SOLUTION: 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 successi

Algebra ->  Probability-and-statistics -> SOLUTION: 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 successi      Log On


   

Question 1207764: 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 blue and the third is red?
Answer by ikleyn(52794) About Me  (Show Source):
You can put this solution on YOUR website!
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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 blue and the third is red?
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The total candies is  12 + 12 + 7 + 13 + 3 + 10 = 57.


Now the probability is  P = %2812%2F57%29%2A%2811%2F56%29%2A%2812%2F55%29 = %2812%2A11%2A12%29%2F%2857%2A56%2A55%29 = 1584%2F175560 = 6%2F665 = 0.009  (rounded).

Solved.

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The structure of the formula makes it SELF-EXPLANATORY.