WEBVTT
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re encounter a challenging word problem. Best strategy is
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first to establish exactly what they've told us and second
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, to figure out exactly what they're asking us to
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find out. So in this case, we've got
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a manufacturer whose producing widgets and bundling them in packets
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of 10. We know that half the time no
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widgets are broken. You call that stage a A
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00 for zero working widgets. We know that's half
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the time we'll call that 0.5. I will be
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the top branch of a tree. They recommended that
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we make a lovely trees. We're gonna go ahead
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and do that. We know that 30% of the
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time one widget is broken. We call that state
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a one that equals 0.3. And we know finally
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that 20% of the time to ridge it's out of
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10 broken, so that call that state a to
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20% of the time. We know also that the
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quality control inspector is going to reach into a random
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bundle wickets and pull out two and then test them
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. When this happens, we know that there are
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three possible states. One state is that no widgets
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are broken one state is that one? Would you
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just broken in one state is that two wickets are
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broken. We're gonna call these states be zero for
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zero broken, the one for one broken B two
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for two broken. And of course, in this
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situation, we're looking at those probabilities given a state
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of A these earlier probability is of different sorts of
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Brokenness part of the P there. So we can
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do these same branches for all the different states of
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a. But we know that this is gonna take
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a little bit of math because, of course,
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we're pulling out to Ridge. It's on you pull
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up to we've got a conditional math problem. Conditional
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probability problem there, as it is. So let's
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make sure that we need to know those things before
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we actually take the time to calculate miscalculations Air a
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little bit complicated. What a reading asked to find
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we know that's ah, once once their co quality
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control inspector has pulled up to which it's untested them
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. That inspector then wants us to tell her what
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the chances are that she drew those widgets from been
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a one a zero or a 21 of the cabins
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, which one is most likely this is drawn from
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given that none of them are broken, were given
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that one of them is broken so we can express
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naps as, ah, probability of having selected from
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been a zero, given none of the widgets are
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broken. So this is the form of the solution
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were pursuing. What are the chances of a given
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be? And they're going to be, of course
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, six versions of this because when nothing is broken
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, we might have pulled for any of the first
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three mins a zero a one or a two.
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And of course we could do the same, but
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a math for whenever we pull out a single broken
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widgets, that's a B now, perhaps kindly.
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They're not asking us to pursue the Final three possibilities
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, which are what if I pulled out two widgets
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and both of them were broken. Now this is
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actually is not a particular kindness. They're not sparing
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us any difficult math. In fact, they're denying
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us the opportunity to the easiest part of the problem
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. And if you want to take just a moment
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to think about it, you don't even have to
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do any arithmetic or anything like that, you can
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just logically reason out. Why would they not bother
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asking? Which kind of been a zero a one
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or two? Which kind of been where they picking
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from when both widgets were broken? It's a pretty
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solid logic there. So when we're when we're given
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a bit of information that allows us to derive fairly
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, simply be given a and then told, what
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we're looking for is a given be where we're transposing
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those two states. What that tells us is that
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we need to use based here. Okay, it's
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a base there, um, written more generically probability
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of a I m given be And you could express
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this as the products of a now the probability product
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of the probability of B n. Given I am
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times the probability of A M invited by probability would
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be then. Now, some of this is,
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ah, quite refreshing because, um, the problem
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is a for any state of a we've already calculated
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we know the green part. Probably it'd be given
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a Oops. Well, that's the second generation of
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our treat. We know that we can figure that
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out. We haven't figured out yet, but we
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know it's within their grasp. Probability of B However
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, this is Ah, this is missing in annex
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in an overt way from the tree. Fortunately,
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bases of a solution to that too ability of be
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eyes equal to the sun of the probability of be
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in given a m. Sorry, given a I
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times nobody of a i for all possible states of
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a which in this case is a zero. A
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one and a two I'm here to, though already
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got the green part already got the blue part So
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this is just going to be the sum of several
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different things that we can already calculate. So we
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know that the, uh gnarly bit of this is
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going to be thinking, figuring out what goes in
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blue part of our tree, the second generation of
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our tree Once we've got fat, it's just a
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matter of plugging that in the base their, um
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their doing so carefully so that we get the right
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answer. Okay, so let's begin the process of
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figuring out what the second generation of our free holds
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for us. I'm now. At no point did
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they ask us to do anything with B two.
