Reputation: 9
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
sum(dbinom(3:5,n,p))
And I got 0.83, but I know that answer is 0.18.
pbinom(5,n,p)-pbinom(2,n,p)
Can someone please help me.
Upvotes: 0
Views: 120
Reputation: 73265
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
# [1] 0.18522
Note, there is really a trap here. It is so easy to assume the answer as
dbinom(3, 5, 0.7)
# [1] 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.
Upvotes: 4