The life expectancy of a component of a machine is $20$ and its variance is $625$. If the component breaks it will be replaced instantanously by a new one. Estimate how many spare parts you need so that the machine runs at leat a time period of $1000$ with a probability of $0.95$.
My approach:
We know that the life of one component or in other words the time until the first failure obeys an exponential distribution, with parameter $\lambda=\frac{1}{20}$. So $P(X\leq t)=\int\limits_{0}^t\lambda e^{-\lambda x}dx$.
The gamma distribution provides the probability of the occurrences of $r$ many failures. If $X_r$ obeys a gamma distribution then $$ P(X_r\leq t)=\int\limits_{0}^{t}\lambda^r\frac{x^{r-1}}{(r-1)!}e^{-x\lambda}\implies P(X_r>t)=1-\int\limits_{0}^{t}\lambda^r\frac{x^{r-1}}{(r-1)!}e^{-x\lambda}. $$ Now we can simply interpret $X_r$ as the time the machine runs until $r$-many failures have occurred, so that
$P(X_r\geq 1000)=1-P(X_r<1000)=1-\int\limits_{0}^{1000}\left(\frac{1}{20}\right)^r\frac{x^{r-1}}{(r-1)!}e^{-\frac{x}{20}}dx>0.95 \implies \int\limits_{0}^{1000}\left(\frac{1}{20}\right)^r\frac{x^{r-1}}{(r-1)!}e^{-\frac{x}{20}}dx <0.05.$
If we solve for $r$ we are done. Unfortunately I can't solve for $r$. Is this approach wrong? Or is there a trick to solve for $r$?