Finding $P(X = 3)$ for Poisson distribution with lambda unknown

Question

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}

Answer 1

If $P(X=1)=0.1$, then $\dfrac{e^{-\lambda}\lambda^1}{1}=0.1$. We have nonlinear equation $$\lambda e^{-\lambda}=0.1.$$ So we must solve numerically (using calculator or numerical method for finding root of nonlinear equation). We have $\lambda=0.1118325592$ or $\lambda=3.577152064$.

If $P(X=2)=0.2$, then $\dfrac{e^{-\lambda}\lambda^2}{2}=0.2$. We have nonlinear equation $$\lambda^2 e^{-\lambda}=0.4.$$ Solving this equality we have $\lambda=0.6052671213$ and $\lambda=4.707937918$.

So we cannot have $P(X=1)=0.1$ and $P(X=2)=0.2$ with the same $\lambda$.

If $P(X=1)=0.1$, then we have $$P(X=3)=\dfrac{e^{-0.1118325592}0.1118325592^3}{6}=0.0002084420217.$$ or

$$P(X=3)=\dfrac{e^{-3.577152064}3.577152064^3}{6}=0.2132669481.$$

If $P(X=2)=0.2$, then we have $$P(X=3)=\dfrac{e^{-0.6052671213}0.6052671213^3}{6}=0.02017557070.$$ or $$P(X=3)=\dfrac{e^{-4.707937918}4.707937918^3}{6}=0.1569312640.$$

Answer 2

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, $$P(X=3)=\dfrac{e^{-\lambda}\lambda^3}{3!}=\dfrac{e^{-4}4^3}{6}=0.1953668148.$$