I need only a hint please : Is there a way to calculate the probability if we have the moment generating function as in the following question:
IF the moment generating function is given by : $$M(t)= e^{4.6(e^t-1)}$$ Find $P(3<X<6)$
What I have tried is :
Suppose $S = \{s_1,s_2,\dots\}$ then
$$e^{4.6(e^t-1)}=M(t)=\sum_{s_i\in S}e^{tx}f(s_i)= e^{ts_1}f(s_1)+e^{ts_2}f(s_2)+\cdots $$
$$e^{4.6e^t}=e^{4.6}\left( e^{ts_1}f(s_1)+e^{ts_2}f(s_2)+\cdots \right)$$
If we write
$$e^{4.6e^t}=1+(4.6e^t)+\frac{(4.6e^t)^2}{2}+\cdots$$
Then
$$1+(4.6e^t)+\frac{(4.6e^t)^2}{2}+\cdots = e^{(ts_1+4.6)}f(s_1)+e^{(ts_2+4.6)}f(s_2)+\cdots$$
So
$$1+4.6e^t+10.58e^{2t}+\cdots = e^{(ts_1+4.6)}f(s_1)+e^{(ts_2+4.6)}f(s_2)+\cdots$$
The requested probability $P(3 < X < 6)$ has nothing to do with how the moment-generating function (henceforth MGF) is defined or any of its properties. These kinds of exercises simply intend to ask you to recognize a common distribution from the given MGF (look-up tables are available in most textbooks).
Having said that, in this particular case, your approach happens to work as a practical solution.
Denote the given MGF as $M(t) = e^{\lambda (e^t - 1)}$ with $\lambda = 4.6$ so that your last equation becomes $$ 1 + \lambda e^t + \frac{ \lambda^2 e^{2t} }{ 2! } + \frac{ \lambda^3 e^{3t} }{ 3! } + \cdots = e^{(ts_1 + \lambda)}f(s_1) + e^{(ts_2 + \lambda )}f(s_2) + e^{(ts_3 + \lambda )}f(s_3) + \cdots $$ From which one can make a first round of guess as the following.
Since the terms $f(s_k)$ of probability mass function (henceforth PMF) don't carry any $t$:
With this guess, the support is $k = 0,1,2,3,\ldots$ the non-negative integers. The desired probability thus becomes $P(3 < X < 6) = P(4 \leq X \leq 5)$.
Note that $s_1 = 0$ together with $1 = e^{(ts_1 + \lambda)}f(s_1)$ gives $f(s_1) = e^{-\lambda}$ at $k = 1$. As for other $s_k~$, we have the ratio between adjacent terms \begin{align} &&\frac{ e^{(t s_k + \lambda )}f(s_k) }{ e^{(ts_{k-1} + \lambda )} f(s_{k-1}) } & = \frac{ \lambda^{k-1} e^{(k -1)t} / (k-1)!}{ \lambda^{k-2} e^{(k-2)t} / (k-2)! } \\ \implies&& \frac{ f(s_k) }{ f(s_{k-1}) } &= \frac{\lambda e^{t(s_k - s_{k-1})} }{k-1} = \frac{\lambda e^t }{k-1} \quad \because s_k - s_{k-1} = 1 ~~ \forall\,k\geq 1 \end{align} Therefore, from $f(s_1)$ we have $f(s_2)$, then from $f(s_2)$ comes $f(s_3)$, and so on. \begin{align} f(s_1) = e^{-\lambda} \implies && f(s_2) &= f(s_1)\frac{\lambda e^{t(s_k - s_{k-1})} }{k-1}\Bigg|_{k = 2} = e^{-\lambda} \\ \implies && f(s_3) &= e^{-\lambda} \frac{ e^{2t} }2 \\ \implies && f(s_4) &= e^{-\lambda} \frac{ e^{3t} }{ 3!} \\ \vdots&&& \end{align} This is the Poisson distribution, and our work is done because the MGF is unique. The first round of naive guess that $s_k = k-1$ turns out to be right on, and there's no need for a second round of improvement or anything.
The requested probability $\displaystyle P(4 \leq X \leq 5) = e^{-\lambda} \frac{\lambda^4}{4!} \bigl( 1 + \frac{ \lambda }5\bigr)$ with $\lambda = 4.6$.