Answer: 0.5898
Step-by-step explanation:
Given : J.J. Redick of the Los Angeles Clippers had a free throw shooting percentage of 0.901 .
We assume that,
The probability that .J. Redick makes any given free throw =0.901 (1)
Free throws are independent.
So it is a binomial distribution .
Using binomial probability formula, the probability of getting success in x trials :

, where n= total trials
p= probability of getting in each trial.
Let x be binomial variable that represents the number of a=makes.
n= 14
p= 0.901 (from (1))
The probability that he makes at least 13 of them will be :-

![=^{14}C_{13}(0.901)^{13}(1-0.901)^1+^{14}C_{14}(0.901)^{14}(1-0.901)^0\\\\=(14)(0.901)^{13}(0.099)+(1)(0.901)^{14}\ \ [\because\ ^nC_n=1\ \&\ ^nC_{n-1}=n ]\\\\\approx0.3574+0.2324=0.5898](https://tex.z-dn.net/?f=%3D%5E%7B14%7DC_%7B13%7D%280.901%29%5E%7B13%7D%281-0.901%29%5E1%2B%5E%7B14%7DC_%7B14%7D%280.901%29%5E%7B14%7D%281-0.901%29%5E0%5C%5C%5C%5C%3D%2814%29%280.901%29%5E%7B13%7D%280.099%29%2B%281%29%280.901%29%5E%7B14%7D%5C%20%5C%20%5B%5Cbecause%5C%20%5EnC_n%3D1%5C%20%5C%26%5C%20%5EnC_%7Bn-1%7D%3Dn%20%5D%5C%5C%5C%5C%5Capprox0.3574%2B0.2324%3D0.5898)
∴ The required probability = 0.5898
42% ................................