Let $X$ be a nonnegative random variable with $\mathbb{E}(X) = 1$ and $\Pr(X \geq 2) \geq 1/3$. Prove that $\Pr(X \geq 10) \leq 1/24$.
The hint suggests generalizing the proof of Markov's inequality. I tried but didn't manage to prove the above. Would appreciate your help.
Edit: I add the proofs of Markov's inequality I learned:
Proof I: Assume that $X$ is continuous, and let $t > 0$. Then
$$\mathbb{E}(X) = \int_0^\infty x f_X(x) \,\mathrm{d}x = \int_0^t x f_X(x) \,\mathrm{d}x + \int_t^\infty x f_{X}(x) \,\mathrm{d}x \geq t\Pr(X \geq t)$$
since $\int_0^t x f_X(x) \,\mathrm{d}x \geq 0$ and $\int_t^\infty x f_X(x) \geq \int_t^\infty t f_X(x) \,\mathrm{d}x = t\int_t^\infty f_X(x) \,\mathrm{d}x = t\Pr(X \geq t)$. Hence $$\Pr(X \geq t) \leq \frac{\mathbb{E}(X)}{t}.$$
A similar proof works for a discrete $X$.
Proof II: Let $t > 0$. Let $I_{X \geq t}$ be the indicator of the event $\{X \geq t\}$ (for any event $E$, $I_E(\omega) = 1$ for $\omega \in E$ and $I_E(\omega) = 0$ for $\omega \notin E$). Note that $tI_{X \geq t} \leq X$. For every $\omega \in \Omega$, if $X(\omega) \geq t$ then $tI_{X \geq t}(\omega) = t \cdot 1 = t \leq X(\omega)$, and else if $X(\omega) < t$ then $tI_{X \geq t}(\omega) = t \cdot 0 = 0 \leq X(\omega)$ since $X$ is nonnegative. Since the expectation is a monotonically increasing function,
$$t\Pr(X \geq t) = t\mathbb{E}(I_{X \geq t}) = \mathbb{E}(tI_{X \geq t}) \leq \mathbb{E}(X)$$
and hence $\Pr(X \geq t) \leq \frac{\mathbb{E}(X)}{t}$.
Best Answer
Note that $$ \mathbb E[X] \ge \mathbb E[X1_{\{X \in [2,10)\}}] + \mathbb E[X1_{\{X \ge 10\}}] \ge 2\mathbb P(X \in [2,10)) + 10\mathbb P(X \ge 10) $$ $$= 8\mathbb P(X \ge 10) + 2 \mathbb P(X \ge 2)$$ where at the second inequality we used pointwise bounds on respective sets, and last equality is due to $\mathbb P(X \in [2,10)) = \mathbb P(X \ge 2) - \mathbb P(X \ge 10)$. Since by assumptions $\mathbb E[X]=1$ and $\mathbb P(X \ge 2) \ge \frac{1}{3}$, after rearraning we get $$ \mathbb P(X \ge 10) \le \frac{\mathbb E[X]-2\mathbb P(X \ge 2)}{8} \le \frac{1-\frac{2}{3}}{8} = \frac{\frac{1}{3}}{8} = \frac{1}{24} $$