Expected frequency of replacements How exactly do you answer the following question?:
"Any given day there is a malfunction on a plane with probability 1/10 which then requires a replacement. Additionally, each plane that has been around for 100 days must be replaced. What is the long-term frequency of replacements?"
If $X$=number of days until replacement then does $X$ follow a geometric distribution up to $99$ ($F_X(x)=1-(9/10)^{-x}$) and $F_X(100)=1$ ? Are we then asked to find $EX$? That's pretty easy, but I am not sure if I am understanding the question correctly.
 A: Yes exactly: "long-term frequency of replacements" means exactly the average time of replacement you will have in a long term of activity.
As you noted, the pmf is truncated in $t=100$. Obviously this truncation does not significantly affect the result, thus in this situation the requested frequency is $10$ without any calculations.
You can try to solve your problem with a different truncation, i.e. $t=20$ in order to appreciate the different results with or without truncation
So I would first calculate the truncated pmf
$$P(X=x)=\frac{1}{1-0.9^{20}}\cdot 0.1\cdot 0.9^{x-1}\cdot\mathbb{1}_{\{1;2;3;\dots;20\}}(x)$$
A: I would regard the interval $t\in\{1,2,\cdots, 100\}$ only. Then you have the same situation vor the following intervals. The random variable $X$ denotes the number of replacements in the interval.

*

*You have one replacement in 100 days, if you have no replacement in
the first 99 days. Thus, for $x=1$, the probability is $0.9^{99}$.


*You have two replacements in 100 days, if you have one replacement in
the first 99 days. Thus, for $x=2$, the probability is $\binom{99}{1}\cdot 0.9^{98}\cdot 0.1$.
$$\vdots$$

*

*You have 100 replacements in 100 days, if you have 99 replacements in
the first 99 days. Thus, for $x=100$, the probability is $\binom{99}{99}\cdot 0.9^{0}\cdot 0.1^{99}$.

From these cases we can deduce a probability function:
$$P(X=x)=\binom{99}{x-1}\cdot 0.9^{100-x}\cdot 0.1^{x-1}, \quad x\in \{1,2,\ldots , 100\}$$
From here you can deduce the expected value.
Remark:
The long-term period of replacement is $p=\frac{100}{\mathbb E(X)}$
