A manufacturer of Christmas tree light bulbs knows that $3\%$ of its bulbs are defective. Find the probability that a string of $100$ lights contains at most four defective bulbs using the Poisson distribution.
Source: Opentextbc.ca
My Work So Far
Using the following
$$\begin{align*} \lambda&=\text{np}\\ &=100(0.03)\\ &=3\\ \\ X &\sim P(3)\\ \end{align*}$$
therefore
$$\begin{align*} \mathbb{P}(x\le4)&=\mathbb{P}(x=0)+\mathbb{P}(x=1)+\mathbb{P}(x=2)+\mathbb{P}(x=3)+\mathbb{P}(x=4)\\[5pt] &=\frac{e^{-3}\times3^0}{0!}+\frac{e^{-3}\times3^1}{1!}+\frac{e^{-3}\times3^2}{2!}+\frac{e^{-3}\times3^3}{3!}+\frac{e^{-3}\times3^4}{4!}\\[5pt] &=e^{-3}\Bigg\{\frac{3^0}{0!}+\frac{3^1}{1!}+\frac{3^2}{2!}+\frac{3^3}{3!}+\frac{3^4}{4!}\Bigg\}\\[5pt] &=e^{-3}\left(\frac{131}{8}\right)\\[5pt] &=\boxed{0.8153}\\[5pt] \end{align*}$$
which matches the solution on the website. I attempted the same as above, but finding the probability that a string of $100$ lights contains at least four defective bulbs, to gain further intuition on this topic
$$\begin{align*} \mathbb{P}(x\ge4)&=1-\mathbb{P}(x\le 3)\\[5pt] &=1-\mathbb{P}(x=0)+\mathbb{P}(x=1)+\mathbb{P}(x=2)+\mathbb{P}(x=3)\\[5pt] &=\frac{e^{-3}\times3^0}{0!}+\frac{e^{-3}\times3^1}{1!}+\frac{e^{-3}\times3^2}{2!}+\frac{e^{-3}\times3^3}{3!}\\[5pt] &=e^{-3}\Bigg\{\frac{3^0}{0!}+\frac{3^1}{1!}+\frac{3^2}{2!}+\frac{3^3}{3!}\Bigg\}\\[5pt] &=0.6472318888\\[5pt] &=1-0.6472318888\\[5pt] &=\boxed{0.3528}\\[5pt] \end{align*}$$
Would both of these methods be correct? The solutions worked in this calculator, but I just wanted to make sure this is the correct process. Would there also be a better method to derive the solutions?