Components of a certain type are shipped to a supplier in batches of ten. Suppose that 50$\%$ of all
such batches contain no defective components, 30$\%$ contain one defective component, and 20$\%$ of contain two defective components. Two components from a batch are randomly selected and tested. What are the probabilities associated with $0,1,$ and 2 defective components being in the batch under each of the following conditions?
(a) Neither tested component is defective.
(b) One of the two tested components is defective.
[Hint: Draw a tree diagram with three first-generation branches for the three different types of batches.]
re encounter a challenging word problem. Best strategy is first to establish exactly what they've told us and second, to figure out exactly what they're asking us to find out. So in this case, we've got a manufacturer whose producing widgets and bundling them in packets of 10. We know that half the time no widgets are broken. You call that stage a A 00 for zero working widgets. We know that's half the time we'll call that 0.5. I will be the top branch of a tree. They recommended that we make a lovely trees. We're gonna go ahead and do that. We know that 30% of the time one widget is broken. We call that state a one that equals 0.3. And we know finally that 20% of the time to ridge it's out of 10 broken, so that call that state a to 20% of the time. We know also that the quality control inspector is going to reach into a random bundle wickets and pull out two and then test them. When this happens, we know that there are three possible states. One state is that no widgets are broken one state is that one? Would you just broken in one state is that two wickets are broken. We're gonna call these states be zero for zero broken, the one for one broken B two for two broken. And of course, in this situation, we're looking at those probabilities given a state of A these earlier probability is of different sorts of Brokenness part of the P there. So we can do these same branches for all the different states of a. But we know that this is gonna take a little bit of math because, of course, we're pulling out to Ridge. It's on you pull up to we've got a conditional math problem. Conditional probability problem there, as it is. So let's make sure that we need to know those things before we actually take the time to calculate miscalculations Air a little bit complicated. What a reading asked to find we know that's ah, once once their co quality control inspector has pulled up to which it's untested them. That inspector then wants us to tell her what the chances are that she drew those widgets from been a one a zero or a 21 of the cabins, which one is most likely this is drawn from given that none of them are broken, were given that one of them is broken so we can express naps as, ah, probability of having selected from been a zero, given none of the widgets are broken. So this is the form of the solution were pursuing. What are the chances of a given be? And they're going to be, of course, six versions of this because when nothing is broken, we might have pulled for any of the first three mins a zero a one or a two. And of course we could do the same, but a math for whenever we pull out a single broken widgets, that's a B now, perhaps kindly. They're not asking us to pursue the Final three possibilities, which are what if I pulled out two widgets and both of them were broken. Now this is actually is not a particular kindness. They're not sparing us any difficult math. In fact, they're denying us the opportunity to the easiest part of the problem. And if you want to take just a moment to think about it, you don't even have to do any arithmetic or anything like that, you can just logically reason out. Why would they not bother asking? Which kind of been a zero a one or two? Which kind of been where they picking from when both widgets were broken? It's a pretty solid logic there. So when we're when we're given a bit of information that allows us to derive fairly, simply be given a and then told, what we're looking for is a given be where we're transposing those two states. What that tells us is that we need to use based here. Okay, it's a base there, um, written more generically probability of a I m given be And you could express this as the products of a now the probability product of the probability of B n. Given I am times the probability of A M invited by probability would be then. Now, some of this is, ah, quite refreshing because, um, the problem is a for any state of a we've already calculated we know the green part. Probably it'd be given a Oops. Well, that's the second generation of our treat. We know that we can figure that out. We haven't figured out yet, but we know it's within their grasp. Probability of B However, this is Ah, this is missing in annex in an overt way from the tree. Fortunately, bases of a solution to that too ability of be eyes equal to the sun of the probability of be in given a m. Sorry, given a I times nobody of a i for all possible states of a which in this case is a zero. A one and a two I'm here to, though already got the green part already got the blue part So this is just going to be the sum of several different things that we can already calculate. So we know that the, uh gnarly bit of this is going to be thinking, figuring out what goes in blue part of our tree, the second generation of our tree Once we've got fat, it's just a matter of plugging that in the base their, um their doing so carefully so that we get the right answer. Okay, so let's begin the process of figuring out what the second generation of our free holds for us. I'm now. At no point did they ask us to do anything with B two. I'm gonna leave all the beaches on here anyway, for reasons that I will reveal as we go. Now, let's think about this. A zero case. What are the chances given that I'm pulling out of have been a zero that none of the witches that pullout are broken? Well, there are no broken widgets in the been a zero, so this is gonna happen all the time. I can't pull broken ones out. So given that I'm pulling from a zero, I will always pull out on broken widgets. So probably a B zero, given a zero is one. The chances of pulling out of the broken ones. Either zero or either one or two broken ones. The chances of that are always zero. Of course. All right, what about the probability pulling out a single? I'm sorry. Pulling zero broken widgets when I've already got a broken wizard in the bin. It's gonna be a little more complicated, right? Um, and we'll do the same thing here for B one and B two. I'm going toe spare you the ah, having me rewrite every single bit of the equation. You get the idea for how that looks there. Um, the chances of pulling out zero Broken Ridge. It's when there's a single one in the box here. We're gonna have to use that multiplication. So it's gonna be a conditional problem. Let's think of these one at a time. What are the chances that I pull out a single unbroken, which it? That's pretty easy. Math, right? There are nine unbroken, which is in there. They're 10 total widgets. So the chances of pulling out a single unbroken one is 9/10. All right, given that I've already pulled out an unbroken one, what are the chances of pulling out another unbroken one? Well, there are nine which is left and eight of those room broken, and I can use the oppression