studyn

Ok, so in case 1, the way the formula works is that it first establishes that you made 3 throws and lost 2 throws. Hence (.80 * .80 * .80) * (.20 * .20 ).

Then it accounts the fact that you could have made the three throws in 10 different orders. So, it's (.80 * .80 * .80) * (.20 * .20 )*10.

What you have done in case 2 is equivalent to:
0.5*0.5*0.5*0.5*0.5*10.

In other words, case 2 assumes that your probability of making each throw is 50%, not 80% as stated in the question.

ALSO:

The case 1 method does not give you the probability of making at least 3 throws. It gives you the probability of making exactly 3 throws. I assume that you want this latter thing, since you're saying your answer was correct.