A snack-size bag of M&Ms candies is opened. Inside, there are 12 red candies, 12 blue, 7 green, 13 brown, 3 orange, and 10 yellow. Three candies are pulled from the bag in succession, without replacement.What is the probability that the first two candies drawn are yellow and the third is blue?
Question
A snack-size bag of M&Ms candies is opened. Inside, there are 12 red candies, 12 blue, 7 green, 13 brown, 3 orange, and 10 yellow. Three candies are pulled from the bag in succession, without replacement.What is the probability that the first two candies drawn are yellow and the third is blue?
Solution
To solve this problem, we need to calculate the probability of each event happening and then multiply those probabilities together because we want all of them to happen.
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First, calculate the total number of candies in the bag. This is the sum of all the different colors of candies: 12 red + 12 blue + 7 green + 13 brown + 3 orange + 10 yellow = 57 candies.
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The probability of drawing a yellow candy first is the number of yellow candies divided by the total number of candies. So, the probability is 10/57.
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After drawing a yellow candy, we don't replace it, so there are now 56 candies left in the bag, and only 9 yellow candies. So, the probability of drawing a yellow candy second is 9/56.
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After drawing another yellow candy, we don't replace it, so there are now 55 candies left in the bag, and 12 blue candies. So, the probability of drawing a blue candy third is 12/55.
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The probability of all three of these events happening is the product of their individual probabilities. So, the probability is (10/57) * (9/56) * (12/55) = 0.0038 or 0.38% when rounded to two decimal places.
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