Probability Modem is Defective 
A store has 80 modems in its inventory, 30 coming from Source A and
  the remainder from Source B. Of the modems from Source A, 20% are
  defective. Of the modems from Source B, 8% are defective. Calculate
  the probability that exactly two out of a random sample of five modems
  from the store’s inventory are defective.
(A) 0.010 (B) 0.078 (C) 0.102 (D) 0.105 (E) 0.125


I believe an approximation is involved.
The weighted average % defective = $\frac{30}{80}\cdot{20} + \frac{50}{80}\cdot{8} = 12.5$
Now apply binomial(5,0.125) which yields P(2 defective) = $\frac{1715}{16384} \approx 0.105$, 
Edit
There are differing thumb rules as to when the binomial approximation to the hypergeometric can be used. Here, it appears, we have to go by the expected number of defectives of 10, and apply the hypergeometric formula to get an answer of 0.102, as Bruce Trumbo has done.
