My solution is symmetrical and only uses the 2 given bits.
The machine starts with the red lever because the red ball can be sent to the interceptor directly if the right bit is pointing left.
If the right bit is pointing right the left bit has to be evaluated also, so a crossover is used to lead the red ball to the blue lever.
The same logic applies to the blue ball; If the left bit is pointing right it should be intercepted. So the board can be mirrored.
This solution also works for the mirrored challenge: A right & B right => blue / A right & B left => blue / A left & B right => blue / A left & B left => blue!