Answer:
Step-by-step explanation:
Given that a basketball player who ordinarily makes about 55% of his free throw shots has made 4 in a row.
Probability for a success in any random throw = 
Since each throw is independent no of successive throws is binomial with
p = 0.55 and q = 0.45
Probability his free throw shots has made 4 in a row
= Prob that continuous succesees in 4 throws
=
which shows that this is a very rare event with 0.55 for each success.
Hence this is unusual and there is a necessity to increase 0.55 or 55%