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

To find the probability that Suzan grabs at most one green marble out of six marbles, we need to consider the two cases separately: one green marble and zero green marble.

Total number of marbles = 4 (red) + 2 (green) + 4 (white) + 2 (purple) = 12 marbles

1. Probability of getting one green marble:
Number of ways to choose 1 green marble out of 2 = \binom{2}{1}
Number of ways to choose 5 remaining marbles out of 10 (excluding green marble) = \binom{10}{5}
Total number of ways to choose 6 marbles out of 12 = \binom{12}{6}
Probability = \frac{\binom{2}{1} \times \binom{10}{5}}{\binom{12}{6}}

2. Probability of getting no green marble:
Number of ways to choose 0 green marble out of 2 = \binom{2}{0}
Number of ways to choose all 6 marbles out of 10 (excluding green marble) = \binom{10}{6}
Probability = \frac{\binom{2}{0} \times \binom{10}{6}}{\binom{12}{6}}

Adding the probabilities of the two cases gives the total probability.

Answer: \frac{\binom{2}{1} \times\binom{10}{5} + \binom{2}{0} \times\binom{10}{6}}{\binom{12}{6}}=\frac{17}{22}