Question

A certain NBA basketball player makes 70% of his free throws.  Suppose this player has three free throws.  What is the most likely number of free throws that he will make?

Answer 1

Solution:

Given ,

p = 70% = 0.70

n = 3

So, X follows Binomial(3 , 0.70)

Using binomial probability formula ,

P(X = x) = (_n C _x) * p^x * (1 - p)^{n - x}   , x = 0 , 1 , 2 , 3 , ....., n

Possible values of x are 0 ,1 , 2 , 3

P(X =0 ) = (₃ C ₀) * 0.7⁰ * (0.3)^3-0 = 0.027

P(X =1 ) = (₃ C ₁) * 0.7¹ * (0.3)^3-1 = 0.189

P(X =2 ) = (₃ C ₂) * 0.7² * (0.3)^3-2 = 0.441

P(X = 3 ) = (₃ C ₃) * 0.7³ * (0.3)^3-3 = 0.343

Observe the probabilities. P(X = 2 ) is greater .

So, answer is 2 .

What is the most likely number of free throws that he will make? Answer is 2