The consecutive integers $1, 2,\ldots, n$ are inscribed on $n$ balls in an urn. Let $D_{r}$ be the event that the number on a ball drawn at random is divisible by $r$.
(a) What are $\textbf{P}(D_{3})$, $\textbf{P}(D_{4})$, $\textbf{P}(D_{3}\cup D_{4})$ and $\textbf{P}(D_{3}\cap D_{4})$?
(b) Find the limits of these probabilities as $n \rightarrow\infty$.
(c) What would your answers be if the $n$ consecutive numbers began at a number $a \neq 1$?
MY ATTEMPT
The book gives the following answers to each case:
(a) It uses the floor function in order to solve this problem. Precisely, we have: \begin{align*} \textbf{P}(D_{3}) = \frac{\displaystyle\left\lfloor\frac{n}{3} \right\rfloor}{n},\quad \textbf{P}(D_{4}) = \frac{\displaystyle\left\lfloor\frac{n}{4} \right\rfloor}{n},\quad \textbf{P}(D_{3}\cap D_{4}) = \frac{\displaystyle\left\lfloor\frac{n}{12} \right\rfloor}{n} \end{align*}
From whence we obtain that
\begin{align*} \textbf{P}(D_{3}\cup D_{4}) = \frac{\displaystyle\left\lfloor\frac{n}{3} \right\rfloor + \left\lfloor\frac{n}{4}\right\rfloor - \left\lfloor\frac{n}{12} \right\rfloor}{n} \end{align*}
(b) I do not know how to handle limits with floor function. But it says that it should be \begin{align*} \lim_{n\rightarrow+\infty}\textbf{P}(D_{3}) = \frac{1}{3}, \quad \lim_{n\rightarrow+\infty}\textbf{P}(D_{4}) = \frac{1}{4}, \quad\lim_{n\rightarrow+\infty}\textbf{P}(D_{3}\cap D_{4}) = \frac{1}{12} \end{align*}
(c) It is the same answer as given at (a).
Is there another approach to this problem? Thanks in advance.
Your answer for part $(a)$ is right. But you don't need to use the floor function to evaluate those limits. Clearly if $n$ is divisible by $3$, $P(D_{3})=1/3$. If $n=3k+1$ for some $k$, then $$P(D_{3})=\frac{k}{n}=\frac{k}{3k+1}$$ and if $n=3k+2$ for some $k$, then $$P(D_{3})=\frac{k}{n}=\frac{k}{3k+2}$$ Notice that the limit of each of these as $n\to\infty$ is clearly $1/3$. Assuming we pick $n$ uniformly at random from the positive integers, then since $1/3$ of the positive integers are each of the above forms, the probability is $$\frac{1}{3}(\frac{1}{3}+\frac{1}{3}+\frac{1}{3})=\frac{1}{3}$$ as you suspected. The argument works the same way for $D_{4}$.
And you are right about the unions and intersections. To make those arguments precise, you can again modify my above argument.