Note that three events $A_0$, $A_1$, and $A_2$ of players receiving 0, 1 or 2 cards of type A are exclusive and partition the entire event space (there are only 2 cards of type A). This means that $ \mathbb{P}(A_0) + \mathbb{P}(A_1) + \mathbb{P}(A_2) = 1 $
There total $\binom{11}{5}$ ways to deal 1 card to each of 5 players. There are $\binom{11-2}{5}$ ways to deal 5 cards with no type A card in it (just throw out 2 type A card out of the deck), that makes $ \mathbb{P}(A_0) = \frac{\binom{11-2}{5}}{\binom{11}{5}} = \frac{\frac{9!}{4! 5!}}{\frac{11!}{6! 5!}} = \frac{9!}{11!} \frac{6!}{4!} = \frac{6 \cdot 5}{11 \cdot 10} = \frac{3}{11} $
To compute $\mathbb{P}(A_2)$, deal type A cards first. What remains is to deal 9 type B cards to 3 remaining players, thus $ \mathbb{P}(A_2) = \frac{\binom{9}{3}}{\binom{11}{5}} = \frac{\frac{9!}{6! 3!}}{\frac{11!}{6! 5!}} = \frac{9!}{11!} \cdot \frac{5!}{3!} = \frac{5 \cdot 4}{11 \cdot 10} = \frac{2}{11} $
Thus $ \mathbb{P}(A_1) = 1- \mathbb{P}(A_0) -\mathbb{P}(A_2) = \frac{6}{11} $