the current will be I=(5 - 0.7 - 1.2V)/(100+400[Ω]) =6.2mA right?
No, that's not quite right. It would be correct if the collector was disconnected from the 5 volt rail but, it isn't.
What we can say (as a fairly reasonable approximation) is that the the BJT emitter will be at around 4.3 volts\$^1\$ and this of course assumes that the base is close to 5 volts. Now, the current into the LED is the voltage across R2 divided by R2 hence, it is 3.1 volts ÷ 400 Ω = 7.75 mA.
And, nearly all of that current comes through the collector to the emitter. The base current will contribute a little bit but maybe only about 1% of the 7.75 mA. This, of course, is due to the current gain of the transistor. We tend to say that when the BJT isn't saturated that the current gain is about 100 hence, we might expect about 77.5 μA through the base.
Looking at the data sheet for the PMBT3904,215 transistor, the current gain will probably be more like 200 typically
So, the volt drop is the 100 Ω resistor (R1) is about 7.75 mV i.e. the base is pretty close to 5 volts thus bolstering our original assumption that assumed the emitter would be around 4.3 volts.
\$^1\$ - the emitter always follows the base but, because the base-emitter junction is a forward biased diode we assume that the voltage at the emitter (for an NPN) is around 0.7 volts lower.