Wow! A picture is worth a thousand words.
Having done this type of thing many times, I take for granted that instructions are hard to give. Although the first step is unnecessary, you can skip it and go directly to the second pic.
Props on the 3/16" drill bit recommendation...
And now I'd like to give a quick word on LED's. Think of an LED as a switch, because it IS a switch. So let's say you have a red diode rated for 1.9 volts forward drop and 20mA. This tells me that below about 1.5 volts, the LED will prevent any current from flowing, the LED will be off. But once you go above 1.9 volts, the LED lets ALL current go through. It does not limit current in any way. So if you are working with 5 volts, you know the LED will use 1.9 volts and wants a max of 20mA. Now you just need to find a resistor that drops (5-1.9)=3.1 volts and allows 20 mA of current.
R=V/I
If I(current) is always chosen to be 20 mA, the formula becomes
R=50*V
R=50*(3.1)
R=155 ohms, the closest value is 150, so use that.
This is assuming a 1.9 volt drop for the diode, 5 volt source power, and 20 mA current.
Please, if you choose to connect an LED directly to a voltage source, do not tell us about it. You are hurting the community by spreading such disinformation. I myself connected a green LED to a 12 volt battery just to see how bad it would be. The LED was shining red, not green, and I measured 250 mA of current. It burnt out in roughly 5 seconds.