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4,35 A production manager knows that 5% of components produced by particular manufacturing some defect: Six of these = process have components_ whose charac teristi...

Question

4,35 A production manager knows that 5% of components produced by particular manufacturing some defect: Six of these = process have components_ whose charac teristics can be assumed to be independent other; are of each examned_ Whatis the probability that none of these nents has defect? compo- b What is the probability that one of these = nents has . compo- defect? Whatis the probability that at least two of these components have defect?

4,35 A production manager knows that 5% of components produced by particular manufacturing some defect: Six of these = process have components_ whose charac teristics can be assumed to be independent other; are of each examned_ Whatis the probability that none of these nents has defect? compo- b What is the probability that one of these = nents has . compo- defect? Whatis the probability that at least two of these components have defect?



Answers

Manufacturing Your manufacturing plant produces aii bags, and it is known that $10 \%$ of them are defective. Five air bags are tested. a. Find the probability that three of them are defective. b. Find the probability that at least two of them are defective.

So we have ah, box of 50 parts, seven of which are defective. And we're calculating two different probabilities. Um, well, they're both the probability of two defective parts strong to defective parts one at a time. However, it's slightly different because the first experiment, we're not replacing the parts. And the second time, we are so doing this experiment twice. So we started the first probability when we're doing it without replacement. So we want to defective parts basically in a row, we pick one defective part, and then we pick another defective part. What is the probability of that happening? Well, on our first draw, um, the boxes full, we haven't taken anything out, So we still have seven defective parts out of 50 little parts will have seven out of 50. Now for our second draw. We are not replacing. So we took a part out, so we now have one less part. So that's how we get 49 in our denominator. And we're assuming we drew a defective part because we want the probability of two defective parts. So we drew already drew a defective parts of those will actually one last effective part as well, so we'll have six out of 40. Nice. There's one less in each, the numerator and the denominator, and then we'll multiply those together. And so our probability of two defective parts without replacement is 42 over 2450. Now, this probability is going to change. Would be are replacing. So now the probability of drawing to defective parts while our first draws still seven out of 50 because we haven't drawn anything yet. So everything is still in there. So we stopped the draw a defective part. And now we put that effective part back, right? We're replacing it. So when we draw it again, everything is still the same. We still have seven parts out of a total of 50. So now our probability becomes 49 over 25. That was Excuse me, 2500. And so we'll see. If we convert these two percentages, we can see how they differ. So this first percentages 0.17 and the 2nd 1 is 10196

This problem, we're told that a firm planes that the proportion of defective units is 5%. And so this means that P. Big .05 We're told the buyer has a standard procedure of inspecting 15 units and some things are in value is 15. And so in general, the probability of acts, There's 15 choose that's mhm Times .5 to the X .95 to the 15 -X. Just using our formula for a binomial distribution, We're told that the buyer found five items affected on a We want to find the probability of this attack of this happening. Yeah, So P of five, if we put on five frets, This is 15, choose five Times .5 to the five Times .95 to the 10. And you know, all we did was just plug in five in for X. And our binomial distribution. And evaluating this Is 5.62 Times 10 to the negative 4th power Or .000562. Now on be if we were the buyer, it would be very upset at this because that's a very low probability. So we were there. I would be very upset because there is a low probability of this happening. Do you expect there to be much less that are defective?

Hello everyone. So we're going to solve problems. 119 from 72 of probability. So in this question we are given there are certain suppliers And this applies the components in a lot of 20 data. 20 components a lot. Now we are given certain probabilities like those Lords can have certain defective components is filled. So we are given probability that there will be zero defective components in that 20 items. It's given us 0.6. Probably that there will be one defective component in that lot. And this has given us 0.3. And probability that there will be two defective confidence. It has given us zero What? No, we're given that Uh huh. Two components from the Lord a randomly selected and tested and we found out that they are modified. Let's treat a as even two items selected and there non defense. No, no credibility. Given that the Lord is zero, what is a fabric that both items will be not effective? As in this Lord, there is no defective item. Susceptibility of food items being non defective will be fun. Similarly from the Lord where there is one defective item availability of even A will be 19 C. Two. We waited by 20 C. Yeah, There is one defective items. Those 2 1950 minutes should be coming from those 19 non defective items solving this will eventually give us So you 2.9. Running probability. Uh huh. Given that the Lord is where there are two defective items, what is the probability to what I'm selected will be non effective On the similar lines. This would be meeting c. two Divided by 20 c. two. Where did you just use the combination function? I will just write the combination function for you and see our yes and factory devoted way R. Factorial. Multiply and minus R. Factorial. So just use this formula to calculate. I calculated it as zero 8053. So the city the property that two items selected. Lebanon effectively selected from The Lord where there are two defective items. Now we have to calculate in the A. Part. What is the probability that zero defective components exist in the Lord? Now we're given that a event has already happened? What is the verbosity it came from? Look zero, we'll use the beast here. Here. This will eventually become and to be a maybe. Mhm. See to fight back some of total probability. This would be B zero plus. Now considering the Lord with one defects. Considering a lot two defects, we have all these property values. We have P zero P one, P two. It does give an indication we have T zero B. Haven and P. Two. We just calculated so we'll just put these values here. B0 is one to still become one and 2. 25 1 and two. 0.6 Bliss 0.9 and 2 0.3. This is 0.9 In 20 three. And the last one would be zero point he 053 Multiply by 0.1. Just will eventually come out to be C 2.6 divided by 0.6 Bliss 0.27 plus 0.0 805 three. This will be I will just use calculator to solve this. That's fine. Zero 8053. Mhm. This would be 0.6 divided by 0.950 five treat. This will eventually come out to be is your point 6312. So this will be a possibility for the A part. Now is coming to be part B&C. Parts are on the similar 9th. So be part is what is everybody that one defective items exist in a lot. Now we have to just staple it. Given that it happened, we have to find a property that it was a lot where one defective item was there. So this will just be 0.27 divided by the total, which we have already calculated. So we are not required to calculate it again. Okay, This will eventually become 27 .95053. So come out to be 0.284. You can point to eight ft Yeah, third parties for Israeli the two defective exist in the Lord for two defective. Again, we will be working on the similar lines, it could be 0.8 and the answer for this would be .080 53 Development .95053. This will be 0.084 or 0.05 two B. What precise? Thank you. Yeah.

So in this problem, the probability or feeling is given. So the probability off failing 0.1 according to the problem on the probability off not feeling is 0.99 So therefore, the required probability in subpart a. So the required probability is one minus 0.99 to the power off for which is 0.39 double for In part B, Daddy acquired Babalu T is 0.99 to depart off four, which is 0.960 fight. And in part C. The required probability is four C 30.0 what deployed with 0.1 which is the probability off feeling multiplied with the probability off not feeling 0.99 So therefore it is 396 in tow tend to depart off minus eight.


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