Urn and probabiblity 
Two urns contain n balls each, numbered from 1 to n. We pick a ball from the first one and then a ball from the second. What is the probability that the number of the second ball is
a) smaller
b) equal to the number of the first ball?

My humble attempt:
b) $\frac{1}{n}$
a) Suppose the number picked is $0< k\leq n$
So we have $k-1$ numbers $< k$ and $n-k$ numbers $> k$.
Isn't the probability $\frac{n-k}{n}$  ?
 A: You are right for part b. For part a, note that the probability that the two balls are the same number is $1/n$, so the probability that they are a different number is $(n-1)/n$. Now both balls are equally likely to be the smaller one of the two, so the probability that the second ball is smaller than the first equals $$\frac{n-1}{2n}.$$
If you do not see this directly, you can indeed condition the probabilities on the outcome of the first draw. We than have \begin{align*}P(\text{ball 2 is smaller})&=\sum_{k=1}^nP(\text{ball 2 is smaller than ball 1}|\text{ball 1 equals }k)P(\text{ball 1 equals }k)\\&=\sum_{k=1}^nP(\text{ball 1 is }k)\cdot P(\text{bal }2\text{ is smaller than }k)\\&=\sum_{k=1}^n\frac1n\cdot\frac{n-k}{n}\\&=\sum_{k=1}^n\frac1n-\sum_{k=1}^n\frac{k}{n^2}\\&=1-\frac{1}{n^2}\frac{n(n+1)}{2}\\&=\frac{n-1}{2n},\end{align*} as found before.
A: A symmetry argument is fine. Though in general it may not always be possible. Here is a more detailed approach.
The number of ordered pairs $(i,j)$ with $i<j$ can be counted as follows:
$$\begin{align} i=1&: j\in \{2,...,n\} &... &\ \ \ \ \ n-1 \text{ pairs }\\
 i=2&: j\in \{3,...,n\} &... &\ \ \ \ \ n-2 \text{ pairs }\\
 i=3&: j\in \{4,...,n\} &... &\ \ \ \ \ n-3 \text{ pairs }\\
\vdots\\
 i=n-1&: j\in \{n\} &...& \ \ \ \ \ 1  \text{ pair }
\end{align}$$
Adding these up, we find the number of simple and equally-likely events satisfying your compound event:
$$\sum_{i=1}^{n-1} (n-i) = \dfrac{n(n-1)}{2}$$
The total number of ordered pairs is $n^2$, thus the probability is
$$\dfrac{n(n-1)}{2}\div n^2 = \dfrac{n-1}{2n}$$
By a similar argument, there are exactly $n$ pairs of the form $(i,i)$. Therefore the probability that the are equal is
$$\dfrac{n}{n^2} = \dfrac{1}{n}$$
