To solve this complicated word problem that we're faced with. We ought first to translate the given information of the world problem into mathematical language and then paused to translate the question itself into mathematical language. And we can go from there to figure out which mathematical formula or serums will be most useful for solving the problem. So you know, we were faced with here is a manufacturer whose producing widgets and shipping them in bundles of 10 half the time Those bundles of tender, flawless have zero broken widgets we're gonna do note that state is a zero and say to the probability of the state A zero is 50% half the time our bundles are flawless. We also know that 30% of the time each bundle of 10 will have a single broken budgets. So we'll call that a one for one Brooke and Ridge. It's probability of a one point free or 30% and then the remaining 20%. What they find is the two widgets in the bundle are broken. Problem also recommends that we set up a tree at least the first generation tree showing these probabilities that I've done for you right here. I've also expanded that to a second generation because we do have the information given in the question to define these conditional probabilities in the second generation of the tree. These, of course, the probabilities. Given that you've said selected from a particular bundle, one of the probability is you will find a certain number of broken widgets. The answer to these probabilities will depend on the final observation to make from the problem. Before we define the actual question were being asked. And that is that we have a quality control officer with sampling from each from a random bundle of widgets and determining how many are broken from that sample. Now this quality control officers selecting two regions at a time, which makes our lives just a little bit more complicated. Since there are two which is being selected, we have three possible states for B B for broken. We might find that zero of those widgets are broken. One of them is broken or two of them were broken. Second, consider, because of these nine conditional probabilities for any given state of the Ben itself, a quality control officer might find three different three different levels of broken rigidness. Now what are we actually being asked to find? Because our next bit here would be a relatively complicated bit of math. Let's make sure we need to discover that before we move on. But the question is actually asking us for is to suppose that our quality control officer doesn't know which been she's selecting from, and she selects her to widgets. She tests them and finds out that none of them are broken. Based on that information, we need to derive the expectation of which Benji most likely took things from. What are the chances that when she finds zero broken, she took them from a zero? Been in a one been or eight to bid, we can, intuitively saying when none of them were broken. More likely that she's picking from in a zero gun for several reasons. We'll come up with the actual percentage she's also found, uh, it's occasionally one of them is broken. So when she finds herself in a state of B one, we need to derive what the likelihood has shed. Selected for a zero a one or a two in mathematical language that looks like this over here. Given b zero, what are the chances of a zero? You can be zero. What a the chances of a one, etcetera. So we're looking for six different answers now. They haven't asked us to find, um, solutions for B two when she pulls out two objects and they're both broken. This may seem at first like they're cutting you some slack, but in fact, this is just eliminating the most trivial possibility s. So I would suggest that you take just a moment. Don't even do any math. Just think through in your head what would be the answer if b two were the descriptive state given be to what? The problem is, a zero a one a two that's Matthew could probably do in your head or indeed a solution you could find before taking a probability schools. All right, so we're looking for these probabilities, and that calls to mind our basic rule of probability of conditional probability, which is the situation that you can find in your textbook problem with. This, of course, is that we don't have a number for B of N and computing. This, uh, this intersection here actually relatively complicated s so what we need to do next is transformed this. Using the role of multiplication into base there based thorium is gonna be exactly what we're looking for. To get a solution of this, and the bass tells us is that the conditional probability of a given be is the conditional probably would be given a times in our ability of a divided by the sum of all conditional probabilities of be given the different states of a times, all the probabilities of different states of a. Now, the good thing about this is we actually now can, uh, come up with solutions for each of these component parts. This looks like a big, bewildering, unending, some at fraction. But if we think about it bit by bets, we can come up with a much more reasonable solution. And we can just grind it out step by step. So we start with the ah, top of attraction here. He of a we know all of our piece of a business there indicated here ah p of be given a That's what we're going to cinsault for here. So I need to come up with those numbers we don't need to find the ones for B two. We'll talk in a second about why we're going to do that anyway. And in the bottom part of our fraction here is just the son of all of the possible top parts. So if we can find the probabilities of the different sets of be given the different states of a, we already know that it probably is the difference. It's a week unplugs all into the equation on come up with a satisfying solution. So let's find these conditional probabilities and be given the different states. A. We're gonna start here with a zero because a zero is really the easiest of all of them to handle. If you take two widgets out of a box of 10 unbroken widgets, what are the chances that both are unbroken? Course? Common sense will tell us. That's 100% of the time. There aren't any broken bridges in here, so it's impossible to find any broken widgets. They don't exist. S o P B zero of a zero is one remaining. Probably lose for the difference. It would be we're all zero. All right, now, once we have actual broken widgets and Arbenz. This becomes a little more difficult. So we're selecting never a one. What are the chances the zero of my widgets have a broken? When I take two of here, we can use the rule of multiplication. If I take the first widgets, there's one broken widget in there. What are the chances that I don't pick that one? Well, there are nine unbroken, Which is that a pen. So the chances that a night dance now our next state, which is dependent on this first state we found no broken, which is the first time. What are the chances with Second, which it is also not broken? Well, there are only nine, which it's left. Eight of them are unbroken. So there are eight out of nine possibilities for that to happen and then weaken grain. This through reduce our fraction. We find that this is 8/10 or you're just destruction down here to ah, aches, tents or for or 0.8. If you look, either one is fine, of course, depends on what your instructor prefers. How you want express these actions now be one is actually a slightly tricky one to compute. Let's compute B two, given a one, even though