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I'm gonna leave all the beaches on here anyway,
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for reasons that I will reveal as we go.
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Now, let's think about this. A zero case
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. What are the chances given that I'm pulling out
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of have been a zero that none of the witches
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that pullout are broken? Well, there are no
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broken widgets in the been a zero, so this
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is gonna happen all the time. I can't pull
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broken ones out. So given that I'm pulling from
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a zero, I will always pull out on broken
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widgets. So probably a B zero, given a
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zero is one. The chances of pulling out of
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the broken ones. Either zero or either one or
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two broken ones. The chances of that are always
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zero. Of course. All right, what about
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the probability pulling out a single? I'm sorry.
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Pulling zero broken widgets when I've already got a broken
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wizard in the bin. It's gonna be a little
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more complicated, right? Um, and we'll do
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the same thing here for B one and B two
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. I'm going toe spare you the ah, having
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me rewrite every single bit of the equation. You
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get the idea for how that looks there. Um
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, the chances of pulling out zero Broken Ridge.
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It's when there's a single one in the box here
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. We're gonna have to use that multiplication. So
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it's gonna be a conditional problem. Let's think of
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these one at a time. What are the chances
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that I pull out a single unbroken, which it
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? That's pretty easy. Math, right? There
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are nine unbroken, which is in there. They're
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10 total widgets. So the chances of pulling out
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a single unbroken one is 9/10. All right,
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given that I've already pulled out an unbroken one,
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what are the chances of pulling out another unbroken one
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? Well, there are nine which is left and
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eight of those room broken, and I can use
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the oppression rule. Cancel this nines. I get
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0.8. Now, what we'll see in a second
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when we get to been a two is that the
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fractions here gets imposing and gnarly. I'm going to
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give most of the answers, hearing decibels to simplify
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that a little bit for us. I'm okay now
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, So we've got B zero, given a 1.8
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, be one given a one is a little bit
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challenging to complement Contemplate. Let's let's make this easier
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on ourselves. We know that the only three possibilities
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are zero broken, one broken or two broken.
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So the some of these three answers needs to be
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one thes. Three answers need to add the one
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because they exhaust into the partition all the possible outcomes
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. So instead of calculating, be one, let's
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calculate probability of B two. Given a one is
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this is easier. What are the chances of pulling
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two broken widgets out of a box with zero with
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only one broken region it there is only one broken
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which in the box. So the chance of pulling
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out two broken ones is zero zero plus 00.8 point
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eight one minus point aches this point to so the
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chances of pulling out a single broken widget point to
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All right, let's think about that. Our final
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situation here, which is the been with two broken
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widgets. Now we need to figure out probability of
