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A certain system can experience three different types of defects. Let $A_{i}(i=1,2,3)$ denote the event that the system has a defect of type $i .$ Suppose that

$P\left(A_{1}\right)=.12 \quad P\left(A_{2}\right)=.07 \quad P\left(A_{3}\right)=.05$

$P\left(A_{1} \cup A_{2}\right)=.13 \quad P\left(A_{1} \cup A_{3}\right)=.14$

$P\left(A_{2} \cup A_{3}\right)=.10 \quad P\left(A_{1} \cap A_{2} \cap A_{3}\right)=.01$

(a) What is the probability that the system does not have a type 1 defect?

(b) What is the probability that the system has both type 1 and type 2 defects?

(c) What is the probability that the system has both type 1 and type 2 defects but not a type 3 defect?

(d) What is the probability that the system has at most two of these defects?

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right. So a certain type of system could experience three types of defects with home defect. 123 And so, uh, the event is actually the event. Uh uh. Having defect one would be a one. That's probability of one, too. Uh, next, we have the probability of B two that the second Deepak is the costs, and that's 20.7 in the 315 Then we have some more information about the unions and the intersections of the So here's all the unions of the two sets in front of your section of all. Probably nice to drawing a little been diagrammed, represent this situation, and then we can put the probabilities into Ben diagram. Help us organize it to answer these questions. So I must have this one be a one. This one be you've got to. This one will be defect three. Now, we know that this probability for the whole circles 0.12 probably. This whole circle is seven. And the probability for everything in this earth was five. Right? Um all right, so, uh, we know that where the all crossed all right, we'll be with the unions in a second Okay, So what's the probability? The system does not have a type one defect case. It does not have a type one defect. So that would be that would be this area. Well, let's let's start filling them with some of these other areas. Okay, So, um, a one together with a 2.13 So we just ignore a three that if we put these two circles together the aim one circle and the A to circle that these two circles together our 0.13 Now, we know that, um, if we do a probability of a one and we add, that probably ate, too. In that double counts this middle right here. Okay. So when we take that out okay of it, then this is the probability of a one and a two. You know that that answer is going to be a 0.13 Okay, so this is 0.19 and minus this probability, people's 0.13 So we know that the probability of these things happening together gave the intersection of these two is gonna be 0.6 That means that this part will be 05 Yeah. Now we're gonna do the same thing or the 0.1 foreign. The 0.1. So for Theatre section of one and three. So to find this area in here, find this area in here we're gonna do is gonna do point. Want to probability that you're in circle, eh? Minus 0.5 Probably union circle for three minus the probability of their intersection. And that's gonna be equal to 0.14 All right, so he's too added together minus the intersections. This is 0.17 minus the probability of their intersection eagles 0.14 step means the probability of their intersection is gonna be 0.3 So that means this little red eyeball thing here is 0.3 This ask you. Consider Row Two. Okay. Good for the last one Over here. Get so for a one and a two and a three. This one right here. So if we add hmm, a three together be, um, 0.7 0.7 plus 0.0 by So that area, the 0.7 plus the 0.5 double counts. The part there. That's, um, common the boat. Okay, so we can So if we subtract the probability that intersection that's gonna equal 0.1. Right. So this is 0.12 minus the intersection goes 0.1 zero p and so you can see that more easily. This has people 02 So this the green eyeball area eerie between two and three. That's 30.21 Oh, let's just finish it up. Right. So so far, we've got 0.8 in a one that means this has a 0.4 So far, we've got 0.0 for in the A three circle. And they stayed with fives one and this A to circle last and upto Windsor of seven in the armory. So zero All right now, the problem should be pretty easy. What's, uh, probability that it does not have a tight one defect? That would be this area here. This area here in this area here comes that's all. Outside of point of the area, one defect. So that answer Thio a would be, um, 102 we'll be is asking, you see, was a probability that the system has both type one and type two defects, so both type one and type two defects that would be this area year People type one and type two? Um, not nothing, man. Just and no type. Agree or not. So if it's type one and type two, but no type three, Um, when you say 0.5 if you just want to know, doesn't have those two defects, then you might say that's actually Quincy six. It's okay to also have a type three, right? How much of the word is that? Here, you see? Oh, okay. The next question's was the probability of this isn't both type one and type TV expert, not type three. Okay, so that means iwas the 30.6 This is, um, sorry about squeezing all this in, but we're almost there. And Alice, there's a heated seats so far. And then, um, D was probably the system as most to defects most to defense. So, at most, two defects means to defects or, um, one defect or no defects. So that's everything except having three defects. And so the only thing that's not included in that problem in party is this area here. Okay, that's three defects. But every other part in here has one or two defects and if it's outside, the circle has no defense. OK, so basically, what we wanna do is or deep we're gonna do one minus 101 So the probability that it has, um, at most two defects? No, that's not three beef accidents. Clean knife. All right.