Q:

Conditioning using more than one distribution

Let N denote the number of automobile accidents on a given stretch of interstate
highway over a specific period. It is known that N has the geometric distribution with pmf 
$f_N(n)=x(1-x)^{n-1}$, $n=1,2,3,...$ 
where x is bounded between 0 and 1, $0< x< 1$ 
for the ith accident $Y_i=1$ if accident contains a fatality, 
$Y_i=0$ if no fatality,
for each i, $P(Y_i=1)=p$, each $Y_i$ is independent
Let T be the total number of accidents with at least one fatality. 
Then, $T=Y_1 + Y_2 +...+Y_N$
Find $E(T)$ and $Var(T)$,
I was thinking that since each $Y_i$ is bernoulli that is dependent on N then sum of each $E(Y_i)$ is $Nxp$ and the variance would be summing $p(1-p)$ which is $Npx(1-px)$
Maybe this is $E(T)=E(E(T|N))=\frac{p}{x}$ then $Var(T)=...$
Find $Corr(N,T)$=
$\frac{E(NT)-E(N)E(T)}{sd(T)sd(N)}$
Find $P(T=0)$
I seem to be missing something in my analysis that would lead me to get the rest of the problem

A:

You have $T\mid N\sim \mathcal{Bin}(N, p)$ and $N\sim\mathcal{Geo}_{1}(x)$  (which means you're guaranteed at least one accident in that period).
Thus we have: $\;\mathsf E\big(\mathsf E(T\mid N)\big) = \mathsf E(N\,p) = \frac p x\;$ as you stated. (A $\color{green}{\checkmark}$ for you.)
Similarly you need to apply: $\;\mathsf{Var}(T) = \mathsf E\big(\mathsf {Var}(T\mid N)\big)+\mathsf {Var}\big(\mathsf {E}(T\mid N)\big)$
(That's the Law of Iterated Variance.)

$\mathsf E(T\mid N) = Np\\ \mathsf{Var}(T\mid N) = Np(1-p) \\ \mathsf E(N) = \frac 1 x\\ \mathsf {Var}(N) = \frac {1-x}{x^2}$

$\begin{align}\mathsf P(T=t) & = \sum\limits_{n=t}^\infty \mathsf P(N=n)\mathsf P(T=t\mid N=n) \\[1ex] & = \sum\limits_{n=t}^\infty x(1-x)^{n-1} \dbinom{n}{t} p^t (1-p)^{n-t} \\[1ex] & = \dfrac{xp^t}{ (1-x) (1-p)^t } \sum\limits_{n=t}^\infty \dbinom{n}{t} (1-x-p+xp)^n \\[1ex] & = \dfrac{ xp^t (1-x)^{t-1} }{(x+p-px)^{t+1}}\end{align}$

