Question 840651
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Hi
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(defective) = .1 & q = .9 
nCx = {{{n!/(x!(n-x)!)}}}
P(of six items two are defective)= 6C2(.10)^2(.90)^4 = .0984
P(of ten items, at least one is defective) = 1 - P(none defective)
 1-(.90)^6 = 1-.5314= .4686