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Problem 53 Hard Difficulty

Reconsider the system defect situation described in Exercise $28 .$
(a) Given that the system has a type 1 defect, what is the probability that it has a type 2 defect?
(b) Given that the system has a type 1 defect, what is the probability that it has all three types of defects?
(c) Given that the system has at least one type of defect, what is the probability that it has exactly one type of defect?
(d) Given that the system has both of the first two types of defects, what is the probability that it does not have the third type of defect?

Answer

(a) .50 $\\$ (b) .0833 $\\$ (c) .3571 $\\$ (d) .8333

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Video Transcript

all right for this problem. We're calling back to a previous problem where it gave us the probability of certain types of defects. And now we have some conditional probability to deal with. So the 1st 1 we want to find the probability of a type to error, given that there's a type one er so h you given a one and for that we're gonna use our law of conditional probability. So that's gonna be the probability of a two intersection of a one over the probability of a one. Now, this problem assumes that you've done the previous problem. Problem number 28. But if you haven't or you just don't want to refer back, I'm just gonna walk through driving. What probability of h two intersect A one is. You just used your addition law, like so. And you just plug in what you know, and we're going to do that and solve for the probability of a one intersect a two. So the probability of the union is 0.13 Probability of a one is open. One to probability of a two is your 20.7 minus probability of a one intersect a two 0.13 equals 0.19 minus the probability of a one intersect eight to or the probability of a one intersect a two equals 0.6 That numerator come 0.6 We divide that by the probability of a one which is 0.12 and this becomes 1/2 or 0.5. All right, same thing over here. Basically, we want to find the probability of a one intersect H two intersect a three given a one. So we're going to use that law again. So we're gonna have a one intersect A to intersect a three all intersected with a one over the probability of anyone. Intersection is community and associative, which means we can just imagine these parentheses aren't here for the time being, and we see that a one is duplicated, so we only need to express a once one wearing an intersection. So it just simplifies down to that and we know both of these. The top is gonna be zero points or one. The denominator is gonna be 0.12 This become 0.833 round 2 to 4 decimal places. All right, we want to find the probability of exactly one defect, given that there is a defect. So we're gonna use a Venn diagram, So I'm getting with the red equal a one green equal. Let's redraw that. Let's move that circle so I can write in. And actually, let's let the green equal the green, not the black, for some reason, equal a two and the blue Be a three. I'm gonna make this. They are There we go. Right. So we know the center part. Based on this piece of information, here is 0.1 If we go back here, we found out that the intersection of a one a 20.6 total and that includes this. So the intersection here without a three is 0.5 If we math up that weaken math out the intersection between H two and a three really quickly using the same method or yeah, we could do a one day three, but let's just do a two and a three just because it doesn't really matter any three minus the intersection of Sorry. Wait, That should be a union. This should be an intersection. There we go. So the union is 0.10 Ah, Probability of a two is 0.7 Probability of a 30.5 Minus the probability that we're looking for, uh, 0.10 equals 0.7 Sorry. Wait. No, Hold on. We have those together. You got 0.12 So therefore, the intersection between a two and a three equals 0.2 We already know that we have a 0.1 here, so we get zero points or one. We know eight's your total should be 0.7 He's already add up to 0.7 So this is 0.0 Uh, eventually, we did have to calculate the intersection shin between a one and a three. So we'll do that real quick. Same setup. And again, if you did the problem that this referred to problem number 28 then you should have this already. I'm just doing again for the sake of refreshing your memory. Ah, let's see. This is 0.14 because 0.120 point 05 minutes. All this skipping some of the algebra, You know that This must be This is seven. This is 70.17 this 0.14. So this is 0.3 We already have a 0.1 in here is that this becomes 0.2 We know the A one must add up to 12. This adds up. The 120.8 are Sorry. 0.12 This adds up to a 0.8 Right now you are missing a 0.4 And this almost add up to five. We currently have four. Or I should say 0.50 points or four. So we're currently missing a 0.1 That's where Venn diagram looks like. I'm gonna delete all this just to clean up our work. Oops. We need that information. You just put that back in the corner here and I'm gonna take a Venn diagram. No, no, no. I don't need to highlight everything. Hold on. There we go. And I'm just gonna move aside for a reference. So we need a probability of exactly one defect to given all this. So using our lot of conditional probability that the probability of let's just call this event e just for shorthand. He given a one union, a two union a three. This adds up by the way to four plus five is 99 plus one is 10 10. Plus. She was 12. Well close. One is 13 13 plus one because 14. So the probability here because 0.14 So the problem. So that's useful because that's our denominator. That's being intersection, not a line. There we go. So we know the bottom 0.14 We need the probability there's exactly one defect given all this. Well, we just add up these probabilities here 0.5 And if we divide that out, we get zero groups point three 571 All right. For the last part, we need to find the probability of not a three given a one intersection a two. Well, immediately, I can say that this is one minus the probability of a three a one over a two, and then we'll use our law. Conditional probability. It's gonna be a probability of a three. No section anyone. Intersection A two divided by the probability a one intersect a two. So that's one minus the top part. It's just 0.1 in the bottom. Part of we referred to heart eh was 0.6 So this is one minus. This reduces down to 1/6. We're just 56 which reduces down. Our ingestible form is 8.333 rounding to four decimal places.