I tried to resolve this issue, but I don’t know if it’s correct, could anyone appreciate the resolution and help me?
The emission of a radioactive source is such that the number of particles emitted in a time period, given in seconds, $X$, has Poisson distribution with $E[X^2] = 12.$ observed the issue during that period, what is the probability to be issued $4$ or more particles?
The key is that is that the mean and variance of a Poisson distribution are both equal to λ. In particular $\lambda=\text{Var}(X)=E[X^2]-(EX)^2=12-\lambda^2\iff\lambda^2+\lambda-12=0 \quad (\lambda>0)$
We have $\lambda = 3$ \begin{align*} P(X=x)=\frac{e^{-\lambda}.\lambda^{x}}{x!} \Rightarrow P(X\geq 4)&= 1-P(X<4)\\ &=1-\left[\frac{e^{-3}\cdot 3^{0}}{0!}+\frac{e^{-3}\cdot 3^{1}}{1!}+\frac{e^{-3}\cdot 3^{2}}{2!}+\frac{e^{-3}\cdot 3^{3}}{3!}\right]\\ &= 1-[e^{-3}\cdot13]\\ &=1-0,6472\\ &=0,3528. \end{align*}
Thanks!
Since $Var(X)=\lambda=E[X^2]-(E[X])^2=E[X^2]-\lambda^2$ you get the quadratic equation $\lambda^2+\lambda-12=0$ hence $\lambda=-4$ or $\lambda=3$, therefore $\lambda=3$. Then $$ P(X\ge 4)=1-P(X=0)-P(X=1)-P(X=2)-P(X=3), $$ which you easily compute with the pmf.