The Fibonacci numbers denoted by $F_i$ for $i\ge1$ are $$1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,\cdots$$ where they satisfy the property $F_{i+2}=F_{i+1}+F_i$.
I have listed the first $15$ numbers of the sequence as they will be useful for reference later.
Now define $\theta_i$ as the concatenation of $1.F_i$ so that $1<\theta_i<2$.
So for example, $$\theta_1=1.1,\quad\theta_6=1.13,\quad\theta_{15}=1.987.$$
Next, consider the following $3\times3$ system of equations: $$\begin{bmatrix}\theta_i&\theta_{i+1}&\theta_{i+2}\\\theta_{i+4}&\theta_{i+5}&\theta_{i+6}\\\theta_{i+8}&\theta_{i+9}&\theta_{i+10}\end{bmatrix}\begin{bmatrix}X\\Y\\Z\end{bmatrix}=\begin{bmatrix}\theta_{i+3}\\\theta_{i+7}\\\theta_{i+11}\end{bmatrix}$$ from which we can solve for $[X\quad Y\quad Z]^T$ but of course this would be very tedious.
When $i=1,2$, both $X$ and $Y$ are negative but $Z$ is positive. When $i=3$, the opposite occurs.
Question: For $i\le n$, what is the probability that exactly one of $X,Y,Z$ will be negative?