An urn contains 10 balls: 4 red and 6 blue. A second urn contains 16 red balls a

Question

An urn contains 10 balls: 4 red and 6 blue. A second urn contains 16 red balls and an unknown amount ofblue balls. A single ball is drawn from each urn. The probability that both balls are the same color is.44. Calculate the number of blue balls in the second urn. An urn contains 10 balls: 4 red and 6 blue. A second urn contains 16 red balls and an unknown amount ofblue balls. A single ball is drawn from each urn. The probability that both balls are the same color is.44. Calculate the number of blue balls in the second urn.

Explanation / Answer

Let there be x blue balls in urn 2 P(blue ball selected from urn 1) = 8/20 = 2/5 as 8 of 20 balls are blue P(red ball selected from urn 1) = 12/20 = 3/5 as 12 of 20 balls are red In urn 2, there are 9 red balls + x blue balls for a total of 9+x balls. Thus, P(blue ball selected from urn 2) = x/(9+x) P(red ball selected from urn 2) = 9/(9+x) Then, the probability that both balls selected are the same color = P(blue ball is selected from both urns) + P(red ball is selected from both urns) = P(blue ball is selected from urn 1)P(blue ball is selected from urn 2) + P(red ball is selected from urn 1)P(red ball is selected from urn 2) = substituting probabilities from above 2/5 * x/(x+9) + 3/5 * 9/(x+9) = 2x/(5(x+9)) + 27/(5(x+9)) = (2x + 27)/(5x+45) Thus, (2x + 27)/(5x+45) = .49 = 49/100 Cross-multiplying, 200x + 2700 = 245x + 2205 2700 - 2205 = 245x - 200x 495 = 45x x = 11 There are 11 blue balls in urn 2 We can check the result easily If there are 11 blue balls in urn 2, then P(blue ball) = 11/(11+9) = 11/20 and P(red ball) = 9/(11+9) = 9/20 Then, P(blue ball is selected from urn 1)P(blue ball is selected from urn 2) + P(red ball is selected from urn 1)P(red ball is selected from urn 2) = 2/5*11/20 + 3/5*9/20 = 22/100 + 27/100 = 49/100 = .49 The answer checks.

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