If what you mean is that during this 10 day window, the onset of rain (like a monsoon "breaking") has a probability of 0.5 on any day, and it will go on raining till the end thereafter,
E(x) = $\sum_{k=1}^{10} 0.5^k\cdot(11-k) = \frac{9217}{1024}$, ≈ 9.001
This is not an integer, but as has been pointed out, that is not a problem !
Denoting the current day as index $i$ (so that yesterday is $i-1$ and the day before yesterday is $i-2$), the previous state comprises two elements, $State_{n-1}=\{S_{i-1},S_{i-2}\}$, where $S_k$ is the state of the weather for day $k$. Thus the current state is $State_{n}=\{S_{i},S_{i-1}\}$
Lettting $R$ be the occurrence of rain on a particular day, so that $\overline{R}$ is when no rain occurs for that day, there are $4$ possible values of the previous state $State_{n-1}$
$$\{\overline{R},\overline{R}\}, \{\overline{R},R\},\{R,\overline{R}\},\{R,R\}$$
For each of these $4$ states, there are only two possible next states:-
$$\begin{align}
State_n=\{\overline{R},\overline{R}\}\text { or }State_n=\{R,\overline{R}\}\text{ given }State_{n-1}=\{\overline{R},\overline{R}\}
\\
State_n=\{\overline{R},\overline{R}\}\text { or }State_n=\{R,\overline{R}\}\text{ given }State_{n-1}=\{\overline{R},R\}
\\
State_n=\{\overline{R},R\}\text { or }State_n=\{R,R\}\text{ given }State_{n-1}=\{R,\overline{R}\}
\\
State_n=\{\overline{R},R\}\text { or }State_n=\{R,R\}\text{ given }State_{n-1}=\{R,R\}
\end{align}$$
With the information given in the question, and based on the constraints of what the next state is based on the current state, you can figure out the structure of the transition matrix.
For example, the information Specifically, suppose that if it has rained for the past two days, then it will rain tomorrow with probability $0.7$, corresponds to
$$P(State_{n+1}=\{\overline{R},\overline{R}\}|State_{n}=\{\overline{R},\overline{R}\})=0.7\\\Rightarrow P(State_{n+1}=\{R,\overline{R}\}|State_{n}=\{\overline{R},\overline{R}\})=1-0.7=0.3$$
and if it rained today but not yesterday, then it will rain tomorrow with probability 0.5 corresponds to
$$P(State_{n+1}=\{\overline{R},\overline{R}\}|State_{n}=\{\overline{R},R\})=0.5\\\Rightarrow P(State_{n+1}=\{R,\overline{R}\}|State_{n}=\{\overline{R},R\})=1-0.5=0.5$$
Best Answer
The probability that the event takes place exactly one time during an infinite sequence of independent experiments is already $1$:
$$P("\text{the number of successes is exactly one}")=p+(1-p)p+(1-p)^2p+\cdots=$$ $$=p\frac{1}{1-(1-p)}=1.$$
The probability that the event takes place at least one time is larger or equal than that but smaller or equal than $1$.
As far as raining: The daily raining events are not independent. So, we cannot use this example to enlighten ourserves.