Question

Suppose a NBA basketball player has 5% chance of picking up a foul every minute he plays. What is the probability that he will play the entire 48 minutes of the game without fouling out (i.e., without picking more than four fouls)?

Answer 1

Answer:

He wont foul out with probability 0.9093

Step-by-step explanation:

The total number of fools he picked is a Binomial ditribution noted by X with parameters p = 0.05 and N = 48. The mean of this random variable is μ = np = 48*0.05 = 2.4 and the variance is σ² = np(1-p) = 2.4*0.95 = 2.28, hence its standard deviation is σ = √2.28 = 1.51.

Note that, if approximate probability is asked, we could just approximate X with a Normal random variable with mean 2.4 and standard deviation 1.51 (this can be done because of the central limit theorem). We will calculate the probability manually. He wont foul out if he picks 0,1,2,3 or 4 fouls, thus

$P(X \leq 4) = P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4) \\= 0.95^{48}+48*0.95^{47}*0.05+{48 \choose 2}*0.95^{46}*0.05^2+\\{48 \choose 3}*0.95^{45}*0.05^3+{48 \choose 4}*0.95^{44}*0.05^4 \\= 0.085+0.215+0.266+0.215+0.127 = 0.9093$

As a consecuence, he wont foul out with probability 0.9093.