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With \(\mu_1 = 10\text{,}\)
\begin{align*} P(X \lt 10) & = F(9)\\ & = \frac{10^0 e^{-10}}{0!} + \frac{10^1 e^{-10}}{1!} + \frac{10^2 e^{-10}}{2!} + ... + \frac{10^8 e^{-10}}{8!} + \frac{10^9 e^{-10}}{9!}\\ & = e^{-10} \cdot ( 1 + 10 + \frac{100}{2} + ... + \frac{10^8}{8!} + \frac{10^9}{9!} ) \end{align*}
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