aekaakee: Probability

Let $X$ have a Poisson distribution with mean $4$. Find

Mean of Poisson distribution $= \lambda = 4$

To find $P \left(X \leq 3\right)$:

$\begin{aligned} P \left(X \leq 3\right) & = \sum \limits_{x = 0}^{3} \dfrac{e^{- \lambda} \times \lambda^{x}}{x!} \\\\ & = e^{- \lambda} \times \left[\dfrac{\lambda^{0}}{0!} + \dfrac{\lambda^{1}}{1!} + \dfrac{\lambda^{2}}{2!} + \dfrac{\lambda^{3}}{3!}\right] \\\\ & = e^{- 4} \times \left[1 + \dfrac{4}{1} + \dfrac{4^2}{2} + \dfrac{4^3}{6}\right] \\\\ & = 0.0183 \times 23.67 = 0.4331 \end{aligned}$
To find $P \left(2 \leq X < 5\right)$:

$\begin{aligned} P \left(2 \leq X < 5\right) & = \sum \limits_{x = 2}^{4} \dfrac{e^{- \lambda} \times \lambda^{x}}{x!} \\\\ & = e^{- \lambda} \times \left[\dfrac{\lambda^{2}}{2!} + \dfrac{\lambda^{3}}{3!} + \dfrac{\lambda^{4}}{4!} \right] \\\\ & = e^{- 4} \times \left[\dfrac{4^2}{2} + \dfrac{4^3}{6} + \dfrac{4^4}{24} \right] \\\\ & = 0.0183 \times 29.33 = 0.5368 \end{aligned}$