rule. Cancel this nines. I get 0.8. Now, what we'll see in a second when we get to been a two is that the fractions here gets imposing and gnarly. I'm going to give most of the answers, hearing decibels to simplify that a little bit for us. I'm okay now, So we've got B zero, given a 1.8, be one given a one is a little bit challenging to complement Contemplate. Let's let's make this easier on ourselves. We know that the only three possibilities are zero broken, one broken or two broken. So the some of these three answers needs to be one thes. Three answers need to add the one because they exhaust into the partition all the possible outcomes. So instead of calculating, be one, let's calculate probability of B two. Given a one is this is easier. What are the chances of pulling two broken widgets out of a box with zero with only one broken region it there is only one broken which in the box. So the chance of pulling out two broken ones is zero zero plus 00.8 point eight one minus point aches this point to so the chances of pulling out a single broken widget point to All right, let's think about that. Our final situation here, which is the been with two broken widgets. Now we need to figure out probability of B zero, given a a to and similarly probability of B one, Give it a two and, um, and add in B two down here for similar reasons. What we talked about with a one all right. So what are the chances of pulling out to unbroken widgets when two of the Richardson had been a broken same principle as probably be zero, given a one different math. So instead of nine and broken widgets, they're eight. So the first widgets when I pull out the first one, there's an eight out of 10 chance that it's unbroken when I reach in for thes second widgets, they're seven unbroken ones remaining, and there are nine widgets total remaining. I could do a little bit of math here. This is point six to heating. If you like the fractions, it's 28 45th. In fact, if you really like competent pork, so you can use commented works to find this problem that Number 45 which you'll find in the denominators of all of these answers for a two that exists for the same because of how competent works for this it's ended order. 10. A magnitude 10 columns of that second coefficient is gonna be 45 all right now, probably probably filling out a single broken one is difficult to complicate contemplate. So let's pull out to broken ones. Chance. Apulia to Brooke ones while they're too broken ones in there out of 10. So when I reach in to grab the first broken when I've got a 2/10 chance of grabbing that first broken one when I reach in for the second broken one, there's only one broken one left and there are nine widgets left. So this is too out of 90 or one out of 45 or 0.0 to repeating now. Uh, just as with a one with a two, these three answers here should also zero. So instead of taking time to calculate the state of B one, I could just say I used a little bit of math. It's 0.35 repeating. So now we've got all of our blue numbers all over second generation of the tree. All of our be given aids good. So next thing we need to dio is calculates Probability of B. No, what we do here I'm for a probability of B zero is we take one times 10.5 plus pointed times point free plus 0.62 times 0.2. So we take basically the top of the second generation tree and multiplied by its corresponding branch of the first generation of the tree and add those three things together. What you should get here is 0.5 plus 0.24 plus zero 0.12 Now these should not add up to one, because these are not the problem. The partition in totality instead thes the chances, regardless of which been pulling from the chances that I called to unbroken widgets. These some toe approximately 0.86 are using fractions. I could be more exact, but for our purposes, these decimals will do just fine. That's my P zero. All right, let's do the same thing, I might say. That's my P F B zero. We do the same thing now for PFP one. And here we grabbed from the middle branch of the second generation and we multiply it by the branch from which it comes. So zero times 00.5 0.2 times 0.3 0.35 times 0.2 I'm so here we get, of course, zero plus 0.6 US zero point 07 We had those up and we get approximately 0.13 I say approximately again because these are approximated, um, from fractions. Good. Now, these two also should not add up to one because there's a whole other state B two that needs to be taken due account, and you can calculate the chances would be too by using, uh, these to some together. All right, so we got ah, the denominators for each of these solutions we need now only calculates the new readers because these first three, the chances of a given b zero, we're all gonna have 0.86 in the denominator. And these bottom three chances of a given be one will have 10.1 free in the denominator. Now, we just needed plug in the correct numerator and the Newman eight enumerators of this are, of course, the product of the second branch of the Tree Times. The first branch of the tree which also, incidentally, happened to be the numbers that we used to find chances of B. So PNB zero is the some of these three numerator get the dominator. So we know that the top here is 0.5. The next one is pointed to four and the next one is 10.12 We know that the numerator here is going to be zero that the other ones will be point of six and 0.7 eso light. Possibly that's as far as you would need to go. If you want to reduce those fractions, you're, of course, invited you. So I'm sure points five 78 which to 7/8 on and went one for four. These three should add upto one, and if you take the time to do it, you'll find that they, in fact, do. They should have add up to one. Because if I've got zero broken ones, there are only three possible then six should take it from These are partitions of the possibilities, so they must sum to one when you look at their probabilities. I mean, what we find, of course, is that when I have no broken widgets, it's most likely the case that I have taken those widgets from one of the bins that doesn't have any broken widgets in it, both because there are more of those bins. And because there are fewer broken widgets in those bids, so are probabilities have borne out our intuition here. If we had positive beginning to say what do I think the answer would be We would expect it to look like this in this next set. So when you've got a single broken one, it's actually difficult to have an intuition about it. We know, of course, if there's a single broken one, we're not taking anything from a zero. But will we get a bigger number from a one or from a two? Is it more likely that we put a single broken? Been a single broken widget? I'm sorry, a single broken widget from the bins that are more numerous but have fewer broken widgets in them, their arm or a one bins. But there are fewer broken widgets in They won. Vince, you can see it's pretty close. I mean, if you grind out the math, you'll get ah point for 58 and twins five. Or to just slightly more likely that you got that broken widget from one of the bins with mawr broken widgets in it. And these three numbers 0.458 and 0.54 to those should also add up one, because they exhaust again all the possibilities when you've got a single broken Richards. Good. I hope you find this a useful demonstration, and I wish you best of luck in your studies