Average number of inspections
Well, let's see. Please point out if I understood the problem model incorrectly.

Suppose we have $n = 5$ elements. Suppose it's sorted in the way $s = [1, 3, 4, 0, 2]$. Then using linear search we are going to with probability $(1/2)^3$ have 2 steps in our search and with probability $(1/2)^4$ have 4 steps in our search.

So, what's the average number of steps we need to find an element giving arbitrary sorting $s$? Let's say $X$ - number of steps needed in our search. Probability to get $X=i$ steps is equivalent to request a search for an element $k \in \{0,1,2,..,n-1\}$ giving we have it at the $i$th position. That is $$P(X = i) = \sum_{k=0}^{n-1} P(search\ for\ k\ |\ the\ element\ k\ is\ at \ ith\ position)P(the\ element\ k\ is\ at\ ith\ position)$$ Then we notice that $P(the\ element\ k\ is\ at \ ith\ position) = \frac{(n-1)!}{n!} = \frac{1}{n}$ because we have an element at a fixed position and $(n-1)!$ possible permutations for other elements out of the total $n!$ possible permutations of our $n$ elements.
Then we notice that $\sum_{k=0}^{n-1} P(search\ for\ k\ |\ the\ element\ k\ is\ at \ ith\ position) = 1$. It means that $P(X=i) = \frac{1}{n}$. So, the expectation is just $\sum_{k=1}^{n} \frac{k}{n} = \frac{n+1}{2}$.

Of course, the second part is much easier to derive, it's just $$E(X|sorted\ in\ decreasing\ probability) = \sum_{k=1}^{n-1} \frac{k}{2^{k}} + \frac{n}{2^{n-1}}$$ (just notice that the elements are going to be from $1$ to $n-1$ and then the element $0$, however $0$ could be the pre-last element because it have same probability as $n-1$). So, we just calculating the expectation of the probability distribution, though this time not for the joint probability but conditional.