Hi,
to get some light in the discussion and darkness of the theory of probabilities...
Assuming that the chances for one mouboo to drop an aqua is 1% (which means 0.01) and that we can deal every head like a single mouboo, which is a good assumtion but has to be proven to be correct...
Now the chances to get at least 1 (in words one) aqua includes the following three cases:
1. every head drops one aqua -> 0.01 x 0.01 x 0.01 = 0.000001
2. two of the heads drop one
and the last one drops none-> 0.01 x 0.01
x 0.99 = 0.000099
this can happen three times (because we don't care which two heads drop the aqua, do we?)
->
0.000099 x 3 = 0.000297
3. just one of the heads drops one and the other two drop nothing -> 0.01
x 0.99 x 0.99
obviously this also can happen in three cases (as stated above

) ->
0.009801 x 3 = 0.029403
Now because we would be happy with one, two or three aquas we add up all three and the result is:
0.029403 + 0.000297 + 0.000001 = 0.029701 which is 2.9701% which is a little bit
less than 3% .
I hope you enjoyed the excourse in probabilities...
Greetings,
Dicekid
P.S.: I don't like to quote myself, but please don't be too serious...
EDIT: I recognised a tiny error in my calculation
(which is no longer existing, so this here shows just another approach to calculate...). One has to calculate the chance that no aqua drops which is
0.99 x 0.99 x 0.99 = 0.970299 and then subtract this from 1 -> 0.029701 -> 2.9701%. I will think about how to get the same result the other way round or where the logical mistake was. But this one is at least correct. Because even if you fight a million mouboos there is a chance that you won't get any aqua from them and this would not be the case with the former calculation (as you can easily see...)
So sorry for the confusion...
2nd EDIT: Sorry... Ok, I corrected the things above,
the red things are new and this is the correct way to calculate the correct possibility.

I hope you don't mind....