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B zero, given a a to and similarly probability
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of B one, Give it a two and,
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um, and add in B two down here for
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similar reasons. What we talked about with a one
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all right. So what are the chances of pulling
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out to unbroken widgets when two of the Richardson had
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been a broken same principle as probably be zero,
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given a one different math. So instead of nine
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and broken widgets, they're eight. So the first
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widgets when I pull out the first one, there's
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an eight out of 10 chance that it's unbroken when
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I reach in for thes second widgets, they're seven
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unbroken ones remaining, and there are nine widgets total
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remaining. I could do a little bit of math
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here. This is point six to heating. If
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you like the fractions, it's 28 45th. In
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fact, if you really like competent pork, so
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you can use commented works to find this problem that
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Number 45 which you'll find in the denominators of all
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of these answers for a two that exists for the
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same because of how competent works for this it's ended
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order. 10. A magnitude 10 columns of that
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second coefficient is gonna be 45 all right now,
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probably probably filling out a single broken one is difficult
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to complicate contemplate. So let's pull out to broken
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ones. Chance. Apulia to Brooke ones while they're
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too broken ones in there out of 10. So
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when I reach in to grab the first broken when
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I've got a 2/10 chance of grabbing that first broken
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one when I reach in for the second broken one
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, there's only one broken one left and there are
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nine widgets left. So this is too out of
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00:12:13.830 --> 00:12:20.850 A:middle L:90%
90 or one out of 45 or 0.0 to repeating
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00:12:22.309 --> 00:12:24.940 A:middle L:90%
now. Uh, just as with a one with
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00:12:24.950 --> 00:12:28.549 A:middle L:90%
a two, these three answers here should also zero
209
00:12:30.440 --> 00:12:33.750 A:middle L:90%
. So instead of taking time to calculate the state
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00:12:33.750 --> 00:12:37.169 A:middle L:90%
of B one, I could just say I used
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00:12:37.169 --> 00:12:39.850 A:middle L:90%
a little bit of math. It's 0.35 repeating.
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00:12:41.389 --> 00:12:43.620 A:middle L:90%
So now we've got all of our blue numbers all
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00:12:43.620 --> 00:12:46.809 A:middle L:90%
over second generation of the tree. All of our
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00:12:46.809 --> 00:12:50.039 A:middle L:90%
be given aids good. So next thing we need
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00:12:50.039 --> 00:12:56.120 A:middle L:90%
to dio is calculates Probability of B. No,
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00:12:56.120 --> 00:13:03.789 A:middle L:90%
what we do here I'm for a probability of B
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00:13:03.789 --> 00:13:11.940 A:middle L:90%
zero is we take one times 10.5 plus pointed times
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00:13:11.940 --> 00:13:16.940 A:middle L:90%
point free plus 0.62 times 0.2. So we take
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00:13:16.940 --> 00:13:20.799 A:middle L:90%
basically the top of the second generation tree and multiplied
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00:13:20.799 --> 00:13:22.840 A:middle L:90%
by its corresponding branch of the first generation of the
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00:13:22.840 --> 00:13:26.740 A:middle L:90%
tree and add those three things together. What you
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00:13:26.740 --> 00:13:46.049 A:middle L:90%
should get here is 0.5 plus 0.24 plus zero 0.12
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00:13:46.440 --> 00:13:48.429 A:middle L:90%
Now these should not add up to one, because
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these are not the problem. The partition in totality
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00:13:52.029 --> 00:13:54.590 A:middle L:90%
instead thes the chances, regardless of which been pulling
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00:13:54.590 --> 00:13:58.230 A:middle L:90%
from the chances that I called to unbroken widgets.
