Given E[X] and Var[X], what is P(X=0)? Let $E[X]=\theta(n^{1.5})$ , $Var[X]=\theta(n^{2.5})$ , $X \ge 0$
What can you say about $P(X=0)$ using Chebyshev's inequality? A: $P(X=0)=O(1/n^2)$ 
B: $P(X=0)=O(1/n^{1.5})$ 
C: $P(X=0)=O(1/n)$ 
D: $P(X=0)=O(1/n^{0.5})$ 
The original question required to find $E[X]$ and $Var[X]$, and i found them, and they're correct.
To solve this part of the question, i tried using the following Chebyshev's inequality:
$${\bf P(X \ge b)=\frac{1}{2}P(|X-E[X]| \ge \alpha \sigma) \le \frac{1}{2}\cdot\frac{1}{\alpha^2}} \space\space,\space\space\alpha\sigma =b-E[X]\space\space,\space\space\sigma=\sqrt{Var}$$
So i said:
$P(X=0)=P(X<1)=1-P(X \ge 1)$
And now i can apply Chebyshev's inequality:
$P(X \ge 1)=\frac{1}{2}P(|X-n^{1.5}| \ge \alpha \sigma) \le \frac{1}{2}\cdot\frac{1}{\alpha^2}$
$\alpha\sigma=1-n^{1.5}$
$\sigma=\sqrt{Var}=n^{1.25}$
$\alpha=\frac{1-n^{1.5}}{n^{1.25}}$
$\frac{1}{2}\cdot\frac{1}{\alpha^2}=\frac{n^{2.5}}{2(1-n^{1.5})^2}=O(n^{0.5})$
Hence:
$$P(X=0)=1-P(X \ge 1) \ge 1-\frac{1}{n^{0.5}}$$
The correct answer should be D. What was wrong in my solution?
Thanks
A: 
$P(X \ge 1)=\frac{1}{2}P(|X-n^{1.5}| \ge \alpha \sigma)$ ... (where) ... $\alpha\sigma=1-n^{1.5}$

No idea why you think this holds. Note that with your choice of $\alpha\sigma$, $|X(\omega)-n^{1.5}| \geqslant \alpha \sigma$ is true for every $n\geqslant1$ and every $\omega$.
On the other hand, $[X=0]\subseteq[|X-E[X]|\geqslant E[X]]$ hence Chebychev inequality applied to the event on the RHS yields $P[X=0]\leqslant\mathrm{Var}[X]/E[X]^2$. By hypothesis, $\mathrm{Var}[X]\leqslant an^{5/2}$ and $E[X]\geqslant bn^{3/2}$ hence $P[X=0]\leqslant c/n^{1/2}$ with $c=a/b^2$. QED.
A: The probability mass $p = P(X=0)$ has a moment of inertia (about the center of mass at $\Theta(n^{1.5})$) that has value $p[\Theta(n^{1.5})]^2 = p\Theta(n^3)$. But since the variance (which is the
total moment of inertia of all the probability masses about the center of mass)
is $\Theta(n^{2.5})$, it must be that $p$ is $O(n^{-0.5})$, since any
larger mass will give a variance that exceeds $\Theta(n^{2.5})$.
