Problem source

Suppose that a solution $(X,Y,Z)$ of the equation
\[X+Y+Z=20,\]
with $X$, $Y$ and $Z$ non-negative integers, is chosen at random (each such solution being equally likely).
Are $X$ and $Y$ independent? Justify your answer.
Show that the probability that $X$ is divisible by $5$ is $5/21$.
What is the probability that $XYZ$ is divisible by 5?

Solution source

They are not independent:

\begin{align*}
&& \mathbb{P}(X = 20 \,\, \cap Y = 20) = 0 \\
&& \mathbb{P}(X = 20 )\mathbb{P}(Y = 20) \neq 0 \\
\end{align*}

\begin{align*}
X = 0: && 21 \text{ solutions} \\
X = 5: && 16 \text{ solutions} \\
X = 10: && 11 \text{ solutions} \\
X = 15: && 6 \text{ solutions} \\
X = 20: && 1 \text{ solutions} \\
5 \mid X: && 55 \text{ solutions} \\
\\
&& \binom{20+2}{2} = 11 \cdot 21 \text{ total solutions} \\
\Rightarrow && \mathbb{P}(5 \mid X) = \frac{55}{11 \cdot 21} = \frac{5}{21}
\end{align*}

\begin{align*}
\mathbb{P}(5 \mid XYZ) &= 3\cdot \mathbb{P}(5 \mid X) - 2\mathbb{P}(5 \mid X, Y, Z) \\
&= \frac{3 \cdot 55 - 2 \cdot \binom{4+2}{2}}{11 \cdot 21} = \frac{35}{77}
\end{align*}