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These some toe approximately 0.86 are using fractions. I
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00:14:03.379 --> 00:14:05.720 A:middle L:90%
could be more exact, but for our purposes,
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00:14:07.009 --> 00:14:09.309 A:middle L:90%
these decimals will do just fine. That's my P
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00:14:09.309 --> 00:14:11.370 A:middle L:90%
zero. All right, let's do the same thing
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00:14:11.370 --> 00:14:11.759 A:middle L:90%
, I might say. That's my P F B
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00:14:11.759 --> 00:14:13.730 A:middle L:90%
zero. We do the same thing now for PFP
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one. And here we grabbed from the middle branch
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00:14:22.429 --> 00:14:26.539 A:middle L:90%
of the second generation and we multiply it by the
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00:14:26.549 --> 00:14:28.450 A:middle L:90%
branch from which it comes. So zero times 00.5
236
00:14:28.840 --> 00:14:35.049 A:middle L:90%
0.2 times 0.3 0.35 times 0.2 I'm so here we
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00:14:35.049 --> 00:14:48.149 A:middle L:90%
get, of course, zero plus 0.6 US zero
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00:14:48.740 --> 00:14:54.769 A:middle L:90%
point 07 We had those up and we get approximately
239
00:14:54.340 --> 00:15:01.049 A:middle L:90%
0.13 I say approximately again because these are approximated,
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00:15:01.539 --> 00:15:03.639 A:middle L:90%
um, from fractions. Good. Now, these
241
00:15:03.639 --> 00:15:07.860 A:middle L:90%
two also should not add up to one because there's
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00:15:07.860 --> 00:15:09.240 A:middle L:90%
a whole other state B two that needs to be
243
00:15:09.429 --> 00:15:11.190 A:middle L:90%
taken due account, and you can calculate the chances
244
00:15:11.190 --> 00:15:16.379 A:middle L:90%
would be too by using, uh, these to
245
00:15:16.389 --> 00:15:20.269 A:middle L:90%
some together. All right, so we got ah
246
00:15:20.279 --> 00:15:28.039 A:middle L:90%
, the denominators for each of these solutions we need
247
00:15:28.039 --> 00:15:33.549 A:middle L:90%
now only calculates the new readers because these first three
248
00:15:35.139 --> 00:15:37.049 A:middle L:90%
, the chances of a given b zero, we're
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00:15:37.049 --> 00:15:52.059 A:middle L:90%
all gonna have 0.86 in the denominator. And these
250
00:15:52.059 --> 00:15:54.450 A:middle L:90%
bottom three chances of a given be one will have
251
00:15:54.450 --> 00:16:08.019 A:middle L:90%
10.1 free in the denominator. Now, we just
252
00:16:08.019 --> 00:16:11.909 A:middle L:90%
needed plug in the correct numerator and the Newman eight
253
00:16:11.919 --> 00:16:15.250 A:middle L:90%
enumerators of this are, of course, the product
254
00:16:15.529 --> 00:16:18.559 A:middle L:90%
of the second branch of the Tree Times. The
255
00:16:18.559 --> 00:16:21.000 A:middle L:90%
first branch of the tree which also, incidentally,
256
00:16:21.009 --> 00:16:22.509 A:middle L:90%
happened to be the numbers that we used to find
257
00:16:22.750 --> 00:16:27.000 A:middle L:90%
chances of B. So PNB zero is the some
258
00:16:27.009 --> 00:16:32.649 A:middle L:90%
of these three numerator get the dominator. So we
259
00:16:32.649 --> 00:16:37.019 A:middle L:90%
know that the top here is 0.5. The next
260
00:16:37.019 --> 00:16:41.789 A:middle L:90%
one is pointed to four and the next one is
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00:16:41.789 --> 00:16:48.370 A:middle L:90%
10.12 We know that the numerator here is going to
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00:16:48.370 --> 00:16:51.399 A:middle L:90%
be zero that the other ones will be point of
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00:16:51.399 --> 00:17:02.549 A:middle L:90%
six and 0.7 eso light. Possibly that's as far
264
00:17:02.549 --> 00:17:03.460 A:middle L:90%
as you would need to go. If you want
