The Problem: Suppose that $X$ and $Y$ are independent, $X\thicksim\text{Geom}(p),$ and $Y\thicksim\text{Geom}(q)$. Find $P(X<Y)$. [Hint: Break the event into smaller pieces.]
My Attempt: Recall that $P(X=k)=p(1-p)^{k-1}$ and $P(Y=n)=q(1-q)^{n-1}$ for all $k,n\in\mathbb N$, where we assume that $0<p,q<1$. Following the hint, we note that $\{X<Y\}=\bigcup_{1\leq k<n,n\in\mathbb N}\{X=k,Y=n\}$ is a union of pairwise disjoint events. To see the set equality, note that if $(\omega_1,\omega_2)\in\{X<Y\}$ then there is $k,n\in\mathbb N$ with $k<n$ such that $X(\omega_1)=k<Y(\omega_2)=n$, so it follows that $(\omega_1,\omega_2)\in\bigcup_{1\leq k<n,n\in\mathbb N}\{X=k,Y=n\}.$ Now let $(\omega_1,\omega_2)\in\bigcup_{1\leq k<n,n\in\mathbb N}\{X=k,Y=n\}$. Then there $X(\omega_1)=k$ and $Y(\omega_2)=n$ for some $k,n\in\mathbb N$ with $k<n$, so that $(\omega_1,\omega_2)\in\{X<Y\}.$ That the sets are pairwise disjoint is clear since for $i\ne j$ we have that $\bigcup_{1\leq k<i}\{X=k,Y=i\}$ and $\bigcup_{1\leq k<j}\{X=k,Y=j\}$ are disjoint because if $(\omega_1,\omega_2)\in\bigcup_{1\leq k<i}\{X=k,Y=i\}$ then $X(\omega_1)=k$ and $Y(\omega_2)=i\ne j$ so that $(\omega_1,\omega_2)\notin\bigcup_{1\leq k<j}\{X=k,Y=j\}$, and similarly for the other direction. By the countable additivity of the probability measure and the independence of the random variables $X$ and $Y$ it follows that \begin{equation*}\begin{split} P(X<Y)&=P\left(\bigcup_{1\leq k<n,n\in\mathbb N}\{X=k,Y=n\}\right)=\sum_{1\leq k<n,n\in\mathbb N}P(X=k,Y=n)\\ &=\sum_{n=1}^\infty\sum_{k=1}^{n-1}P(X=k)P(Y=n)=\sum_{n=1}^\infty\sum_{k=1}^{n-1}p(1-p)^{k-1}q(1-q)^{n-1}\\ &=pq\sum_{n=1}^\infty (1-q)^{n-1}\frac{1-(1-p)^{n-1}}{p}=q\sum_{n=1}^\infty(1-q)^{n-1}(1-(1-p)^{n-1})\\ &=1-q\sum_{n=1}^\infty[(1-q)(1-p)]^{n-1}=1-\frac{q}{1-(1-p)(1-q)}\\ &=\frac{1-(1-p)(1-q)-q}{1-(1-p)(1-q)}\\ &=\frac{p(1-q)}{p+q-pq}. \end{split}\end{equation*}
Do you agree with my attempt at a solution to this problem?
Any feedback is much appreciated. Thank you for your time.
This is an alternative route based on the hint.
Let $B_1,B_2,\cdots,C_1,C_2,\dots$ be independent random variables such that $B_i\sim\mathsf{Bernoulli}(p)$ and $C_i\sim\mathsf{Bernoulli}(q)$.
Then let $X:=\min\{n\mid B_n=1\}$ and $Y:=\min\{n\mid C_n=1\}$ so that $X$ and $Y$ are independent with $X\sim\mathsf{Geom}(p)$ and $Y\sim\mathsf{Geom}(q)$.
Then: $$\{X<Y\}=\{B_1=1,C_1=0\}\cup\{B_1=0=C_1,X<Y\}$$ so that:$$P(X<Y)=P(B_1=1,C_1=0)+P(B_1=0=C_1)P(X<Y\mid B_1=0=C_1)=$$$$p(1-q)+(1-p)(1-q)P(X<Y)$$and leading to:$$P(X<Y)=\frac{p(1-q)}{1-(1-p)(1-q)}$$
Used is that: $$P(X<Y\mid B_1=0=C_1)=P(X<Y)$$