Consider a random variable $X \sim \operatorname{Poisson}(\lambda)$

Question

I think plugging it in the mass probability function would give the answer but I seem to get nowhere. I know the answer is $4.37$ but I don't know how to get to it.

Find $\lambda$ given that $\mathbb P(X = 1) + \mathbb P(X = 2) = \mathbb P(X = 3)$.

Answer 1

$$ \frac{\lambda^1e^{-\lambda}}{1!} + \frac{\lambda^2 e^{-\lambda}}{2!} = \frac{\lambda^3 e^{-\lambda}}{3!} $$ Divide both sides by $\lambda e^{-\lambda},$ getting $$ 1 + \frac\lambda 2 = \frac{\lambda^2} 6. $$ Multiply both sides by $6,$ getting $$ 6 + 3\lambda = \lambda^2. $$ That is a quadratic equation.

Answer 2

The equation is

$$e^{-\lambda}\cdot \frac{\lambda^1}{1!}+e^{-\lambda}\cdot \frac{\lambda^2}{2!}=e^{-\lambda}\cdot \frac{\lambda^3}{3!}$$

$$e^{-\lambda}\cdot \frac{\lambda^1}{1!}+e^{-\lambda}\cdot \frac{\lambda^2}{2!}-e^{-\lambda}\cdot \frac{\lambda^3}{3!}=0$$

Factoring out $e^{\lambda}\cdot \lambda$. This factor is unequal to $0$.

$$e^{-\lambda}\cdot \lambda\cdot \left( \frac{1}{1}+\frac{\lambda}{2}- \frac{\lambda^2}{6}\right)=0$$

$$ \frac{1}{1}+\frac{\lambda}{2}- \frac{\lambda^2}{6}=0$$

Multiplying the equation by $6$

$$-\lambda^2+3\lambda+6=0$$

This is a quadratic equation which can be solve easily.