An urn contains 14 balls indentical in every aspect except color. Th.pdf

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An urn contains 14 balls indentical in every aspect except color. There are 7 green, 5 red and 2 white balls in the urn. a) If you draw a ball, record the color, replace it, draw a second and record the color, what is the probability of choosing a green followed by a red ball? b) If you draw a ball, record the color then draw a second, what is the probability of choosing two white balls? Solution There are total 14 balls, out of which 7 are green so the probability of choosing green ball is P(green)= 7/14 5 are red balls so the probability of choosing red ball is P(red)=5/14 2 are white balls so the probability of choosing red ball is P(white)=2/14 (a) Since balls are drwan with replacement so probability of choosing any color ball remain same as defined above. So the probability of choosing a green followed by a red ball is P(green)P(red)= (7/14)*(5/14)=5/28 (b) Here it is not given that after first draw ball has been replaced so assuming that drawn are without replacement. So the probability of choosing white ball in first draw is P(first white)=2/14 After that there is only one white ball remaining and total 13 ball remaining in the urn. So the probability of choosing second white ball is P(second white)=1/13 Therefore the probability of choosing two white balls is P(first white)P(second white)=(2/14)*(1/13)=1/91.

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