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