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 $c)$ 

HINT
$N_i$ is the trial at which $i$ first appears.  So $N = \min(N_1, N_2)$ is the trial at which $1$ or $2$ first appears.  In other words, before the $N$th trial, only $3$ has been appearing.  In other words, $N>n$ is the event that the first $n$ trials are all $3$s.
(a) I would directly calculate $P(N>n \mid N_1 < N_2)$ and $P(N > n \mid N_2 < N_1)$ and show that they are equal.  Do you see why this implies the required independence?
(b) Consider $P(N_1 > n \mid N_1 < N_2)$.  Can you see why this is exponential?
(c) is just an application of (b)
