Let $X$ be a random variable that is uniformly chosen from the set of positive odd numbers less than $100$. The expectation $E[X]$ is:

The positive odd numbers less than 100 are $\{1,3,5,\dots ,99\}$.This is an arithmetic progression with $a=1,d=2$, and last term $l=99$.Number of terms $n$: $99=1+(n-1)2\Rightarrow 98=2(n-1)\Rightarrow n-1=49\Rightarrow n=50$.Since the selection is uniform, the expectation is the arithmetic mean of these numbers.
$$E[X]=\frac{\text{Sum of numbers}}{n}$$
$\text{Sum}=\frac{n}{2}(a+l)=\frac{50}{2}(1+99)=25\times 100=2500$.
$$E[X]=\frac{2500}{50}=50$$
Alternatively, for a symmetric uniform discrete distribution, the mean is the average of the first and last terms: $\frac{1+99}{2}=50$.