Question

A basketball player’s free-throw average is calculated as the average number of made shots relative to total attempts. Suppose Shaquille currently has a 25% free-throw average and is a very consistent (if not a very good) shooter. During today’s game, this player will be at shoot exactly nine free throws (a) What is the probability that he ends up with three made shots? (b) What is the probability he ends up with no made shots? (c) What is the probability he ends up with at least four made shots? (d) What is the probability he ends up with fewer than three made shots?

Answer 1

Solution:

This is a binomial probability question. So, using combinations: P(X = x) = C(n,x)*p^x*(1-p)^n-x

Free-throw average = average number of made shots/total attempts = 0.25

So with consistency, probability of Success, p = 0.25, and of failure = 1-0.25 = 0.75

Total number of free-throws received, n = 9

a) Probability of 3 made shots:

P(X=3) = C(9,3)*(0.25)³*(0.75)^9-3

P(X=3) = 9!/(3!*6!)*0.0156*0.18

P(X = 3) = 0.2359, so required probability is 0.24 (approximately)

b) Probability of 0 made shots:

P(X = 0) = C(9,0)*(0.25)⁰*(0.75)^9-0

P(X=0) = 9!/(0!*9!)*1*0.075

P(X = 0) = 0.075, so required probability is 0.075

c) Probability of at least 4 made shots:

P(X = at least 4 made shots) = P(X = 4) + P(X=5) + P(X=6) + P(X=7) + P(X = 8) + P(X=9)

Also, probability of at least 4 made shots is same as 1 minus probability of at most 3 shots, so

P(X = at least 4 made shots) = 1 - [P(X = 0) + P(X=1) + P(X=2) + P(X=3)] ... (*)

Of this, we already know that P(X = 0) = 0.075 (from part (b)), and P(X = 3) = 0.24 (from part (a))

For remaining, P(X = 1) = C(9,1)*(0.25)¹*(0.75)^9-1 = 9!/(1!*8!)*0.25¹*0.75⁸ = 0.225

P(X=2) = C(9,2)*(0.25)²*(0.75)^9-2 = 9!/(2!*7!)*0.25²*0.75⁷ = 0.3003

So, using (*), we have P(X = at least 4 made shots) = 1 - [0.075 + 0.225 + 0.3003 + 0.24]

P(X = at least 4 made shots) = 1 - 0.8403 = 0.1597

d) P(X = fewer than 3 made shots) = ?

Fewer than 3 shots could possibly carry made shots either 0, 1 or 2

So, P(X = fewer than 3 made shots) = P(X = 0) + P(X = 1) + P(X = 2)

P(X = fewer than 3 made shots) = 0.075 + 0.225 + 0.3003 = 0.6003

So, required probability is 0.6 (approximately).