Calculating probability that $P(X\geq Y+1)$, where $X$ and $Y$ are independent random variables 
Let $X$ and $Y$ be independent random variables with probability functions given by
$$
p_{X}(x)=\left\{\begin{array}{cc}
\frac{2^{x}}{15}, & x \in\{0,1,2,3\} \\
0, & \text { otherwise }
\end{array}\right. \text { e } p_{Y}(y)=\left\{\begin{array}{cc}
\frac{3^{y}}{40}, & y \in\{0,1,2,3\} \\
0, & \text { otherwise }
\end{array}\right.
$$
Determine $P(X\geq Y+1)$.

By my calculations, we have
$$ \begin{aligned}
P(X \geq Y + 1) &= \sum_{i=0}^{2} P(X \geq i+1)P(Y=i) \\
&=P(X \geq 1)P(Y=0)+P(X \geq 2)P(Y =1) + P(X \geq 3)P(Y = 2) \\
&=(P(X=1) + P(X=2)+P(X=3)) \cdot \frac{1}{40} + (P(X=2)+P(X=3))\cdot \frac{3}{40}+P(X=3)\cdot \frac{9}{40} \\
&=\left(\frac{2}{15}+\frac{4}{15}+\frac{8}{15}\right)\cdot \frac{1}{40}+\frac{12}{15}\cdot \frac{3}{40} + \frac{8}{15} \cdot \frac{9}{40} \\
&=\frac{61}{300}
\end{aligned}
$$
Can someone check my work? I realise it's a pretty simple problem but I'd just like to know that I didn't make any mistakes.

We have $$\{X\geqslant Y+1\} = \bigcup_{i=1}^3\bigcup_{j=0}^{i-1} \{X=i\}\cap\{Y=j\}, $$
and the union is of disjoint events. By independence, then
\begin{align}
\mathbb P(X\geqslant Y+1) &= \mathbb P\left(\bigcup_{i=1}^3\bigcup_{j=0}^{i-1} \{X=i\}\cap\{Y=j\}\right)\\
&= \sum_{i=1}^3\sum_{j=0}^{i-1}\mathbb P(X=i)\mathbb P(Y=j)\\
&= \frac1{15\cdot40}\sum_{i=1}^3\sum_{j=0}^{i-1} 2^i3^j\\
&= \frac{61}{300},
\end{align}
which agrees with your result. Indeed, a simple problem, just somewhat of a tedious computation.
