Suzan sees a bag of marbles, she grabs a handful at random. She has seen a bag containing three red marbles, two green ones, thr
ee 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.
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<span>. 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/ </span>11*10= 3/55 <span> The probability for </span><span>at least one green one: 1-3/55= 52/55</span>