The question is:
Probability of scoring $1$ out of $3$ shots when you have 80% throw rate.
I solved this problem in the inverse way:
P(At least one basket) = $1$ - P(No basket) = $1$ - ($.20 \times .20 \times .20) = 0.992 ~ or ~99.2 \% $
I wanted to know how would I go about solving this the other (which I realize is not the best way)..