Hi, 

Note: The probability of x successes in n trials is: 

.

P = nCx* * where p and q are the probabilities of success and failure respectively. 

In this case p = .06 and q = .94 

nCx = 



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 

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