WEBVTT
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alright for this problem, We are considering customers buying
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a full set of tires from a particular tire store
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. Vent A is where the tires were made in
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the USA Event B is where the purchaser had the
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tires balanced immediately and events see where the purchaser requests
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a front end alignment. And then there are a
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prime b prime and see prime, which are the
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the knots of each of those events. We also
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have these given probabilities and conditional probabilities. I'm not
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gonna bother reading all those out. So for part
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A, we want to construct a tree diagram for
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the 1st, 2nd and 3rd generation branches and plays
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an event label and appropriate probability next to each branch
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. So we know we're going to have our starting
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point at our first decision. We're gonna have to
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our second decision. We are going to have four
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. They're not for sorry. We are going to
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have eight. One second. 123 Okay, I'm
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messing that up. Yeah, for our second decision
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, we should have total four. And then for
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our third decision, we should have eight. So
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I'm gonna zoom in quite a bit here because it's
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going to get a little nasty. So we start
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off and then we have event A with probability of
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a is 0.75 and we have a prime. The
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probability of a prime is going to be one minus
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10.75 or 0.2 to 5, then here for this
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branch, we have B where the probability of be
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given a is equal to 0.9 and we have be
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prime where the probability of the prime given a is
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equal to 0.1 Up here we have C probability of
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see, given B and A is equal Thio 0.8
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. And this event here that event is um yeah
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, that event is Yeah. No, that's the
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best way of labeling that actually. Sorry, but
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my hesitation there we have c prime down here.
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Probability of C prime given B and a equal to
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0.2 than this branch have see ability of see given
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be prime and a is equal to that 0.6 probability
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of C prime. Just this branch down there ability
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of C prime given be prime and a is equal
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thio 0.4 Then we have offer a prime branch here
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B and B prime again. Where we have probability
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of be given a prime is equal 2.8 probability of
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be prime given a prime is equal Thio point to
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we have C see Prime and C and C prime
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Where you gonna zoom? In further probability of C
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given a prime and be is equal 2.7 the probability
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of yes ability of see prime given a prime nb
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equal 2.3 then lastly or the last pair you have
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a probability of C given a prime and be prime
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which is that is going to be 0.3. What
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and we have done here the probability of see prime
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given a prime and be prime which is equal 2.7
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And then these events at the end correspond to so
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that is see given a or rather not given that
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is A and B and C A and be and
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see prime A and be prime and see a and
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B prime and see prime um, a prime and
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B and C a prime and be prime and or
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sorry a prime and be and see prime. Then
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we have a prime and be prime and see.
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And we have a prime and be prime and see
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prime. We can calculate the probabilities of each of
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those by multiplying down our branches. So I'm going
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to do that off screen here. Actually, my
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mistake calculating all of those eyes going to be a
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bit unnecessary. So we now have the proper branches
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and event labels for everything. That's all we need
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for part a part B. Yes. So this
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is the total tree there, Part B. We
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want to find the probability of A and B and
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C, which we confined just by following our tree
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. That is 0.75 times 0.9 times 0.8, which
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comes out to 0.54 part C. We want to
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find the probability of B and C happening, so
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that means that it's going to be mhm probability of
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B and C and a plus the probability of B
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and C and not a which we already have.
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BNC and a keep in mind that for the ANZ
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, the order doesn't matter. Forgiven for the conditioning
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it does, but for and you're ordering doesn't matter
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. It was 0.54 plus 0.25 Excuse me, M
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0.8 time 0.7. That gives us 0.68 for D
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. We want to find the probability of see occurring
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. So that is going to be the probability of
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B and C plus the probability of not be NC
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. Uh huh. You know that the probability of
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B and C 0.68 the probability of not being C
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going to be so that this event can be broken
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down into actual do it off to the side there
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. Keep in mind that that's going to be essentially
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the probability of a and not B and C plus
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the probability of not a and not be and not
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be and see not be so probability of a and
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not B N c his 0.75 times 0.1 time 2.6
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and the probability of not a and not B and
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C is 0.25 times 0.2 times times 0.3. So
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adding those all together we get 0.74 is the result
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. Lastly for E, we want to find the
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probability of a given B and C you can find
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this using Bayes Theorem, which tells us that that
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probability is going to be the same thing as the
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probability of be all right. Same thing is the
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probability of A and B and C divided by the
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probability of B and C, so we already found
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each of those pieces. Previously, ability of A
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and B and C is 0.54 Probability of BNC is
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0.68 so we end up with the probability of a
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given B and C 0.79