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 green?
To calculate the probability, we need to determine the total number of candies in the bag and the number of ways to select three candies in total.
Total number of candies in the bag = 12 (red) + 12 (blue) + 7 (green) + 13 (brown) + 3 (orange) + 10 (yellow) = 57 candies
Number of ways to select 3 candies from 57 = 57C3 = 22,610 ways
Now, let's calculate the number of ways to pick the first two candies as yellow and the third as green.
Number of ways to pick the first two yellow candies = 10C2 = 45 ways
Number of ways to pick the third green candy = 7C1 = 7 ways
Therefore, the total number of ways to pick the first two yellow candies and the third green candy = 45 * 7 = 315 ways
So, the probability of selecting a yellow candy, then another yellow candy, and then a green candy = 315 / 22,610 ≈ 0.0139
Therefore, the probability that the first two candies drawn are yellow and the third is green is approximately 0.0139 or 1.39%.