Question 443534
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Hi
A manufacturing machine has a 1% defect rate. 
4 items are chosen at random
P(at least one will have a defect)
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 = .01 and q = .99
nCx = {{{n!/(x!(n-x)!)}}}
 P(at least one defective) = 1 - P(none defective) = 1 - (.99)^4 =.0394 
P(at least one defective)= .0394