How are Transistors in the Cut-off Region in Double Reverse Bias?
Here both BE and BC junctions have reverse external biasing voltage. There's no current except leakage as long as the breakdown voltages of the junctions are not exceeded.

Turn the polarity of the source in the left. Then the base current starts and the transistor can also sink some collector current from the source in the right. But now as drawn the transistor is in the cut-off state because there's no current except leakage.