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Numerade Educator

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Problem 56 Hard Difficulty

One box contains six red balls and four green balls, and a second box contains seven red balls and three green balls. A ball is randomly chosen from the first box and placed in the second box. Then a ball is randomly selected from the second box and placed in the first box.
(a) What is the probability that a red ball is selected from the first box and a red ball is selected from the second box?
(b) At the conclusion of the selection process, what is the probability that the numbers of red and green balls in the first box are identical to the numbers at the beginning?

Answer

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Video Transcript

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