Compound Experiment with $k+1$ Possible Results Some experiment has $k+1$ possible outcomes, numbered $\{0,,.k\}$, and for $ i \in \{0,,.k\}$ the probability the result is $i$ is $p_i$ while $\sum_{i=0}^{k}p_i = 1$.
Repeat this experiment infinite times, with the assumption each trial's result is independent of the rest.
Let $X$ be a random variable whose result is the number of trials needed for the outcome to $not$ be zero.
Let $Y$ be a random variable whose result is the outcome of the first $non$ zero trial; $Y(0003) = 3 = Y(03)$.
I need to calculate $P_X(n)$ and $P_Y(i)$ for $n\geq 1$ and $1\leq i \leq k$.
To get $P_X(n)$: the probability that the first $n$ trials yield zero is $p_0^n$ hence $P_X(n) = 1-p_0^n$.
I am not sure how to proceed with finding $P_Y(i)$ without proving $X,Y$ are independent first, and not sure how to prove they are independent without finding $P_Y(i)$. A direction please?
 A: As lulu pointed out in the comments, your calculation of $P_X$ is wrong, what you are calculating is the probability that at least one of the first $n$ results is not $0$. You need the probability that all $n-1$ results are $0$ or $p_0^{n-1}$ times the probability of getting a non zero result on the $n$th trial or $(1-p_0)$.

For $Y$ you know that $Y$ cannot be $0$, and because all trials are independent, you should just compute the conditional probability of every outcome from $1$ to $k$ given that it is not outcome $0$, or $$\mathbb{P}(O_k | k\ne 0)=\frac {\mathbb{P}(O_k \cap k \ne 0)}{\mathbb{P}(k\ne 0)}=\frac {p_k}{1-p_0}$$
A: The probability distribution of the outcome of the first nonzero trial is simply the re-weighted distribution of the nonzero outcomes; i.e., if $\Pr[\text{Outcome} = i] = p_i$, then $$\Pr[Y = i] = \frac{p_i}{1-p_0}.$$
A: As pointed out in @lulu's comment,
$$
P(X=n)=p_0^{n-1}(1-p_0).
$$
The distribution of $Y$ can be given by conditionning on the fact that the n-th outcome is non-zero. Call $Z_k$ the outcome of the $k$-th experiment:
$$P(Y=i)=P(Z_n=i|Z_1=Z_2= ...\ =Z_{n-1}=0, \ Z_n\neq0)
=P(Z_n=i| Z_n\neq0)=\frac{P(Z_n=i)}{P(Z_n\neq0}=\frac{p_i}{1-p_0}
$$
