What bound does Chebychev’s inequality give for $P(Y ≥ 3 \text{ or } Y ≤ 1)$? Suppose you flip four fair coins. Let Y be the number of heads obtained.
(a) What bound does Chebychev’s inequality give for $P(Y ≥ 3 \text{ or } Y ≤ 1)$?

$E(Y) = \sum_{y=1}^{4}yP(Y = y) = \sum_{y=1}^{4}y \frac{1}{2} = \frac{1}{2}(1+2+3+4) = 5 \neq 2$
$E(Y^2) = \sum_{y=1}^{4}y^2P(Y = y) = \frac{1}{2}(1^2 +2^2 + 3^2+4^2) = 15$
$Var(Y) = 15 - 25 = -10 \neq 1$
(a) $P(Y ≥ 3 \text{ or } Y ≤ 1) = P(|Y -E(X)| \geq ?)$
Where did I go wrong? 

Bas bas, $P(Y=y) \ne \frac 1 2$...$Y$ is binomially distributed with parameters $n=4, p = \frac 1 2$ soooo $$P(Y=y) = \binom{4}{y}(\frac12)^{4} $$ In fact,
$$E[Y] = np = 2, Var[Y] = np(1-p)=1$$
By Chebyshev's,
$$P(|Y-2| \ge k(1)=k) \le \frac{1}{k^2}$$
You have the right idea choosing $k=1$.
$$\therefore, P(|Y-2| \ge 1) \le 1$$
