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I tried this problem using Chebysev's inequality but I failed. One more thing I realized later on that Chebysev's inequality will not work here because after all its an inequality.

Any help will be appreciated. Thanks

Guide:

$$frac1{216}{(5+e^t)^3}=left(frac{5+e^t}{6}right)^3=left( left(1-frac{1}{6}right)e^{tcdot 0}+frac{1}6e^{tcdot 1}right)^3$$

Notice that $left( left(1-frac{1}{6}right)e^{tcdot 0}+frac{1}6e^{tcdot 1}right)$ is a Bernoulli distribution.

Hopefully you can interpret what happens when we cube it.

Let us assume that $f$ is the density of our random variable $X$. From
$$ M_X(t) = mathbb{E}[e^{tX}]=int_{-infty}^{+infty}e^{tx}f(x),dx $$
$$ mathbb{P}[X>1] = int_{-infty}^{+infty}mathbb{1}_{x>1}(x) f(x),dx $$
we have that the question is solved by expressing $mathbb{1}_{x>1}(x)$ in terms of exponentials. $mathbb{1}_{x>1}(x)$ is the inverse Laplace transform of $frac{1}{s e^s}$ and $frac{1}{t e^t}=int_{-infty}^{-1}e^{tx},dt$, bingo. This gives, in general,

$$ mathbb{P}[X>1] = 1 - mathcal{L}^{-1}left(frac{M_X(-t)}{t}right)(1).$$
In our case
$$ frac{M_X(-t)}{t} = frac{1}{6^3 t}(5+e^{-t})^3 $$
has an inverse Laplace transform which is piecewise-constant and reveals the obvious: $X$ takes the value $0$ with probability $frac{5^3}{6^3}$, the value $1$ with probability $frac{5^2cdot 3}{6^3}$, the value $2$ with probability $frac{5cdot 3}{6^3}$ and the value $3$ with probability $frac{1}{6^3}$, so the answer is $frac{16}{216}=frac{2}{27}$.