Question 637647
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Hi, 
From a lot of 12 missiles, 5 are selected at random and fired. 
If the lot contains 3 defective missiles {{{1/4}}}th are defective
P(5 will fire) = (3/4)^5
P(at most 2 will fire) = P(0 fire) + P(1 fire) + P(2 fire)
                       = {{{(1/4)^5 + 5(3/4)^1(1/4)^4 + 10(3/4)^2(1/4)^3}}}

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 = 3/4 & q = 1/4
nCx = {{{n!/(x!(n-x)!)}}}