Let $X \sim \text{Poisson}(\lambda)$. If $P(X = 1) = 0.1$ and $P(X = 2) = 0.2$, what is the value of $P(X = 3)?$
I've tried to expand the given equalities, finding $\lambda = 4$, which is clearly wrong if you try to evaluate $P(X = 1)$ or $P(X = 2)$. I'm stuck at this point; I don't know how to know the value of $\lambda$ given that
\begin{cases}e^{-\lambda}\cdot\lambda &= 0.1\\
e^{-\lambda}\cdot\lambda^2 &= 0.4
\end{cases}
Best Answer
Given $X\sim \text{POI}(\lambda)$, so $X$ have p.d.f. $$P(X=x)=\dfrac{e^{-\lambda}\lambda^x}{x!}, x=0,1,\ldots.$$ So, we have \begin{eqnarray} P(X=1)=0.1\iff \dfrac{e^{-\lambda}\lambda}{1}=0.1\\ P(X=2)=0.2\iff \dfrac{e^{-\lambda}\lambda^2}{2}=0.2. \end{eqnarray}
Now we have $e^{-\lambda}\lambda=0.1$ and $e^{-\lambda}\lambda^2=0.4$. Divide two last equations, $$\dfrac{e^{-\lambda}\lambda^2}{e^{-\lambda}\lambda}=\dfrac{0.4}{0.1}=4,$$ we have $\lambda=4$. So, $$P(X=3)=\dfrac{e^{-\lambda}\lambda^3}{3!}=\dfrac{e^{-4}4^3}{6}=0.1953668148.$$