Model marbel toy with finite automata I'm resolving this question of Hopcroft and et al Book. Figure 1 below is a marble rolling-toy. A marble is dropped at A or B. Levers $x_1,x_2$ and $x_3$ cause the marble to fall either to the left or to the right. Whenever a marble encounters a lever, it causes the lever to reverse after the marble passes, so the next will take the opposite branch.



I got resolve the question ...

a)Model this toy by finite automaton: To model this toy, a state is represented as sequence of three bits followed by r or a (previous input rejected or accepted). For instance, 010a, means left, right, left, accepted. Then the Transition table is 



b)Informally describe the language of the automaton

but not b) question. How I will be able to resolve this question?

EDIT
I see a solution here I modify my question using suggestions and that solution. According that solution exist a case:

Penultimate configuration interrupts is * \ / , where * means / or \; and the ending is 1 and X mod 4 = 0, or (X-3) mod 4 = 0 (X is the numbers of 0's and 1's).

For my question, from this case I get the subcase 

Penultimate configuration interrupts is * \ / , where * means / or \; and the ending is 1 and X mod 4 = 0.

Then, the restrictions are: the number of 1's, without the ending, is even and the number of 0's is impar. I will be able to verify that configuration with that restrictions for few instances, but How I will be able to demonstrate that the $x_2$ interrupt configuration is \?
Count $A$'s and $B$'s. The next $A$ will only exit at $D$ iff the first two levers are in right position. When does that happen? The first lever toggles at every $A$, but the second lever interacts with the $B$'s.

At the same time, the situation for $B$ is symmetric. The next $B$ will not exit $D$ only if both levers are left. When does that happen?