What bound does Chebychev’s inequality give for $P(Y ≥ 3 \text{ or } Y ≤ 1)$?

Question

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

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$E(Y) = \sum_{y=1}^{4}yP(Y = y) = \sum_{y=1}^{4}y \frac{1}{2} = \frac{1}{2}(1+2+3+4) = 5 \neq 2$

$E(Y^2) = \sum_{y=1}^{4}y^2P(Y = y) = \frac{1}{2}(1^2 +2^2 + 3^2+4^2) = 15$

$Var(Y) = 15 - 25 = -10 \neq 1$

(a) $P(Y ≥ 3 \text{ or } Y ≤ 1) = P(|Y -E(X)| \geq ?)$

Where did I go wrong?

Answer 1

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 $k=1$.

$$\therefore, P(|Y-2| \ge 1) \le 1$$