Binomial coefficient question? Part D? 
Jinbao produces tubas and ships them in lots of twenty. 
Suppose that 60% of all such lots contain no defective tubas, 30% 
  contain one defective, and 10% contain two defectives.  Now suppose 
  that a lot is inspected, with two tubas being selected from it at 
  random, and neither is found to be defective.
a) What is the probability that there are no defectives in that lot?
b) What is the probability that there is one defective in that lot?
c) What is the probability that there are two defectives in that lot?
d) Suppose that the inspected lot is from a shipping container that 
  contains 10 lots, and the other 9 lots were not inspected.  What is 
  the probability that there are no defectives in that container?

I've done part a,b,c. Any idea how to do part d?
 A: You are trying to find the probability of 19 good given 10 good. An idea might be to draw a tree diagram and investigate each arm.
A: 
d) Suppose that the inspected lot is from a shipping container that contains 10 lots, and the other 9 lots were not inspected. What is the probability that there are no defectives in that container?

Since the probabilities of finding defectives in the other containers are conditionally independent of finding defectives in the first, then this is simply: the probability of the tested lot being good, given that its samples were both good, times the probabilities that each of the remaining lots are good.
Let $D_x$ count the number of defectives in lot of $20$ tubas, for lot number $x$, and $T_x$ count the number of defectives in the sampled two tubas.
$$\begin{align}\operatorname{P}(\mathop{\cap}_{x=1}^{10} D_x=0\mid T_1=0) & = \operatorname{P}(D_1=0 \mid T_1=0)\times\prod_{x=2}^{10} \operatorname{P}(D_x=0) \\ ~ & = \operatorname{P}(D=0\mid T=0)\times(\operatorname{P}(D=0))^9\end{align}$$
Where $P(D=0\mid T=0)$ the the answer to part (a). (The probability of no defectives in a lot given no defectives in the sample.)
