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 $\Pr(X \geq t) \leq \frac{\mathbb{E}(X)}{t}$.
