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Atacertain gas station; tne prooability tnatan automobile needs an cil change 0.36, tne probability tnatit needs new oil filter 0.46_ and the prooability that ooth ...

Question

Atacertain gas station; tne prooability tnatan automobile needs an cil change 0.36, tne probability tnatit needs new oil filter 0.46_ and the prooability that ooth the cil and the filter needs changingis 0.2 Given that a automobile dces NOT need oil change; what isthe conditional probability tnat it needs new oil filter?

Atacertain gas station; tne prooability tnatan automobile needs an cil change 0.36, tne probability tnatit needs new oil filter 0.46_ and the prooability that ooth the cil and the filter needs changingis 0.2 Given that a automobile dces NOT need oil change; what isthe conditional probability tnat it needs new oil filter?



Answers

The probability of an automobile being filled with gasoline also needing an oil change is 0.25 ; the probability that it needs a new oil filter is $0.40 ;$ and the probability that both the oil and the filter need changing is 0.2 . (a) If the oil has to be changed, what is the probability that a new oil filter is needed? (b) If a new oil filter is needed, what is the probability that the oil has to be changed?

Hello everyone. So we'll be going to solve problems 106 from Chapter two probability. Here we are, given that in a service station A 30 minute period. How many how the gases filled in certain number of cars? So how many number of cars the gases failed? There are different probabilities. Number of cars it's given to you one too three, four it's like and there are different probabilities that are given gas is 30. So in zero cards the verbosity of filling a gas 0.03, there is a 20 minute period that has been given In one car had a 0.1 A. And two cars. The 0.24 And three cuts of 0.2. It Okay And four cards. 0.10 Five cars had a C 0.17. This is nothing but the data that has been given indication. Now we have to calculate in the a part you have to calculate the gas has been filled in two Plus cars, started more than two cars. So we will more than two cars. We will choose some. The guy's been filled in three cars too. Mhm Plus C 2.1 Seattle plus zero point 1 7. Basically the some of these three. So this will come out to be 0.55. So the probability of filling a gas in more than two cars. Travel day 0.55 in people. We have to tell gasping, filled in at most four cars, so we'll simply subtract The property of filling a gas in five cars 1 -0.17. This comes out to be zero 8 3. So filling gas and At most four cars, The prerogative is 0.83. Now there is C. Part, the gas is being filled in more food or more cars. So this is the son of these two C 2.1, so you don't plus 0.17 This comes out to be 0.27. The probability of feeling guessing four or more cars Will be 3.27. Thank you.

So this problem is asking us to find the probability of when John is driving to school with company. Now, let's start with the three main important information before we go into the answer. So the first important information is that the probability of John driving probability of John driving is 0.37. Okay. Ah. And I'm gonna put here John driving Now. That other probability is that we have a probability of John writing with some friends. Told school being 0.23. Okay. And I'm here, gonna put writing with friends, were at up writing with friends, writing with friends, and that's but at least they give us a probability of John being taken by their parents. Okay, so and as a probability of zero point for okay. And I'm gonna put here parents to make it easier for us. No, they give us a very important data, and it's or fact. And it's that when John is driving his alone So John driving and we have to put here alone, and they want us to find what is the probability of young writing to school? Doesn't matter if he's driving or not. John. Right into school with, right with company, whichever that person or persons might be company. So which of this three facts right? We have to think about the three facts. Which of the three fax It's jump with company. So I know that John. Oh, sorry about that. Yon travels with company, right? John is with company in two of the facts, and I'm gonna market with Red. OK, so when he's writing with friends to school and when he's with his parents, not when he's driving. Because when his driving his driving what alone? He can only be alone. So that means that I'm gonna take the probability of John writing with friends, right? Ah, which is 0.23. Those because it's with right. And that works as an end. The probability of John writing with its parents, which is 0.4. So a point for and I'm gonna add them. And the total probability of your own driving with a company is of 0.60 three. And that's our answer.

Starting with problem A. We want to find the probability that no more than four cars will be serviced. So we want to find the probability the X Is greater than or equal to four. Yeah, we do this by finding the probability of three which is point 12 plus The probability of x being equal to four. Or the how many cars are serviced on such a given day. So that would be p of 0.1 night. So X would equal for So this is x equals three And this is x equals four, which would equal 0.31 So to find the probability that a mechanic Services no more than four cars, you would take the probability of X equals three plus the probability of X equals four. For the next problem, you want to Find the probability that he will serve as fewer than eight cars. So you would do the probability of X greater than eight equals. And for this problem you want to do one minus the probability of X being greater Being less than or equal to eight. And all you're going to do for this problem is you're going to take all of your values and add them together from 1-7 and you'll get a number or you can do another thing. And this is easier. You can take the last number that fits in this parameter and the only number that fits in this parameter Is Equal 2.07. So you do 1 -17 equals 0.93 And that's what you're doing for the first numbers too. You're fighting the probability of some arbitrary parameter which is x equals three And x equals four for this one for part A So for the last question, part C We want to know the probability that we want to know the probability That he will service three or four cars, which is equal to be X equals three plus the probability of X equals for which is 0.12 plus 0.19 equals 0.31

Okay for this problem. We have one moment here. Yeah, we are considering a gas station where 40% of customers use regular gas. We call that, uh, event a one that has a probability of 0.4 35% use mid grade gas, so probability of a to equal to 0.35 and 25% use premium gas. So you have a three is equal to 0.25 I also know that of those using regular gas, only 30% filled their tanks. So we know that the probability of B given a one is equal to 0.3. You know that the probability that or of the customers using mid grade gas, 60% filled their tanks probability of be given a to 0.6 and the probability of, or rather 50% of those who get premium gas will fill their tanks The probability of be given a three equal to 0.5 Now for part A. We want to find the probability that the next customer will request mid grade gas and fill the tank. So we want to find the probability a to and be so probability of a two and B is equal to the probability of be given a to times the probability of a to so that is going to be 0.6 times 0.35 that is going to equal 0.21 for Part B. We want to find the probability that a customer will fill their tank, so that is the probability of event be. What we can do is we can use the law of total probability. Here. She tells us that the probability of event be where a one in this case, a one to a three, our exclusive and exhaustive. They have to get regular mid grade or premium and their, uh, you can't get regular and grade or any other combination of them so we can use the We'll have total probability, so probability of be given a one times probability of a one, plus the probability of be given a to times the probability of a to plus the probability of be given a three. I'm the probability of a three, which we should recognize that each of these are the same as that's the same as probability of B and a one so on, so that plugging in what we have there we get 0.3 times 0.4 plus 0.6 times 0.35 plus 0.5 I'm 0.25 something. All that up, we get 0.455 Next for part C, the next customer fills the tank. What is the probability that regular gas is requested mid grade gas or premium gas? So that is the probability of a one given B. And similarity is going to be a probability of a two, given the A three given, but can calculate this using I believe it would be. Think it's Bae's here? Um, yes, Bayes Theorem 1.5 in the text. So using base serum that is going to be equal to the probability of a one times the probability of be given a one divided by the probability of B, which in our case is going to be 0.4 times 0.3, divided by 0.455 using the result from Part B. Um, that gives us a 0.27 doing a little bit of rounding. There than probability of a two given B equal to the probability of a to times probability of E given A to divided by the probability of B that is going to give us 0.35 times 0.6 divided by 0.455 is going to equal 0.46 and, lastly, probability of a three given B is equal to. I'm going to skip writing that general form, but we know what we're looking at. Just walk up the 18th for a threes that's going to be 0.25 times 0.5 divided by 0.455 So that is going to be 0.27


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