Our Discord hit 10K members! 🎉 Meet students and ask top educators your questions.Join Here!

Like

Report

Numerade Educator

Like

Report

Problem 60 Hard Difficulty

Three plants manufacture hard drives and ship them to a warehouse for distribution. Plant I produces 54$\%$ of the warehouse's inventory with a 4$\%$ defect rate. Plant II produces 35$\%$ of the warehouse's inventory with an 8$\%$ defect rate. Plant III produces the remainder of the warehouse's inventory with a 12$\%$ defect rate.
(a) Draw a tree diagram to represent this information.
(b) A warehouse inspector selects one hard drive at random. What is the probability that it is a defective hard drive and from Plant II?
(c) What is the probability that a randomly selected hard drive is defective?
(d) Suppose a hard drive is defective. What is the probability that it came from Plant II?

Answer

Check back soon!

Discussion

You must be signed in to discuss.

Video Transcript

So for this problem, we have three plants manufacturing hard drives and shipping them to a warehouse for distribution. Plant one produces 54% of inventory I for inventory and has 4% defect rate land to not pant plant to has or produces 35% of the inventory and has an 8% defect rate. Plant three produces the remainder, so that's 100 percent minus 54%. Minus 35% is equal to 11% of inventory, and it has a 12% the fact rate. So first thing that we want is we want to draw a tree diagram that represents this information. So we start off going to make sure I give myself a lot of space here. Okay, so we start off, then we could go plant one. I'm going to call that event a one. And that has a probability of 0.54 then. Yeah, once we have determined that we were from plant one, then we have the possibility that it is defective. All say that defective is event Be so we have probability of defective given it came from plant one. So that is defective. That is going to be 4% or 0.4 going to zoom in a little bit and I'm going to make my pan a little bit smaller here. So that is 0.4. Then we have not defective, going to say not defective his event be prime. And I'm just going to say Good gets us there. Alternatively, could come from Plant one Key A to which is equal to E. A to Waas. Where is it? 35% or 0.35 that is plant to. Similarly, we have probability of be given a to which is equal to Oh, I just realized I made a mistake. That was 0.4 there. That should be 0.4 and that probability of not defective up there should be 0.96 We have probability of effective given it came from Plant to is 0.8 which means then that's the probability that it's not defective, given it came from plant to equal to 0.92 then we have a probability came from Plant three is equal to 0.11 and the probability that that is defective probability of be given a three is equal to 0.12 and the probability it is not defective, given a came from Plant three is equal. Thio zero point eight eight. So that catalogs all the information that we were given for the next problem. We want to find the probability that a warehouse inspector select one hard drive at random and finds that it is both defective and from plant to So we can figure that out well, in in terms of interpreting that that is the event or were wanting to find the probability that event a two occurs and event be it occurs. So it was from plant to, and it is defective that we can get just thinking about the logic of going down our, um, tree diagram. It would be a 0.35% chance of going down this branch. So a 0.35% chance that it was from plant to and then given that there's a 0.8% chance that it is defective to arrive at the conclusion that both of those events have happened without it being conditional, we basically just multiply together the two probabilities that 0.35 on 0.8 Actually, that should be an ax to so make a little bit easier to read. And I'll note that, of course, that is the same thing as probability that be occurs, given a two occurs times the probability that a two occurs so 0.35 times 0.8 35 times 0.8 is going to be so it is a 0.28 or a 2.8% chance that it occurs than Part C. We are asked, what is the probability that a randomly selected hard drive is defective? The probability that a randomly select the hard drive is defective, that is, overall, that is the probability of event. Be happy here. We can use the law of total probability because the A events a one, a two and a three are mutually exclusive and exhaustive. So it's either from plant one plant to or plant three. That's the exhaustive and mutually exclusive. It can't be from both Plant one and from plant to. So the law of total Probability here tells us that the probability it's defective, equal to the probability that it is effective, given that it came from 81 times the probability that it came from a one plus probability of be given a to times probability of a to plus the probability of be given a three, given that it is from Plant three. So the probability of we can look back at what we have up here, so we had be given a one is 0.4 Yeah, your 0.4 times ability of a one is 0.54 plus probability of be given a to 0.8 times probability of a to was 0.35 plus probability of be given a three 0.12 time's up. That should have a P out front 0.12 times probability that it came from Plant three, which is 0.11 putting all of that together, throwing it into a calculator. Get that? The total probability of something being effective is 0.6 to 8 or 6.28%. Andi, for the last part, we suppose that a hard drive is defective. We want to determine what the probability is that it came from plant to So here we want the probability of a to given event be which is obviously going in a bit of a different direction. We know probability of be given a two, but not a two given b. So for this problem, we can apply Bayes theorem. So for our specific case, we could have that the probability of a two given B is equal to the probability of be given a to times the probability of a to divided by the sum from K or sorry from I equals one up to K, where K is the number of different possible events. So we have cables three. If the sum from I equals 123 the probability of be given a sub i times the probability of a I so that would give us the be given A to that was p of A to divided by be of be given a one on P of a one plus p of be given a to Times eight or probability of a to plus p of be given a three times probability of a three. I'm going to pause and throwing the numbers here so we would have on top 0.8 times 0.35 The bottom. We have 0.4 times, 0.54 plus 0.8 times, 0.35 plus 0.11 times 0.12 So the result of plugging all of that in I calculated it off screen. Result of calculus of playing all that in going to be 0.44 um six. Or you can approximate that to a 0.45 That is the probability that a defect if we have a defective product, then it came from plant to