Question 1200689
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The number of all possible triples from 100 CDs is  

    {{{C[100]^3}}} = {{{(100*99*98)/(1*2*3)}}} = 161700.


The number of triples CDs, that contain exactly 2 of 6 defective CDs is 

    {{{C[6]^2*C[94]^1}}} = 15*94 = 1410.


The desired probability is the ratio of these numbers

    P = {{{1410/161700}}} = 0.00872   (rounded).    <U>ANSWER</U>
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Solved.