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00:17:03.460 --> 00:17:06.380 A:middle L:90%
to reduce those fractions, you're, of course,
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00:17:06.390 --> 00:17:14.089 A:middle L:90%
invited you. So I'm sure points five 78 which
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00:17:14.089 --> 00:17:26.269 A:middle L:90%
to 7/8 on and went one for four. These
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00:17:26.269 --> 00:17:27.690 A:middle L:90%
three should add upto one, and if you take
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00:17:27.690 --> 00:17:29.779 A:middle L:90%
the time to do it, you'll find that they
270
00:17:29.779 --> 00:17:32.430 A:middle L:90%
, in fact, do. They should have add
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00:17:32.430 --> 00:17:34.259 A:middle L:90%
up to one. Because if I've got zero broken
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00:17:34.259 --> 00:17:37.440 A:middle L:90%
ones, there are only three possible then six should
273
00:17:37.440 --> 00:17:40.589 A:middle L:90%
take it from These are partitions of the possibilities,
274
00:17:40.640 --> 00:17:41.470 A:middle L:90%
so they must sum to one when you look at
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00:17:41.470 --> 00:17:44.740 A:middle L:90%
their probabilities. I mean, what we find,
276
00:17:44.740 --> 00:17:45.599 A:middle L:90%
of course, is that when I have no broken
277
00:17:45.599 --> 00:17:48.730 A:middle L:90%
widgets, it's most likely the case that I have
278
00:17:48.730 --> 00:17:52.329 A:middle L:90%
taken those widgets from one of the bins that doesn't
279
00:17:52.329 --> 00:17:53.619 A:middle L:90%
have any broken widgets in it, both because there
280
00:17:53.619 --> 00:17:57.099 A:middle L:90%
are more of those bins. And because there are
281
00:17:57.099 --> 00:18:00.970 A:middle L:90%
fewer broken widgets in those bids, so are probabilities
282
00:18:00.970 --> 00:18:03.430 A:middle L:90%
have borne out our intuition here. If we had
283
00:18:03.430 --> 00:18:06.990 A:middle L:90%
positive beginning to say what do I think the answer
284
00:18:06.990 --> 00:18:08.380 A:middle L:90%
would be We would expect it to look like this
285
00:18:10.240 --> 00:18:11.289 A:middle L:90%
in this next set. So when you've got a
286
00:18:11.289 --> 00:18:15.369 A:middle L:90%
single broken one, it's actually difficult to have an
287
00:18:15.369 --> 00:18:17.769 A:middle L:90%
intuition about it. We know, of course,
288
00:18:17.769 --> 00:18:18.829 A:middle L:90%
if there's a single broken one, we're not taking
289
00:18:18.829 --> 00:18:22.349 A:middle L:90%
anything from a zero. But will we get a
290
00:18:22.349 --> 00:18:26.569 A:middle L:90%
bigger number from a one or from a two?
291
00:18:26.579 --> 00:18:27.990 A:middle L:90%
Is it more likely that we put a single broken
292
00:18:27.990 --> 00:18:30.329 A:middle L:90%
? Been a single broken widget? I'm sorry,
293
00:18:30.579 --> 00:18:33.829 A:middle L:90%
a single broken widget from the bins that are more
294
00:18:33.829 --> 00:18:37.180 A:middle L:90%
numerous but have fewer broken widgets in them, their
295
00:18:37.180 --> 00:18:38.450 A:middle L:90%
arm or a one bins. But there are fewer
296
00:18:38.450 --> 00:18:41.369 A:middle L:90%
broken widgets in They won. Vince, you can
297
00:18:41.369 --> 00:18:44.559 A:middle L:90%
see it's pretty close. I mean, if you
298
00:18:44.569 --> 00:18:48.759 A:middle L:90%
grind out the math, you'll get ah point for
299
00:18:48.829 --> 00:19:00.819 A:middle L:90%
58 and twins five. Or to just slightly more
300
00:19:00.819 --> 00:19:03.160 A:middle L:90%
likely that you got that broken widget from one of
301
00:19:03.160 --> 00:19:06.640 A:middle L:90%
the bins with mawr broken widgets in it. And
302
00:19:06.640 --> 00:19:10.710 A:middle L:90%
these three numbers 0.458 and 0.54 to those should also
303
00:19:10.710 --> 00:19:12.529 A:middle L:90%
add up one, because they exhaust again all the
304
00:19:12.529 --> 00:19:18.230 A:middle L:90%
possibilities when you've got a single broken Richards. Good
305
00:19:18.250 --> 00:19:21.039 A:middle L:90%
. I hope you find this a useful demonstration,
306
00:19:21.130 --> 00:19:22.049 A:middle L:90%
and I wish you best of luck in your studies