Double BJT circuit with one BJT off and one in saturation
You are overthinking this. 

The emitter current is 4.8mA using your 200mV drop for a transistor in saturation. The emitter voltage is 4.8V. 

The base current is (10V-(4.8V+0.7V))/10K =  450uA

The sum of all three currents flowing into the transistor must be ??