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Two balls are randomly selected from a bag that contains 3 red balls, 5 green balls, and 2 yellow balls. If the first ball is replaced before the second ball is selected, determine the probability that you select a red ball and then a green ball. Express your answer as a fraction in simplest form. If the first ball selected is not replaced before the second ball is picked, determine the probability that you select a red ball and then a green ball. Express your answer as a fraction in simplest form.

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Final answer:

The probability of selecting a red ball and then a green ball with replacement is 3/20, while without replacement it is 1/6.

Step-by-step explanation:

The probability of selecting a red ball and then a green ball with replacement is calculated by multiplying the probability of selecting a red ball on the first draw and the probability of selecting a green ball on the second draw. There are 3 red balls out of 10 total balls, so the probability of selecting a red ball is 3/10. Since the ball is replaced, there are still 5 green balls out of 10 total balls, making the probability of selecting a green ball 5/10. Therefore, the combined probability with replacement is (3/10) * (5/10) = 15/100, which simplifies to 3/20.

Without replacement, after a red ball is selected, there are now 9 balls left in the bag. The probability of selecting a green ball has now changed to 5 green balls out of 9 total balls, hence 5/9. The combined probability without replacement is (3/10) * (5/9) = 15/90, which simplifies to 1/6.

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