SOLUTION: The probability that a player makes a free throw is 0.60. In 3 free throws, find the probability that the player makes a. exactly 3 free throws b. at least 1 free throw c. an od

Hi

The probability that a player makes a free throw is 0.60. In 3 free throws, find the probability that the player makes:

Note: The probability of x successes in n trials is: 

P = nCx*  where p and q are the probabilities of success and failure respectively. 

In this case p = .6 & q =.4

nCx = 

a. exactly 3 free throws = .6^3 = .216

b. at least 1 free throw = 1 - .4^3 = 1 -.064 = .936

c. an odd number of free throws (1 '0r' 3) = 3*.6*.4^2 + .6^3 = .288 +.216

d. if X=number of free throws a player can successfully make. Compute the value of the following:

i. Mean = .6X                      |mean = np

ii. Variance = .6*.4*X = .24X      |variance = npq