Show that $N$ is independent of $\{N_1 < N_2\}$ A random experiment has exactly three possible
outcomes, referred to as outcomes $1, 2,$ and $3,$ with probabilities $p_1 > 0, p_2 > 0,$ and $p_3 > 0,$ where $p_1 +p_2 +p_3 = 1.$ We consider a sequence of independent
trials, at each of which the specified random experiment is performed.) For
$i = 1, 2,$ let $N_i$ be the number of trials needed for outcome $i$ to occur, and
put $N := N_1 \wedge N_2.$
(a) Show that $N$ is independent of $\{N_1 < N_2\}.$
(b) Evaluate $E[N_1 \mid N_1 < N_2].$
(c) Roll a pair of dice until a total of $6$ or $7$ appears. Given that $6$ appears before $7,$ what is the (conditional) expected number of rolls?
The answer to $b)$ is $1/(p_1+p_2)$ and $c)$ is  $3.272727$ but I'm unsure of even where to start for $a)$ or the steps involved in arriving at the answers for $b)$ and $c)$ 

Here is a brute force way to prove (a). I'm sure there is a simpler way to do this, and hopefully someone will post it.
We have the joint pdf
\begin{align}
\mathbb{P}(N_1=k, N_2 = j) &= 1_{k<j} \;p_3^{k-1}p_1(p_1+p_3)^{j-k-1}p_2 + 1_{k>j}\; p_3^{k-1}p_2(p_2+p_3)^{j-k-1}p_1\\
&= p_1p_2p_3^{k-1}\left[ 1_{k<j} \;(p_1+p_3)^{j-k-1} + 1_{k>j}\;  (p_2+p_3)^{j-k-1} \right]
\end{align}
Using the above, one can show that 
$$ \mathbb{P}(N=n, N_1<N_2 ) = p_3^{n-1}p_1$$
$$ \mathbb{P}(N=n) = p_3^{n-1}(p_1+p_2)  $$
$$\mathbb{P}(N_1<N_2 ) = \frac{p_1}{p_1+p_2}$$
And therefore, 
$$\mathbb{P}(N=n, N_1<N_2 ) = \mathbb{P}(N=n)\cdot \mathbb{P}(N_1<N_2 )$$
proving independence of the events.
