theorem, which tells us that that probability is going to be the same thing as the probability of be all right. Same thing is the probability of A and B and C divided by the probability of B and C, so we already found each of those pieces. Previously, ability of A and B and C is 0.54 Probability of BNC is 0.68 so we end up with the probability of a given B and C 0.79