WEBVTT
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all right for this problem. We're calling back to
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a previous problem where it gave us the probability of
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certain types of defects. And now we have some
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conditional probability to deal with. So the 1st 1
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we want to find the probability of a type to
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error, given that there's a type one er so
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h you given a one and for that we're gonna
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use our law of conditional probability. So that's gonna
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be the probability of a two intersection of a one
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over the probability of a one. Now, this
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problem assumes that you've done the previous problem. Problem
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number 28. But if you haven't or you just
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don't want to refer back, I'm just gonna walk
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through driving. What probability of h two intersect A
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one is. You just used your addition law,
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like so. And you just plug in what you
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know, and we're going to do that and solve
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for the probability of a one intersect a two.
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So the probability of the union is 0.13 Probability of
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a one is open. One to probability of a
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two is your 20.7 minus probability of a one intersect
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a two 0.13 equals 0.19 minus the probability of a
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one intersect eight to or the probability of a one
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intersect a two equals 0.6 That numerator come 0.6 We
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divide that by the probability of a one which is
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0.12 and this becomes 1/2 or 0.5. All right
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, same thing over here. Basically, we want
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to find the probability of a one intersect H two
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intersect a three given a one. So we're going
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to use that law again. So we're gonna have
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a one intersect A to intersect a three all intersected
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with a one over the probability of anyone. Intersection
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is community and associative, which means we can just
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imagine these parentheses aren't here for the time being,
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and we see that a one is duplicated, so
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we only need to express a once one wearing an
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intersection. So it just simplifies down to that and
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we know both of these. The top is gonna
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be zero points or one. The denominator is gonna
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be 0.12 This become 0.833 round 2 to 4 decimal
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places. All right, we want to find the
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probability of exactly one defect, given that there is
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a defect. So we're gonna use a Venn diagram
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, So I'm getting with the red equal a one
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green equal. Let's redraw that. Let's move that
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circle so I can write in. And actually,
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let's let the green equal the green, not the
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black, for some reason, equal a two and
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the blue Be a three. I'm gonna make this
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. They are There we go. Right. So
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we know the center part. Based on this piece
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of information, here is 0.1 If we go back
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here, we found out that the intersection of a
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one a 20.6 total and that includes this. So
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the intersection here without a three is 0.5 If we
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math up that weaken math out the intersection between H
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two and a three really quickly using the same method
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or yeah, we could do a one day three
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, but let's just do a two and a three
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just because it doesn't really matter any three minus the
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intersection of Sorry. Wait, That should be a
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union. This should be an intersection. There we
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go. So the union is 0.10 Ah, Probability
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of a two is 0.7 Probability of a 30.5 Minus
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the probability that we're looking for, uh, 0.10
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equals 0.7 Sorry. Wait. No, Hold on
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. We have those together. You got 0.12 So
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therefore, the intersection between a two and a three
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equals 0.2 We already know that we have a 0.1
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here, so we get zero points or one.
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We know eight's your total should be 0.7 He's already
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add up to 0.7 So this is 0.0 Uh,
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eventually, we did have to calculate the intersection shin
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between a one and a three. So we'll do
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that real quick. Same setup. And again,
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if you did the problem that this referred to problem
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number 28 then you should have this already. I'm
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just doing again for the sake of refreshing your memory
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. Ah, let's see. This is 0.14 because
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0.120 point 05 minutes. All this skipping some of
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the algebra, You know that This must be This
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is seven. This is 70.17 this 0.14. So
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this is 0.3 We already have a 0.1 in here
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is that this becomes 0.2 We know the A one
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must add up to 12. This adds up.
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The 120.8 are Sorry. 0.12 This adds up to
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a 0.8 Right now you are missing a 0.4 And
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this almost add up to five. We currently have
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four. Or I should say 0.50 points or four
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. So we're currently missing a 0.1 That's where Venn
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diagram looks like. I'm gonna delete all this just
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to clean up our work. Oops. We need
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that information. You just put that back in the
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corner here and I'm gonna take a Venn diagram.
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No, no, no. I don't need to
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highlight everything. Hold on. There we go.
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And I'm just gonna move aside for a reference.
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So we need a probability of exactly one defect to
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given all this. So using our lot of conditional
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probability that the probability of let's just call this event
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e just for shorthand. He given a one union
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, a two union a three. This adds up
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by the way to four plus five is 99 plus
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one is 10 10. Plus. She was 12
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. Well close. One is 13 13 plus one
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because 14. So the probability here because 0.14 So
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the problem. So that's useful because that's our denominator
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. That's being intersection, not a line. There
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we go. So we know the bottom 0.14 We
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need the probability there's exactly one defect given all this
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. Well, we just add up these probabilities here
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0.5 And if we divide that out, we get
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zero groups point three 571 All right. For the
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last part, we need to find the probability of
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not a three given a one intersection a two.
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Well, immediately, I can say that this is
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one minus the probability of a three a one over
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a two, and then we'll use our law.
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Conditional probability. It's gonna be a probability of a
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three. No section anyone. Intersection A two divided
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by the probability a one intersect a two. So
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that's one minus the top part. It's just 0.1
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in the bottom. Part of we referred to heart
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eh was 0.6 So this is one minus. This
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reduces down to 1/6. We're just 56 which reduces
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down. Our ingestible form is 8.333 rounding to four
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decimal places.