Use Chebyshev’s inequality to find an upper bound for P(Y ≥ 14) Let $Y$ be a geometric random variable with parameter $p =\frac{1}{6}$
Use Chebychev’s inequality to find an upper bound for $\mathbb{P}(Y \geq 14)$.
I'm stumped on this question and the formula seems quite abstract. Any help?
$\mathbb{P}(\mid x-\mu\mid \geq k)\leq\frac{\sigma^2}{k^2}$
 A: Markov's inequality seems to be more applicable here. (Chebychev's ineqequality is a special case of this inequality.) Markov's inequality tells us that for any non-negative random variable $X$, and any $\alpha >0,\, q > 0$ one has
$$
P(X\ge \alpha) \le \frac{E(X^q)}{\alpha^q}.
$$
In particular,
$$
P(Y\ge14) \le \frac{E(Y^2)}{14^2}. \tag{1}
$$
How to find $E(Y^2)$? 
Note that 
$$
\frac{1-p}{p^2}=Var(Y) = E(Y^2)-(E(Y))^2, 
$$
So that 
$$
E(Y^2) = \frac{1-p}{p^2} + (E(Y))^2 = \frac{1-p}{p^2} + \frac{1}{p^2} = \frac{2-p}{p^2}.
$$
Plugging this into $(1)$ you get
$$
P(Y\ge14) \le \frac{E(Y^2)}{14^2} = \frac{2-p}{p^2}\frac{1}{14^2} = \frac{2-1/6}{(1/6)^2}\frac{1}{14^2} = \frac{33}{98}.
$$
A: Since $Y$ is a geometric random variable, it has mean $\frac{1-p}{p}=\frac {5/6}{1/6}=5$
and variance $\frac{1-p}{p^2}=\frac{5/6}{1/6^2}=30$.
Chebyshev's inequality says
$$\begin{split}\Pr(|Y-5|>8)&\le\frac{\operatorname{Var}(Y)}{64}\\
\Pr(Y>13)&\le\frac{30}{64}\end{split}$$
since $Y$ does not take on negative values. Since $Y$ is discrete, $\Pr(Y\ge14)\le30/64$.
