A candy store grab bag contains 10 pieces of sour candy, 12 pieces of chocolate candy, and 6 pieces of sweet candy. If you select 1 piece of candy, replace it, and then select another piece, what is the probability that the first piece of candy taken out is a sour candy and the second piece of candy taken out is not a sour candy?
Question
A candy store grab bag contains 10 pieces of sour candy, 12 pieces of chocolate candy, and 6 pieces of sweet candy. If you select 1 piece of candy, replace it, and then select another piece, what is the probability that the first piece of candy taken out is a sour candy and the second piece of candy taken out is not a sour candy?
Solution
To solve this problem, we need to calculate two separate probabilities and then multiply them together.
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The probability of drawing a sour candy first: There are 10 sour candies out of a total of 28 candies (10 sour + 12 chocolate + 6 sweet = 28). So, the probability of drawing a sour candy first is 10/28 or 5/14.
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The probability of not drawing a sour candy second: Since the first candy is replaced, there are still 28 candies in the bag. The number of candies that are not sour is 18 (12 chocolate + 6 sweet = 18). So, the probability of not drawing a sour candy second is 18/28 or 9/14.
Finally, we multiply these two probabilities together to get the overall probability. So, the probability that the first piece of candy taken out is a sour candy and the second piece of candy taken out is not a sour candy is (5/14) * (9/14) = 45/196.
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