Markov inequality with bounded random variable A test is graded on a scale of 0 to 80 points. Because the grading is completely random, the
grade of a student can be represented by an integer-valued random variable X (0 ≤ X ≤ 80)
and E[X] = 60
a) What is the maximum possible probability that X = 80?
b) Suppose that we change the bounds to 20 ≤ X ≤ 80, but E[X] is still 60. Now what is the
maximum possible probability that X = 80?
Part a is just basic application of Markov inequality
part b however is confusing. What I did is that
E[x] = $\sum_{x=20}^{80}xP(x) \geq 80P(x=80) +20P(x=20)$
as I assume that P(x=80) would be maximize with fixed E[x]=60 if the score is allocated to 80 (highest score) and 20 (lowest score only), which mean I'd have :
$\left\{\begin{matrix}
80P(x=80) +20P(x=20) = 60\\ 
P(x=80)+P(x=20) =1
\end{matrix}\right.$
Is this correct?
 A: You have to prove what you wrote (which is correct and excellent thinking to boot, especially if you have a multiple choice question and don't need to justify working).
If you do need to justify working, then what you need to do is use the "best possible" Markov inequality available here.
That is, because $20\leq X \leq 80$, we know that $0 \leq X-20 \leq 60$. Now, $X-20$ is as tight as it gets : so we should be applying Markov to this, to get :
$$
\mathbb P[X=80] = \mathbb P[X \geq 80] = \mathbb P[X-20 \geq 60] \leq \frac{\mathbb E[X-20]}{60} = \frac{\mathbb E[X]-20}{60} = \frac 23 
$$
Now, can this be attained? Using your logic : yes, by the random variable $X$ with $$
X = \begin{cases}
20 & \text{ with probability $\frac 13$} \\
80 & \text{ with probability $\frac 23$}
\end{cases}
$$
which satisfies the conditions and attains the bound.

To justify the idea that the maximum of $\mathbb P[X=80]$ is likely to be attained by a random variable that only takes values $20$ and $80$ (call it $X$ as above), one needs to use abstract arguments.
Indeed, every probability distribution on $\{20,21,\ldots,80\}$ can be represented by a $61$-tuple of positive real numbers $X \sim (p_{20},p_{21},...,p_{80})$ adding up to $1$, where $p_i = \mathbb P[X=i]$. If we now ask for $\mathbb E[X] = 60$ then this forces $\sum_{x=20}^{80} xp_x = 60$. Thus, the set of all random variables in question is the following set in $\mathbb R^{61}$ :
$$
S = \left\{P = (p_{20},...,p_{80}) : p_i \geq 0 , \sum_{i=20}^{80} p_i = 1,\sum_{x=20}^{80} xp_x = 60 \right\}
$$
This is actually a convex set : one can verify that if $P,Q \in S$ then $(1-\lambda)P+\lambda Q \in S$ for all $\lambda \in [0,1]$. Then , one sees that the function $P \to p_{80}$ on $S$ is quite obviously a convex function (in fact, a linear one). It follows from Bauer's theorem, that the maximum of $p_{80}$ on $S$ is attained at an extreme point i.e. a point that isn't a strict linear combination of two other points in $S$.
It is easy to see that any point that has three of the $p_i$ non-zero cannot be an extreme point, since if $p_i,p_j,p_k$ are non-zero then we can use a strict convex combination of two points , one with $p_i,p_j$ non-zero and $p_k=0$ and the other with $p_i,p_k$ non-zero and $p_j=0$(and the rest of the coordinates equals to that of $P$), to see that such a point cannot be extreme.
Thus, the set of extreme points is contained in the set of all points with at most two coordinates non-zero. If we're maximizing $p_{80}$, then it's obvious by a simple comparison argument with other two-coordinate distributions that the maximizing distribution must have only $p_{20},p_{80}$ non-zero, and then you can find this distribution by solving the simultaneous linear equations given in $S$.
That's an abstract proof of why this "extreme" behaviour maximizes what we need. It can be adjusted to provide for other linear functionals as well, which is why it's really useful.