they don't ask you this, this very easy thing to compute and can actually make it much simpler to solve for B one, given it one. The tents, of course, of B two. Given a 10 there's only one broken thing in the den, so there's no way I could ever pull up to Bergen Things. And because we know that B zero, B one and B to exhaust all possible outcomes, we know that the probably be one given a one has to add up one. When we added to the problem of the zero give name. What is this? Has a 2.2. You can skip all the fancy mathematics now the last one is the most complicated of them. The probability of a of B 01 or two, given a two, has a bit of a sidekick here. If you're really more comfortable with competent works, you can use combinatorics. Resolve this problem as long as you're careful. So let's solve first for B zero, I have two Broken ridge. It's in the bin, and I'm pulling to I'm too broken, which has been I'm pulling to a random out. What are the chances that the first of those two is unbroken there, eight unbroken ones in there. That's eight out of 10 and again conditional probabilities and multiply role here when I reach in for my second one, there are seven on Brooklyn's left and pulling out, uh, seven out of mine, Sebastian's of being what we look at their will solve for that fraction, a swell. We got 28 45th. So the 45 eyes that is a call back there to that combinatorics principle is all this comment works. The significance of 45 will be meaningful to you. Let's skip PB one again. Because again, that's the more complicated of the three states. What about probably of be too well when I reach in and I grabbed my first with its cancers of that one being broken. There are two broken Richards at a pen, said Stuart, him reach back into my second one. Now there are only nine wickets left, and only one of them is broken. So it's a 1/9 chance and by and that's to get one out of 45 is Ah, that's my That's my probably have grabbing two broken wickets, having one every 45 times on. And I could do a little bit of math here, 16 45th remaining of that middle at a great for beat one. What remains here is just do a whole bunch of math plug in. All of the different numbers that we need for this for the first three they were looking for appear for these first three answers. We have the same denominator for our Bayes theorem equation so we can calculate that real quick and what we do to do that is we take all of the second generation branches associated with B zero and multiply them by their first generation matches, and we add up those totals. This is going to be probability of B zero. So it actually is the denominator of this original equation right here on this is the trick of bays law. Basically, it's a lot of fancy mathematics and hand waving, but really, the finding the denominator is just about finding the overall probability of the state that you're looking for other the conditional state. What are the chances of P zero, which is the condition of our very problem here. So these air, we're gonna take the Beezer of state of the second generation That's going one multiplied by the first generation associated with it. He has a 0.5 second generation for B zero and a one is 0.8. And we're going to multiply that by the first generation X 0.3 and then finally third generation 28 45th. So I didn't want Teoh reduce that fraction because we will actually lose accuracy if we do that. But playing that by a point to if you're more comfortable, you could, of course, convert this all to either Approximate decimals 45 here is gonna make that approximate It's got nine involved there. Or you can convert everything fractions on my all straight. Both solutions. Here you grind to the mathematics. You find that the reduction distraction is 389 45th or a little over 86% of the time, and that that's our overall chances of finding a state of B zero on that makes up. Half of the batches have no broken things, and even the broken gadgets are mostly unbroken. Let's go and find ah p of B one as well, because that's gonna be our denominator. For these second three equations. The second few responses. The topic, of course, RB one state has a zero probability. So we don't need to worry about a zero in this case. Well illustrates today. One probability of B one given a one is to we multiply that by its first generation, which is 0.3. And then we find the same information for the state. A two, which was 16 year 16 45th. I am too. And again you can express. This is a fraction. Where you can express it is a repeating decimal, and what you should come up with is these two answers, which are essentially equivalent 59 450 deaths or a 0.1311 Now again, what this is these are the sums that we're getting at the bottom of days there. We're talking to them affront because we're going to need them for all three of these answers. And if we calculate them up front, we reduce the risk of forgetting what they are forgetting a calculated them and therefore reinventing the wheel over and over again to find out three individual responses. Then we just take this final answer here and we multiple. We divide, um, are given numerator by this answer here. Of course, the numerator we're going to need are the three entries in, uh, our discovery of the probably of easier and be one. So the probably of a zero given b zero is going to be one times 10.5, which is, of course, one half. I did buy this some which we found out to be. 389 450 deaths. Eso Why did thereby which was absolutely beautiful. Ah, really gnarly bit of fraction work to be to be got over there you should find is 225 389th, which is approximately equal 0.578 So a little less than 60% of the time. If you find no broken ridge, it's probably drawing from been a from one of the easier events I would go through into the same mathematics for the next ones. That's what you get is he's really just quite horrendous fractions, which I've taken the liberty here producing to approximate decibels. So a little over a quart of the time. If you have nothing broken, you've picked from a one bin, uh, little more than an eighth of the time you a picture of a tube in, uh so in this follows basically where we would have expected the proportion of a zero is a little bit ballooned because there are more a zero bins and there are fewer broken widgets. They're in on important final thing to adhere to check you work. These three numbers should add up to one. It was a little bit of a fraction error here because we're changing the decimals. As long as you're extremely close, you should feel confident If you're gonna take the time to compute these horrifying fractions into ah fraction answer, you should find an exact ah arrival at one problem. Because, of course, there are only three possibilities for a zero a 12 So these are all these partition the possible and then finally have grinded through the last couple of fractions here for you to find solutions for the problems of a zero Anyone, any to given the one that we've got one broken object? Of course, there should be zero for what are the chances of selecting from a zero when you've got one broken object? Well, there's nothing. There are new broken objects. In interesting here, though, is that even though their arm or a one bins because there are more broken objects in a two, it is actually slightly more likely when you've picked a single broken object that you're selecting from, uh, the A to been okay, s. So this is our final answer right over here on. I hope you enjoy this.