Answer:
P(made 2nd attempt|made 1st attempt)=P(made 2nd attempt)
Step-by-step explanation:
Here given that a basketball player that shoots 80% from the free throw line attempts two free throws.
If x is the no of shoots he makes (say) then we find that each throw is independent of the other.
In other words, because he made successful first attempt, his chances for second attempt will not change
Prob for success in each attempt remains the same as 0.80
Hence I throw is independent of II throw.
When A and B are independent,then we have
P(A/B) = P(A)
Hence answer is
P(made 2nd attempt|made 1st attempt)=P(made 2nd attempt)
