Insurance problem The following is the type of problem that I am dealing with.

An insurer makes $n=4$ driving errors, each independently resulting in an accident with probability $p=.3$.  Each accident results in a loss which is exponentially distributed with mean 0.5.  Losses are mutually independent and independent of the number of accidents. The insurance company pays 70% of each loss due to an accident. Calculate the variance of the total unreinbursed loss the insurer experiences due to accidents resulting from these driving errors.

So far this is what I understand.
1), The driving errors are independent from each other will the same probability, so I know that this is a binomial distribution with $n=4$ and $p=.3$.  So if $N$ represents the number of errors made, $$E[N]=np=1.2, \space Var[N]=npq=.84$$
2), The loss is independent from the number of accidents, so simply each accident will result in a loss with pdf $$f(x)=\frac{1}{2} e^{\frac{-x}{2}}$$ where $X$ represents the amount of loss given an accident.  $$E[X]=\frac{1}{2}, \space Var[x]=\frac{1}{4}$$
3), Letting $Y$ the amount paid by the insurance company, I know that $$E[Y]=.7E[X], \space$Var[Y]=.49Var[X]$$ due to independence.
So, the thing that I want to deal with the amount the insurer will not get paid, and I want to say $X-Y$ so I am looking for $$Var[X-Y]=Var[X]+Var[Y]$$ which does not give me the right answer.  
So it would be most helpful if these questions were included in the explanation.
a), I am having trouble seeing how the number of accidents comes into play for this problem.
b), The book uses a formula that kind of looks like $$Var[E[X|S]]+E[Var[X|S]]=E[S]$$
and I would like to know more about this.  So can you lead me to where I can understand this?
 A: The formula your book is using is called the Law of Total Variance. 
Also, assuming you are interested in the total amount the insurance company is out, you want to look at the total amount they have to pay, which is 
$Y = \sum_{i=1}^N 0.7 X_i$ where $X_i$ is the loss amount of the $i^{th}$ accident. Notice that this is a mixture process. Note that for $k$ fixed $\sum_{i=1}^k X_i$ has a Gamma($k,\lambda$) distribution:
$$
  {\bf E}\left[ \sum_{i=1}^k X_i \right]
= k \lambda, \ \ 
  Var\left[ \sum_{i=1}^k X_i \right]
= k\lambda^2.
$$
 Then
$$
  {\bf E}[Y]
= 0.7{\bf E}\left[{\bf E}\left[\sum_{i=1}^N X_i \mid N \right] \right]
= 0.7 {\bf E}[N \lambda]
= 0.7\lambda np
$$
and 
\begin{align}
   Var[Y]
&= Var[{\bf E}[Y \mid N]] + {\bf E}[Var[Y \mid N]] \\
&= Var\left[0.7{\bf E}\left[\sum_{i=1}^N X_i \mid N \right] \right] + 
   {\bf E}\left[0.7^2 Var\left[\sum_{i=1}^N X_i \mid N \right]\right] \\
&= Var[0.7 N \lambda] + {\bf E}[0.7^2 N \lambda^2] \\
&= 0.7^2 \lambda^2 (Var[N] + {\bf E}[N ] ) \\
&= 0.7^2 \lambda^2 (npq + np ) \\
&= 0.7^2 \lambda^2 np(q + 1 ).
\end{align}
