Coffee machine breaking with probability 0,01 - check if two random variables are independent and find bound for $P(X > n)$ There is a coffee machine that makes a cup of coffee with probability $0.9$, makes a cup of hot water with probability $0.09$ and breaks forever with probability $0.01$.

Let $X$ be the number of cups of coffee the machine makes before it breaks and $Y$ be the number of cups of hot water before it breaks.

Find the least $a &gt; 0$ such that $P(X &gt; n) = O(a^n)$.
Are random variables $X$ and $Y$ independent?

I tried something like this:
$$P(X &gt; n) = \sum_{i = n+1}^{+\infty} P(X = i)$$
$$P(X = i) = \sum_{k=i+1}^{+\infty} P(X=i|Z=k) \cdot P(Z=k)= \sum_{k=i+1}^{+\infty} 0.99^{k-1} \cdot 0.01 \cdot {k - 1 \choose i} \cdot 0.9^i \cdot 0.09^{k-1-i}= 0.01 \cdot 0.9^i \sum_{k=i+1}^{+\infty} 0.99^{k-1} \cdot 0.09^{k-1-i} {k - 1 \choose i} $$

I can't go any further with this and I think it is important to count this to be able to check if $X$ and $Y$ are independent.
The joint distribution of X,Y

$$f_{XY}(X=n,Y=m) = {(n+m)\choose(n)} (0.9)^{n}(0.09)^{m}(0.01)$$

$$f_X(X=n) = \sum_{m=0}^{\infty} {(n+m)\choose(n)}(.9)^{n}(.09)^{m}(0.01)$$

$$f_X(X=n) = \frac{1}{91} \left(\frac{90}{91}\right)^n\tag 1$$

$$P(X\gt n) = \sum_{i=n+1}^{\infty} \frac{1}{91}\left(\frac{90}{91}\right)^i$$

$$P(X\gt n) = \frac{90}{91}.\left(\frac{90}{91}\right)^n$$

Thus the least $$a = \frac{90}{91}$$

and X and Y are not independent from the joint distribution as $$f_{XY}(X=x,Y=y) \ne f_X(X=x)f_Y(Y=y)$$

Remark:
I got the result of (1) from wolfram
The joint distribution of X,Y

$$f_{XY}(X=n,Y=m) = {(n+m)\choose(n)} (0.9)^{n}(0.09)^{m}(0.01)$$

$$f_X(X=n) = \sum_{m=0}^{\infty} {(n+m)\choose(n)}(.9)^{n}(.09)^{m}(0.01)$$

$$f_X(X=n) = \frac{1}{91} \left(\frac{90}{91}\right)^n\tag 1$$

$$P(X\gt n) = \sum_{i=n+1}^{\infty} \frac{1}{91}\left(\frac{90}{91}\right)^i$$

$$P(X\gt n) = \frac{90}{91}.\left(\frac{90}{91}\right)^n$$

Thus the least $$a = \frac{90}{91}$$

and X and Y are not independent from the joint distribution as $$f_{XY}(X=x,Y=y) \ne f_X(X=x)f_Y(Y=y)$$

Remark:
I got the result of (1) from wolfram