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You randomly draw two marbles from a bag that contains 14 green, 7 blue, and 9 red marbles.What is the probability that the first marble is blue and the second marble is not blue if(a) the draws are made with replacement?(b) the draws are made without replacement?

Question

You randomly draw two marbles from a bag that contains 14 green, 7 blue, and 9 red marbles.What is the probability that the first marble is blue and the second marble is not blue if(a) the draws are made with replacement?(b) the draws are made without replacement?

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Solution

(a) With Replacement: The probability of drawing a blue marble is 7/30 because there are 7 blue marbles and 30 total marbles. Since the draws are made with replacement, the total number of marbles remains the same for the second draw. The probability of drawing a marble that is not blue is 23/30 because there are 23 marbles that are not blue (14 green + 9 red). Therefore, the probability that the first marble is blue and the second marble is not blue is (7/30) * (23/30) = 161/900.

(b) Without Replacement: The probability of drawing a blue marble is still 7/30 for the first draw. However, since the draws are made without replacement, the total number of marbles decreases to 29 for the second draw. The probability of drawing a marble that is not blue is now 23/29 because there are still 23 marbles that are not blue. Therefore, the probability that the first marble is blue and the second marble is not blue is (7/30) * (23/29) = 161/870.

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