Question 839297
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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)!)}}}
If 6 calculators are picked at random, what is the probability that 2 are defective
 P = 6C2 (.1)^2(.9)^4 = .0984