Determine the probability that the first candy drawn is brown, the second is orange, and the third is yellow.

Question

A snack-size bag of M&Ms candies is opened. Inside, there are 12 red candies, 7 blue, 7 green, 17 brown, 4 orange, and 10 yellow. Three candies are pulled from the bag in succession, without replacement.

Determine the probability that the first candy drawn is brown, the second is orange, and the third is yellow. (Enter your probability as a fraction.)

Answer 1

12 red + 7 blue + 7 green + 17 brown + 4 orange + 10 yellow = a total of 57 candies

First draw (drawing a brown): 17/57

2nd draw (drawing orange): 4/56 (or 1/14)

3rd draw (drawing yellow): 10/55 (or 2/11)

The probability of all three happening: (17/57)(1/14)(2/11) = 34/8778 = 17/4389