Suzan sees a bag of marbles, she grabs a handful at random. She has seen a bag containing three red marbles, two green ones, three white ones, and three purple ones. She grabs eight of them. Find the probability of the following event, expressing it as a fraction in lowest terms. She has at least one green one.

The answer to this question would be: 52/55

In this question, Suzan is taking 8 marble from a bag containing 3 red marbles, 2 green ones, 3 white ones, and 3 purple ones. The total marble in the bag should be: 3+2+3+3 = 11. The question is how much the probability to get at least one green one. This means the opposite probability of no green at all. The green marble is 2 of 11 so the chance to not getting green would be 9/11 for the first attempt and then 8/10 for the second.

Then the probability for no green marble would be:

(9!/9-8!) / 11!/8! = 9*8*7*6*5*4*3*2 / 11*10*9*8*7*6*5*4 = 3*2 / 11*10 = 3/55

The probability for at least one green one: 1-3/55 = 52/55