Question 364644
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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 = .06 and q = .94 
nCx = {{{n!/(x!(n-x!))}}}

P(at least 2 turn out to be defective)
1 - P(0 def) - P(1 defect)
P(0 def) = .94^8 = .610
P(1 defect) = 8C1*.06^1 .94^7 = 8*.06^1 .94^7 = 8*.06*.648 = .311
1 - .921 = .079