I'm a sample bag of m& m's there are 5 brown, 6 yellow, 4 blue, 3 green, and 2 orange. What is the probability of getting 3 yellow m& m's if 3 are taken at a time?

There are (5+6+4+3+2) = 20 M&Ms in the bag.

The probability of the first one being yellow is 6 out of 20 = 6/20.

If it is then the probability of the second one being yellow is 5 out of 19.

If it is then the probability of the third one being yellow is 4 out of 18.

The probability of all three being yellow is then

(6/20) x (5/19) x (4/18) = 120/6,840 = 3/171 = 1/57 = 1.75% (rounded)

There are a total of 5+6+4+3+2=20 M&Ms in the bag. Since there are 6 yellow M&Ms in the bag, there's a 6/20 probability that the first M&M will be yellow. Then, there are 5 more yellow M&Ms and 19 more total M&Ms in the bag, so there is a 5/19 chance the next M&M will be yellow. Finally, there are 4 more yellow M&Ms in the bag and 18 total M&Ms, so there is a 4/18 chance the last M&M will be yellow.

Multiplying these together, we have:

(6*5*4) / (20*19*18)

This simplifies to:

1/57

Therefore, there is a 1/57 probability that all M&Ms will be yellow.