Transistors as switches
Chill out Riddler,
The resistors you require are easy to work out ...
See attached drawings
The spec for the RED LED says 20mA with typically 1.9v dropped across it.
For the GREEN and BLUE it says 10mA with typ. 3.5/3.6v dropped.
The important thing here is the current - the voltage is a by-product of the LED's internal resistance.
OK so I can assume your power supply is more than 3.6v.
I guess the supply is 5v and do some sums (Ohm's Law V=I*R)
For the RED LED...
If the LED drops 1.9v there will be (5-1.9) 3.1v left for the resistor.
R=V/I so R = 3.1v/0.02A = 155 Ohms (R1 in my diagrams)
For the others (about the same as each other)
the LED drops 3.5v so there is (5-3.5) 1.5v left for the resistor.
R=V/I so R = 1.5v/0.01A = 150 Ohms (both R2 and R3)
For NPN switching transistors I like BC107, BC108, BC109 (easy for me to get), any small, low power transistor will do.
A positive bias is required on the base to turn it on, as these transistors have a current gain (hfe) of around 500 and you want them to pass 20mA (for the red LED) you must supply at least 0.02A/500 = 40uA to the base.
Assuming you have a 5v signal to turn the transistor on and the base-emitter junction drops 0.7v when working... (more sums)
R=V/I = (5-0.7) / 0.00004A = 107k.
As this is the MINIMUM and the transistor can cope with 5v into the base (it just wastes power!) I would use 47k or so for all the base resistors.
For PNP transistors the sums are different but only because we are now using the transistors differently (the first cct was common emitter, this is common collector or emitter follower)!
R1, R2 and R3 are calculated the same but we should now include a further 0.7v dropped across the transistor's base-emitter junction, leaving 2.4v for the red LED and 0.8v for the others.
R1 is therefore 120 Ohms
R2 and R3 are 80 Ohms
R4, R5 and R6 can be dumped - emitter-followers are self regulating!
So now you have two ways of doing it, depending on your switching signal ...
Positive signals will turn on the NPN transistors
Negative (connect to 0v) will turn on the PNP transistors