WEBVTT
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okay for this problem. We have two boxes.
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Box one is filled with six red balls and four
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green balls. Box two is filled with seven red
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balls and three green balls. We are going to
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pick one from Box one, pick a single ball
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and place it into box, too. Then we're
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going to pick a single ball from box to and
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place that into Box one. For a or for
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the first part, we want to find what the
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probability we want to find the probability that a red
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ball is selected from the first box and a red
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ball is selected from the second box. So I'm
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going toe label. The events are one, uh
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, oops are one is we get a red ball
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from one are to same thing for two and G
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one and G two similarly for green. So what
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part a is ultimately asking for is our one intersected
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with our two? What's the probability that both of
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those events happened? The catch is that are to
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the event that we take a red ball from box
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to is dependent on the result of the first step
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dependent on whether we get are one or G one
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, so we don't actually know the intersection of these
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two immediately. But what we can do is we
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can logic out what the probability of getting are.
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Two is given that we get event are one.
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So if we assume that we've picked a red ball
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from the first container, then that means that when
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we put it into the second container, there's going
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to be a total of eight red balls and 11
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balls total in the container. So the probability that
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we get a red ball, given that we've gotten
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a red or we get a red ball from two
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, given that we've gotten a red ball from one
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is going to be eight in 11 which is equal
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to eight and 11 is about 0.73 So we know
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events are one happens then there is a 73% chance
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that event are, too. Happens then can use
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the formula that the probability of the intersection of events
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R one and R two is equal to the probability
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of our two. Given are one times the probability
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of event are one. So that means that probability
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of R one and R two equal to 0.73 times
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already is an ex their times 0.6. So that
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comes out to 0.438 for part b of the problem
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. We want to find the probability that at the
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conclusion of the selection process, the same number of
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red and green balls are in box one, as
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there were at the beginning and I had a little
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bit too much on screen there. So I'll go
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through the logic here through the selection process. The
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different possibilities are we take read out of box one
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and read out of box too. We take read
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out of box one green out of box, too
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. We take green out of box one green out
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of box, too, and we take or rather
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we take green out of box one and red box
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too. In the first case we take. So
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we take a red moving to box to. Then
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we take a red from box to move it back
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to the box 13 end with six red four green
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Santas. We started here. We remove a read
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out of green, so we will have five red
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five green. Here we remove a green at a
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green. We'll have six red for green. Here
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we remove a green at a red. We'll have
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seven red three green. All of these referring to
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what would be in the first box. So we
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end up with the situation that we want an unchanged
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number if we get R one R two or if
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we get G one G two. So it's going
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to be the probability this probability there is going to
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be the probability of R one and R two happening
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, plus the probability of G one and G to
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happening. I'm going to just continue this down below
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. So, uh, we already figured out the
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probability of R one and R two happening. That
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was 0.438 Then the probability of G one and G
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to happening. We could use the same process as
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we used before. That's going to be the same
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thing as probability of G two. Given G one
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divided by or not divided by, Rather, it's
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going to be multiplied by multiplied by the probability of
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G one, then that is going to give us
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0.438 plus the probability of G two given g one
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. So we need to think about that for a
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second, knowing that G one happened. So we
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transferred a green from box 12 box, too.
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That means that we'll end up having total 11 balls
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and we'd have four green. So the probability of
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G two given g one is going to be four
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in 11 times the probability of G one happening,
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which was for in six or 0.4, not four
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and six. Rather sorry, four in 10 which
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is zero. So you get 0.438 plus 4/11 is
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0.36 repeating, and that is multiplied by 0.4.
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We get 0.438 Plus, here we get 0.145 doing
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a little bit of rounding. So in total we
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get 0.583 is the probability that after the selection process
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, we end up with same number of red and
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green balls in Box one. As we started