Question 415151
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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* {{{p^x*q^(n-x)}}} where p and q are the probabilities of success and failure respectively. 
In this case p = .6 & q =.4
nCx = {{{n!/(x!(n-x)!)}}}
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