The given MGF is that for a random variable following a Poisson distribution with parameter \(\lambda=3\).
This means \(\mathbb E(X)=\mathbb V(X)=\lambda\), and \(X\) has PMF
\(f_X(x)=\begin{cases}\dfrac{3^xe^{-3}}{x!}&\text{for }x\ge0\\0&\text{otherwise}\end{cases}\)
So, the desired probability is
\(\mathbb P\left(\lambda-\lambda^2
This is equivalent to
\(\displaystyle\sum_{x=0}^2\mathbb P(X=x)=\sum_{x=0}^2\frac{3^x}{x!e^3}=\frac{17}{2e^3}\approx0.4232\)
