So I was looking through my questions and the answers on those questions for feelings of nostalgia when I came across $\text{this}$ question of mine which had proposed the following question (for context, this is not the question that I am going to solve, this is just the question that I had proposed my solution for):
$2$B.$9\quad$Guaranteed life of a machine
The lifetime in hours of a certain component of a machine has the continuous probability density function$$\color{black}{f(x)=\dfrac{1}{1000}e^{-x/1000},x\geq0}$$The machine contains five similar components, the lifetime of each having the above distribution. The makers are considering offering a guarantee that not more than two of the original components will have to be replaced during the first $1000$ hours of use. Find the probability that such a guarantee would be violated, assuming the components wear out independently, and that if a component does fail then the replacement used of particularly high quality and will certainly last for the $1000$ hours.
My solution was correct ($\approx73.6403\%$, therefore it would be unwise to offer the guarantee, for anyone who would want to know), however, @heropup thought that it would be a great idea to deepen my understanding on the question, which is because while I did get the answer correct, it seems that I didn't really have that great of a grasp on the question, so they gave me $2$ bonus questions:
To further your understanding:
$1$ . The original question made a claim of $1000$ hours. What would be the maximum time $t$ that the manufacturer can claim such that the guarantee that no more than $2$ components will fail within $t$ hours will be true with at least $95\%$ probability?
$2$ . The original question used $n=5$ components. What would be the minimum number of components $n$ such that a guarantee at least $3$ original components will still be operational after $1000$ hours will be true with at least a $95$ probability?
For this problem, I will be focusing on bonus question #$2$ only.
$$\text{Here is my attempt at solving bonus question #2:}$$
We know that to do this more easily, we will probably have to write this as a sum to make this more easy to solve. We can do this by writing what we have as$$\sum_{j=0}^{n-3}\binom n{3+j}\left(e^{-1}\right)^{3+j}\left(1-e^{-1}\right)^{n-3-j}$$Which can be simplified to$$\sum_{j=0}^{n-3}\binom n{3+j}\left(e^{-3-j}\right)\left(1-e^{-1}\right)^{n-3-j}$$Which can be further simplified to$$\sum_{j=0}^{n-3}\dfrac{n!e^{-3-j}(1-e^{-1})^{n-3-j}}{(3+j)!(n-3-j)!}$$Which then I can put into Desmos and play around with sliders to get that$$\operatorname{S_{lidernumber1}}(10)\approx77.5\%$$$$\operatorname{S_{lidernumber1}}(11)\approx83.3\%$$$$\operatorname{S_{lidernumber1}}(12)\approx87.7\%$$$$\operatorname{S_{lidernumber1}}(13)\approx91\%$$$$\operatorname{S_{lidernumber1}}(14)\approx93.5\%$$$$\operatorname{S_{lidernumber1}}(15)\approx95.3\%$$Where$$\operatorname{S_{lidernumber1}}(x)$$is the number that I am plugging in for $n$ $$\therefore\text{ }15\text{ machine components is the minimum number of}$$$$\text{components needed to attain a guarantee that }3\text{ components}$$$$\text{will still be operational after }1000\text{ hours with a }95\%\text{ guarantee}$$
$$\mathbf{\text{My question}}$$
Is the solution that I have attained correct, or what could I do to attain the correct solution/attain it more easily?
$$\mathbf{\text{Mistakes I might have made}}$$
- Representing it as a sum incorrectly