2018 Fermat Problem 24

Wayne has 3 green buckets, 3 red buckets, 3 blue buckets, and 3 yellow buckets. He randomly distributes 4 hockey pucks among the green buckets, with each puck equally likely to be put in each bucket. Similarly, he distributes 3 pucks among the red buckets, 2 pucks among the blue buckets, and 1 puck among the yellow buckets. Once he is finished, what is the probability that a green bucket contains more pucks than each of the other 11 buckets?

\textbf{(A)}\ \frac{97}{243}\quad \textbf{(B)}\ \frac{89}{243}\quad \textbf{(C)}\ \frac{93}{243}\quad \textbf{(D)}\ \frac{95}{243}\quad \textbf{(E)}\ \frac{91}{243}

If there are no answer choices shown, enter a numerical answer.

Full credit to this problem is given to the CEMC, you may view all Fermat contests here.