Harry is a basketball player.His probability of making a free throw is 0.70.
-> what is the probability that Harry makes his third free throw on his fifth shot?
things I have tried so far; given p=0.70 n = 5
If he gets 3rd free throw on 5th shot, then
Note, these two events are independent, while the probability of the first is
dbinom(2, 4, 0.7), and the probability of the last is 0.7. So
dbinom(2, 4, 0.7) * 0.7 #  0.18522
Note, there is really a trap here. It is so easy to assume the answer as
dbinom(3, 5, 0.7) #  0.3087
But that only means we have 3 free throws out of 5 shots, which does not guarantees the 5th shot is a free throw.