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Problem 2 A batch of 500 Johnson rods contains five that are defective Two are selected at random from the batch. Do you think the probability of the selection of d...

Question

Problem 2 A batch of 500 Johnson rods contains five that are defective Two are selected at random from the batch. Do you think the probability of the selection of defective (or nondefective, for that matter) Johnson rods is independent? Why? b. What is the probability that the second one selected is defective given that the first one was defective? What is the probability that both are defective? d. What is the probability that both are nondefective?

Problem 2 A batch of 500 Johnson rods contains five that are defective Two are selected at random from the batch. Do you think the probability of the selection of defective (or nondefective, for that matter) Johnson rods is independent? Why? b. What is the probability that the second one selected is defective given that the first one was defective? What is the probability that both are defective? d. What is the probability that both are nondefective?



Answers

A batch of 500 containers for frozen orange juice contains 5 that are defective. Two are selected, at random, without replacement from the batch. (a) What is the probability that the second one selected is defective given that the first one was defective? (b) What is the probability that both are defective? (c) What is the probability that both are acceptable? Three containers are selected, at random, without replacement, from the batch. (d) What is the probability that the third one selected is defective given that the first and second ones selected were defective? (e) What is the probability that the third one selected is defective given that the first one selected was defective and the second one selected was okay? (f) What is the probability that all three are defective?

Hello everyone. So we'll be going to solve problem 96 In this question. We have been given that there is a badge of 500 containers for orange shoes Out of which five are defective. Now it's given that two are selected at random without replacement knowing that. Mm So we have 500 continues how just size are defective on the A path. We are being asked what is a property that the second one selected as defective? Given that the first one was defective. The first one I was defective and everybody of second one were defective would be food Out of 499 has only four defective I left now uh and 499 grand dinners in total are left. So this will be the property and the second container will affect the first one is different coming to the b part. What is the probability that both are defective, Probability that both are defective would be five out of 500. Multiply fold out of 4 99. So the solution property that both the containers would be defective coming to see part but it's apparently that both are acceptable. Both acceptable property for these people. The First One is September 495 5 500. No 2nd 1 is also acceptable. 494 divided by 499. So this will be the remedy that both will be acceptable Coming to depart notice given the three containers selected at random without replacement, Probability that the 3rd 1 is selected effective and 1st and 2nd ones were defective. Now the first to contain this. Both have been effective. Probably the 3rd 1 would be defective a little bit three 4 98. E part property that the first that the third one is selected as effective given that first one was defective and second one was selected was okay. Now, for the victims are left as only one was affected over the first two for her container. Dude effective. Given the conditions, this will be the probability. Now coming to the fight. What is it possible to get? All three are defective. It would be five divided by 500, Went to full, divided by 499 into three, Divided by 4 98. So just look at everything that all the three would be defective. Thank you. Have a good day.

We're looking at a sample size 100 or 20 or bad. And we're selecting two at a time. What is the probability that the first one we get is bad, it's gonna be 20%. And then what is the probability that the second one selected is defective, given that the first one was defective? Um So we got this right here, we could say that it's the conditional probability formula like this, but if we think about it, That would mean that if the first one was bad, but there's only 19 bad ones left And there's only 99 to choose from. Mhm. So that's gonna be a conditional probability book. Now, in part C, what is the probability that both are ineffective? So that is going to be the probability of a union be and we can solve using this formula. So that's just going to be the probability of B given a. Which is this number right here, Times the probability of a. So we'll do this under here 19 over 99 times point to Or just 1/5. That might be better. Right? Instead of mixed fractions and all that. So that is simply going to be 19 over 495. Thank you. And then for part D. How does the answer to be changed if chips selected were replaced prior to the next election? Um instead of 99 we'll be looking at 100. So the probability of be given A is just going to be the probability of drawing a faulty or ineffective Thing. So that's just gonna be 20%.

In this exercise, we are told that the manufacturing of semiconductor chips produces 2% defective chips. So let's say that probability of success is .02. We're also told that the chips are independent And a lot contains 1000. So let's say a lot is our sample and were asked two questions about the probability that the number of defective chips as a certain number in this lot. So we can say that X is a binomial random variable. Now, part is to approximate the probability that more than 25 chips are defective. What is the probability that X exceeds 25? Now note that n times P is equal to 20 Which is greater than five And End Times 1 -2 is 980, which is also greater than five, which means that we can use the normal approximation with reasonable accuracy. Now this is equal to 1 -1 the probability that X Is at most 25. And if we apply the continuity correction factor, It's equal to 1- the probability that X Is less than or equal to 25.5. Now using the normal approximation Approximately equal to 1- the probability that zero is less than or equal to 25.5 minus n times p over the square root of n times p Times 1 -2. That is 1- the probability that said is less than or equal 1.24-3. And this comes out to an approximate probability Of .1071. Then for part B We are asked to approximate the probability that between 20 and 30 chips are defective. So I'm assuming that includes the number is 20 and 30, So this is equal to the probability that X is at most 30 Minour the probability that X is strictly less than 20. If we apply the continuity correction factor, Probability that X is less than or equal to 30.5 Minour the probability that X is less than or equal to 19.5 and now, if we use our normal approximation, the probability that said is less than or equal to you 30.5 minus n times p just 20 divided by the square root of the variance. This comes out to the probability that set is less than or equal to 2.317, Sorry, Minour the probability that is it is less than or equal to zero 1129. And this comes out to the probability of .5361

Okay, So for this problem, we're talking about defective lightbulbs. And typically there's a 0.5% chance the bowls are defective. So our P is equal to 0.5 That's for 5%. And we're looking at a crate of 100 bulbs. So any close 100 we want to find the probability of such things happening. And so, for part, a motherly of again in effect here. So we're not actually working with Ben equals 100 were actually just working with n equals three because we have a crate of 100 but we're only testing out of three light bulbs. So then Aaron is actually be unequal story and I and equals 100. We wanted to find the probability that all three bulbs are defective. So then this would be saying so. This is p of our equals three, and that is going to be three. Choose three, which will be one. Then we'll be times ng r zero point 0005 cube times. There's 0.9995 to the zeroth power. So really, we're just doing 0.5 cube. That's equal to 1.25 times 10 to the negative 10th. It's a very small number that should be expected. Then when we do be what are more, a bulb is defective. So them rather we're doing one minus p of our equals zero. So we're saying this is a probability that we have at least one defective bulb. It's gonna be the total probability, which is always what should always be willing. My, this probably that we have no defective bulbs, so we'll have one minus. And so we'll have to shoot 3 to 0, which is gonna do you want, and then we'll have times. The probability of success is raised to the zeroth power, which seemed the woman in the probability of failures, which is 0.9995 cubed. And so then, which will be 0.9985 And so then the probability at least one failure is zero for meat. 0015


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