Question: 
What I tried. Well for part i the probability of any face being landed on is 1/4. I'm not sure if the way I explanation for the approach I took is correct because it gets the same answer but... $$ {\binom {3}{3}}*({\frac 1 4} )^3$$
My reasoning is that out of the you choose all 3 tetrahedrons to land on black and then 1/4 is to actually land it on the black?