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5~1oXRIn a certain assembly plant, three machines BL_ B2 , and B3, make 30%. 4S%. and 25%, respectively; of the products. It is known from past experience that [%, ...

Question

5~1oXRIn a certain assembly plant, three machines BL_ B2 , and B3, make 30%. 4S%. and 25%, respectively; of the products. It is known from past experience that [%, 2%/, and 3% of the products made by each machire, respectively. are defective Now. suppose that finished product is randomly seleciqd, thenthe probability that it is defective iS 0.0195 195 0.9805G0 -"5daje41

5~1oX R In a certain assembly plant, three machines BL_ B2 , and B3, make 30%. 4S%. and 25%, respectively; of the products. It is known from past experience that [%, 2%/, and 3% of the products made by each machire, respectively. are defective Now. suppose that finished product is randomly seleciqd, thenthe probability that it is defective iS 0.0195 195 0.9805 G0 - "5daje 41



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.

The problem it. We have three machines that are producing springs of the same lands The three Machines. Machine one two and three produces 1% 4% And 2% of the effective springs. And of the total production machine one produces 30% Machine to produces 25 And machines reproduces 45 for party. If one Spring is selected, we want to calculate the probability that this is bring is the victim. We can represent that effective spring as an event. A then to get the probability of aid, it equals the probability of being effective From machine one, then it's probability of a given machine one plus the probability of being the victim. Given it's machine too plus probability of being effective Given its machine three, then it equals the probability of being defective of machine one was deployed by The Probability of Machine one. I've missed something important. Let's repeat it, sorry. It equals the probability of being defective from machine one Given machine one Multiplied by the total production of machine one. Plus this is plus the probability of being effective given its machine too, multiplied by the total production of machine too. Plus the probability Of being effective. If it's machine three deployed by the total production of machine three then equals 1%. Multiplied by 30%. Multiply by 25% plus 2% to fly by 45%. 4.41 drive by going to see plus a point of food Are deployed by 25 0.25 Plus 4.2 Multiply by 4.45. Then it equals this is not A .45, it's a .45 And it's all .02 or 11 Divided by 500 for barbie. It's given that selected his bring is defective. You want to find the probability that it was reduced by machine too. Then we want to find the probability of machine too, given it's defective equals the probability of the intersection and intersection can be re read to be the probability of machine. Two mark employed by the probability of being effective. Given that the machine took like this part, divided by the probability of the second event of being different equals this part 4.25 multiplied by a point offer, divided by Or point or two, Then it equals five, divided by element. Then this is the final answer of party and this of an answer of party.

This problem, we are told that we have a box of 25 and we're taking a sample of three of them to test for defective today. We want to find the probability that a box containing three defectives will be shipped. And so that means we have three defectives And 22 that are not defectives. Now in order for it to be shipped. We don't want to find any defectives at all. And so that means when we choose three of them, we want to choose only from the 22 that are not defective. So 20 to 20 to choose three is the number of different ways in order for that to happen. And then we're gonna divide that by 25 shoes three. Because that's the total number of different ways that we can choose three items from. The total of 25. And so we evaluate this probability And this gives us .66, I'm sorry. .6 696 You and me. We want to find the probability that a box containing only one defective will be sent back for screaming. This this means we have 24 good ones and one defect. We want to find the probability that we find at least one defective. Well, there's your compliment. All this is April 1- the probability that we don't find it any of the defectives. So this is one 24-3 Over 25, choose three For the 24-3. That's the number of different ways that we can choose only good ones From the 24 that we're told there did. And then the 25 choose three is the number of total ways. And evaluating this gives us .12, exactly.

In this question, it is told that there are 10 items and out of them today uh and get back to you and so on. Are not effective. Or we can say that they are safe and we have to select three items from them. And it is also told that we have to find the probability that oh Lord, big thanks to let us say that A. D. North getting up the fed you items. So it can be a bitterness Probability of the 50 for 1st, 2nd and 3rd. Now we can say that these are not independent because replacement is not done there. So you can write it as probability of way, then probability away after a and probability of a after a. N. A. Now, first of all, the ability of a means getting a defective item that is three are defective, so three by 10. And now after the effective. Once again we need effective. So now there are remaining two defectives out of 9th or two x 9. And similarly, at last, after two defectives, we want again defective. So now we have one defective item and we remain with a total items, so one divided by it. After solving it, we can see that We got here for one upon or 120. So when upon 120, it's the answer border required question here. Thank you.

In this question, we're told that a box contains £25 on off £25.3 but effective on the other 22. I'm no defective. So the first thing we need to do is put this into probabilities of being effective, so Snus to find the as defective. So our probability of D is three out of 25. Now you can't keep it as a fracture, but it's often nice that as a decimal, so a probability of being defective is no 0.12 We're now told if two parts of selected without replacement find the following probabilities, So look apart. A. It says both a defective on here. It's important that that we don't have replacement, so it's that they're dependent. So for both to be defective, we need three out of 25. But now, if you take that three out of 25 you know now we got to defective parts left on Day 24 in total. So these are the two probabilities you're going to be multiplying together. So that is that. No 20.12 on Day two of 24 is no 0.8 North eight three, three So on the 2nd 1 you've actually got a lower chance of picking the defective one out because a lower proportion of the parts is not effective as you've already taken the 1st 1 So if we most tried these two together, we get no point, not one. So this is saying there's a 1% chance that if you pick up two parts without replacement, that they will both be defective, which is really low on Part B is saying exactly one is defective. So for that to happen, we need one of two things, and that's important. We they need the 1st 1 to be defective on the 2nd 1 not to be defective. So that would be the 22 Remember on the second time is only 24. So that's the first possibility. All we could do it the other way around on the 1st 1 we pick isn't defective, but the 2nd 1 we pick is defective. Remember, seven times out 24 this is three head because we haven't picked a defective part out yet. So gone No point, Monty, and we're times in that. But 20 to over 24 she's not 0.9167 and then here we have 22 out of 25 which is no 0.88 and the times and that by three out of 24 which is no point. Want to fuck so we can work each bit individually. So no 0.12 times at no 0.9167 is no 0.1 worth. And what? Adding that that no 0.8 times no 0.1 to 5, which is also no 0.1 month. The over probability is no point to to so moving on to part C fancy and asking us the probability that neither is the fact. So for this we need to pick out a non defective one on the first try, which is 20 to 25 which is that No. 188 and then we need to pick out another non defective one. But now that I would be 21 of those left on 24 in total, which has a decimal No. 1.875 So if you multiply those together we get No. 2.77


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