A3. The probabilities that the letters $A, B, C$ or $I$ fall on one die are sequentially $\frac{1}{3}, \frac{1}{3}, \frac{1}{6}$ or $\frac{1}{6}$. If we assume that Klara rolls the dice in such a way that they are arranged in a row, then the probability that the word $B A B I C A$ falls is $\frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{3}=\left(\frac{1}{3}\right)^{4}\left(\frac{1}{6}\right)^{2}$. Any outcome where the letters in this word are shuffled is favorable. There are a total of $\frac{6!}{(2!)^{2}}$ such outcomes, as this is the number of permutations with repetition of 6 elements, where the letters $A$ and $B$ each appear 2 times. Therefore, the probability that Klara can form the word $B A B I C A$ from the obtained letters is $\frac{6!}{(2!)^{2}}\left(\frac{1}{3}\right)^{4}\left(\frac{1}{6}\right)^{2}$.
A3. Klara has 6 identical cubes. On the side faces of each cube, she wrote the letters $B, A, B, I, C$ and $A$, one letter on each side face. Then she rolled all 6 cubes at once. What is the probability that she can form the word $B A B I C A$ from the letters that land on the top faces of the cubes?
(A) $\frac{6!}{(2!)^{2}}\left(\frac{1}{3}\right)^{4}\left(\frac{1}{6}\right)^{2}$
(B) $\frac{6!}{(2!)^{2}}\left(\frac{1}{6}\right)^{6}$
(C) $\left(\frac{1}{3}\right)^{4}\left(\frac{1}{6}\right)^{2}$
(D) $\left(\frac{1}{6}\right)^{6}$
(E) 1