Assume that the box contains 11 balls: 4 red, 5 blue, and 2 yellow. As in the text, you draw one ball, note its color, and if it is yellow replace it. If it is not yellow you do not replace it. You then draw a second ball and note its color. (2) What is the probability that the second ball drawn is red?

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Assume that the box contains 11 balls: 4 red, 5 blue, and 2 yellow. As in the text, you draw one ball, note its color, and if it is yellow replace it. If it is not yellow you do not replace it. You then draw a second ball and note its color. (2) What is the probability that the second ball drawn is red?

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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There are 2 cases case1 the first ball drawn is red case 2 the first ball drawn is not red
ok
there are 4 red balls and 22 balls in total the chance you are in case 1 is 4/22 change you are in case 2 is 18/22 when you are in case 1 you have a 3/21 chance of picking up a red ball the 2nd time when you are in case 2 you have a 4/21 chance of picking up a red ball the 2nd time Total prob of picking up a red is therefore 4/22*3/21 + 18/22*4/21

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another way to view this problem is you are solvint these 2 cases
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(2) There is a 2/11 probability of the first ball being yellow, in which case there would be a 4/11probability of the second ball being red. There is a 4/11 probability of the first ball being red, in which case there is a 3/10 chance of the second ball being red. There is a 5/11 probability of the first ball being blue, in which case there is a 4/10 probability of the second ball being red. So the probability that the second ball drawn is red is (2/11)(4/11) + (4/11)(3/10) + (5/11)(4/10) =
216/605?
yes :)
in percentage 36%
thank you so much!

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