SOLUTION: Arriving late at a Bake Sale , there are only 6 donuts, 4 cupcakes and 2 banana breads left. If Michael asks for two treats at random, what is the probability that he will choose

Question 1161145: Arriving late at a Bake Sale , there are only 6 donuts, 4 cupcakes and 2 banana breads left. If Michael asks for two treats at random, what is the probability that he will choose
a) 2 donuts (b) A donut and a cupcake or a donut and a banana bread
Thank you in advance for any help:)
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
a is (6/12)(5/11), the probability of choosing a donut with non-replacement
this is probability 30/132 or 5/22. You are correct,
for b, there are 12C2 ways to pick 2 items
for a donut it is 6C1 and for the others it is 6C1
This is 36/66=6/11.
A second way is that a donut for the first is 6/12 and the cupcake is 4/11=24/132
The other is 6/12*2/11=12/132
But each of those may be chosen in reverse order so the result is 72/132 or 6/11.
I am assuming that the order does not matter.
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