Question

During the 2015-16 NBA season, J.J. Redick of the Los Angeles Clippers had a free throw shooting percentage of 0.901 . Assume that the probability J.J. Redick makes any given free throw is fixed at 0.901 , and that free throws are independent. Correct answer iconCorrect. (a) If J.J. Redick shoots 14 free throws in a game, what is the probability that he makes at least 13 of them? Round your answer to four decimal places.

Answer 1

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 :

$P(X=x)^nC_xp^x(1-p)^{n-x}$

, 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 :-

$P(x\geq13)=P(x=13)+P(x=14)$

$=^{14}C_{13}(0.901)^{13}(1-0.901)^1+^{14}C_{14}(0.901)^{14}(1-0.901)^0\\\\=(14)(0.901)^{13}(0.099)+(1)(0.901)^{14}\ \ [\because\ ^nC_n=1\ \&\ ^nC_{n-1}=n ]\\\\\approx0.3574+0.2324=0.5898$

∴ The required probability = 0.5898