Probability of $ X(T) = 1 $ for exponentially distributed switches between $ 0 $ and $ 1 $ Let $ X(t) $ be a random variable that may assume the states $ 0 $ and $ 1 $. Every once in a while it switches its value, as the time $ t $ goes on. The probability of a switch from $ 0 $ to $ 1 $ and $ 1 $ to $ 0 $ are governed, respectively, by exponential distributions with parameters $ \lambda_{01} $ and $ \lambda_{10} $.
Is this information enough to calculate the probability that $ X $ equals $ 1 $ at any given time $ T $? i.e. $ P(X(T) = 1) $
Surely some information about the initial state is needed, I would like to see the difference between assuming:
1) The initial state is $ X(0) = x_0 $ for a known $ x_0 $. 
2) $ X(0) $ a random variable with 50-50 chances of being $ 0 $ or $ 1 $.
Just to mention an idea that may potentially work -but I really don't know- I considered that the event $ X(T) = 1 $ may occur as an infinite union of disjoint events: the first state is one and never changed for $ t<T $; the first state was $ 0 $ and changed one time; etc... so that $ P(X(T) = 1) $ may be written as an infinite sum. But still, if I take in this (hypothetically correct) infinite sum, the term corresponding to event $ A $: "the first state was $ 0 $ and changed one time", the change may have happened at any instant of the continuous interval $ (0, T] $, therefore $ P(A) $ should be written as a double integral, right? One for integrating the exponential pdf and calculating the probability of switching before some particular time $ t = \tau < T $ and another one for integrating out the $ \tau $ variable.
Nevertheless, even if this weird idea was ok, I wouldn't know how to consider all possible cases in the infinite sum. 
 A: I will address both questions together. Your idea would work in concept. However, the integrals would quickly become very difficult to compute. Maybe you could use some sort of induction to get a recursive formulation of the terms in your infinite sum, and maybe the sum would be solvable, but I'm not sure. I'm going to introduce a different approach.
Note: I'm assuming you are parameterizing exponential random variables by the inverse mean. So the mean of an exponential random variable with parameter $\lambda$ should be $1/\lambda$. 
Suppose we know that $P(X(t) = 1) = p(t)$. What can we say about $p(t+\Delta t)$ when $\Delta t$ is small? Let's do something similar to what you suggested.
$$P(X(t + \Delta t) = 1|X(t)=1) = P(X\text{ changes an even number of times in }(t,t+\Delta t]|X(t)=1)$$
$$=\sum_{i=0}^\infty P(X\text{ changes } 2i \text{ times in }(t,t+\Delta t]|X(t)=1))$$
Let's do some quick calculations. Using the exponential cdf,
$$P(X\text{ changes } 0 \text{ times in }(t,t+\Delta t]|X(t)=1)) = e^{-\lambda_{10}\Delta t} = 1 - \lambda_{10}\Delta t + O((\Delta t)^2).$$
For further calculations, let $\lambda = \max\{\lambda_{10},\lambda_{01}\}$. Let $Y_i$ be i.i.d. exponential random variables with parameter $\lambda$. Can you see why,
$$P(X\text{ changes more than } 1 \text{ time in }(t,t+\Delta t]|X(t)=1) \leq P\left(\sum_{i=1}^2 Y_i < \Delta t\right)?$$
If you aren't familiar with the relationship between exponential distributions and Poisson distributions, I recommend looking it up. However, the point is we can easily compute the probability on the right:
$$P\left(\sum_{i=1}^2 Y_i < \Delta t\right) = \sum_{i=2}^\infty e^{-\lambda\Delta t}\frac{\lambda^i(\Delta t)^i}{i!} = O((\Delta t)^2).$$
Similarly, 
$$P(X(t+\Delta t) = 1|X(t) = 0) = P(X\text{ changes an odd number of times in }(t,t+\Delta t]|X(t)=0)$$
A quick calculation shows (I use the memoryless property to simplify this a bit):
$$P(X\text{ changes once in } (t,t+\Delta t]|X(t)=0) = \int_0^{\Delta t}\lambda_{01}e^{-\lambda_{01}s}\int_{\Delta t-s}^\infty \lambda_{10}e^{-\lambda_{10}r}dr\,ds $$
$$=\int_0^{\Delta t} \lambda_{01}e^{-\lambda_{01}s}e^{-\lambda_{10}(\Delta t - s)}\,ds = e^{-\lambda_{10}\Delta t} \int_0^{\Delta t} \lambda_{01}e^{-s(\lambda_{01} + \lambda_{10})}\,ds = \frac{\lambda_{01}e^{-\lambda_{10}\Delta t}}{\lambda_{01} + \lambda_{10}}(1 - e^{-\Delta t(\lambda_{01} + \lambda_{10})})$$
$$= \frac{\lambda_{01}}{\lambda_{01}+\lambda_{10}}(\Delta t (\lambda_{01}+\lambda_{10})) + O((\Delta t)^2) = \lambda_{01}\Delta t + O((\Delta t)^2)$$
Using a similar method to what we did above we can show,
$$P(X\text{ changes more than } 2\text{ times in }(t,t+\Delta t]|X(t)=0) = o((\Delta t)^2).$$
Putting it all together,
\begin{align*}
p(t+\Delta t) &= P(X(t+\Delta t) = 1)\\
&= \sum_{i=0}^1 P(X(t+\Delta t) = 1|X(t) = i)P(X(t) = i)\\
&=p(t)(1 - \lambda_{10}\Delta t) + \lambda_{01}\Delta t(1 - p(t)) + O((\Delta t)^2)
\end{align*}
Then,
\begin{align*}
\frac{dp}{dt}(t) &= \lim_{\Delta t \searrow 0} \frac{p(t+\Delta t) - p(t)}{\Delta t}\\
&= \lim_{\Delta t \searrow 0}-(\lambda_{01} + \lambda_{10})p(t) + \lambda_{01} + O(\Delta t)\\
&= -(\lambda_{01} + \lambda_{10})p(t) + \lambda_{01}
\end{align*}
This is a separable ODE, and we can explicitly solve it:
$$p(t) = \left(p(0) - \frac{\lambda_{01}}{\lambda_{01} + \lambda_{10}}\right)e^{-t(\lambda_{01}+\lambda_{10})} + \frac{\lambda_{01}}{\lambda_{01} + \lambda_{10}}.$$
To answer your first question, just plug in $p(0) = x_0$. Then,
$$P(X(T) = 1) = p(T) = \begin{cases}
- \frac{\lambda_{01}}{\lambda_{01} + \lambda_{10}}e^{-T(\lambda_{01}+\lambda_{10})} + \frac{\lambda_{01}}{\lambda_{01} + \lambda_{10}} &\text{ if } x_0 = 0\\
\frac{\lambda_{10}}{\lambda_{01} + \lambda_{10}}e^{-T(\lambda_{01}+\lambda_{10})} + \frac{\lambda_{01}}{\lambda_{01} + \lambda_{10}} &\text{ if } x_0 = 1
\end{cases}.$$
For your second question, simply set $p(0) = 1/2$. Then,
$$P(X(T) = 1) =  \frac{\lambda_{10} - \lambda_{01}}{2(\lambda_{10} + \lambda_{01})}e^{-T(\lambda_{01}+\lambda_{10})} + \frac{\lambda_{01}}{\lambda_{01} + \lambda_{10}}.$$
