For a 40 percent free-throw shooter, 0 points are most likely. For an 80 percent shooter, 2 points are likely. If the probability of making a free throw is 40 percent, the probability of missing is 60 percent.
P(0 points) = .60If the probability of making a free throw is 80 percent, the probability of missing is 20 percent.
P(1 point) = .40 × .60 = .24
P(2 points) = .40 × .40 = .16
P(0 points) = .20
P(1 point) = .80 × .20 = .16
P(2 points) = .80 × .80 = .64