# 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?

$\mathbb{P}(\mid x-\mu\mid \geq k)\leq\frac{\sigma^2}{k^2}$

• I found the E[X] which is 6 since E[X] = 1/6 = p. I also found Var(X) which is 32/3 since Var(X) = ((14-6)^2)*(1/6). But I don't understand how to use the formula – Chance Gordon Oct 12 '17 at 17:36
• Do you know Chebyshev's inequality by any chance? – Dionel Jaime Oct 12 '17 at 17:36

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 $$P(Y\ge14) \le \frac{E(Y^2)}{14^2} = \frac{2-p}{p^2}\frac{1}{14^2} = \frac{2-1/6}{(1/6)^2}\frac{1}{14^2} = \frac{33}{98}.$$