Question 408149
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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 = .05 & q = .95 
nCx = {{{n!/(x!(n-x!))}}}

a) P(exactly 3 of 20 being defective computers) = 1140(.05)^3(.95)17 = .0596
(b) P( test at least 5 computers to find 2 defective ones)
 P = 1 - (.05)^0*(.95)^5 + 5*(.05)^1*(.95)^4 = .0226