2 Poisson distributions questions An emission from a radioactive source is such that the number of emitted particles in a 10 seconds interval is well modeled by a random variable $X$, following Poisson's Law is such that $E(X^2)=6$. Calculate: 1. The probability of the emission of 2 particles in a 10 seconds period 2. The probability of the emission of at least 3 particles in a 10 seconds period My doubt is the $E(X^2)=6$, I don't understand it, I already did the first one with a result of $4.46\%$. I made like this: $P(K)=\frac{e^{-\lambda}.\lambda^{k}}{K!}=$ $P(2)= \frac{e^{-6}.6^{2}}{2!}$ $\frac{0,002478.36}{2}=0.04464=4,46$ Am i missing something?

The key is that is that the mean and variance of a Poisson distribution are both equal to $\lambda$. In particular $$ \lambda=\text{Var}(X)=EX^2-(EX)^2=6-\lambda^2\iff\lambda^2+\lambda-6=0 \quad (\lambda>0) $$ Solve for $\lambda>0$ (hence you will reject one root) and note that in question 1 we seek $$ P(X=2)=\frac{\lambda^2e^{-\lambda}}{2} $$ while in question 2 we seek $$ P(X\ge3)=1-P(X=0)-P(X=1)-P(X=2) $$