Calculating probability matrix Suppose we have 2 switches.
The transition matrix for the first switch is:
$$
        \begin{bmatrix}
        p_{00}^1=0.45 & p_{01}^1=0.55 \\
        p_{10}^1=0.25 & p_{11}^1=0.75 \\
        \end{bmatrix}
$$
Which means, for example, probability that it will be on (1) given that it was off (0) at previous time instant is $$ p_{01} = 0.55 $$
Similarly, for the second switch,
$$
        \begin{bmatrix}
       p_{00}^2= 0.55 & p_{01}^2=0.45 \\
        p_{10}^2=0.2 & p_{11}^2=0.8 \\
        \end{bmatrix}
$$
Then how to find the probabilities-
$$
P_{10}, P_{11}, P_{12}
$$
Where, for example, $P_{12}$ means the probability that both switches will be on given that only one of them was on at previous time instant?
I have tried:
$$
P_{10} = p_{00}^1 * p_{10}^2 + p_{10}^1 * p_{00}^2 (=0.2275)
$$
$$
P_{11} = p_{00}^1 * p_{11}^2 + p_{01}^1 * p_{10}^2 + p_{10}^1 * p_{01}^2 + p_{11}^1 * p_{00}^2 (=0.9950)
$$
$$
P_{12} = p_{01}^1 * p_{11}^2 + p_{11}^1 * p_{01}^2 (=0.7775)
$$
Which doesn't seem quite right because their sum should be 1
 A: Hint (assuming independent and stationary Markovian switch processes)
Both switches operate as stationary Markov processes. Both have their stationary state probabilities. Let $\pi_1^1,\pi_1^0$ denote the probabilities that the first switch is on or off, respectively. For the other switch let the same be denoted by $\pi_2^1$ and $\pi_2^0$. These probabilities can be calculated based on the two state transition matrices given.
Let's denote the state of the switches at the $n^{th}$ moment by $X_n^1$ and $X_n^2$. Both random variables may take the values $0$ and $1$.
Hint to calculate $P_{1,0}$
$$P_{1,0}=P(X_{n}^1=0\cap X_{n}^2=0\mid X_{n-1}^1=1\cap X_{n-1}^2=0\cup X_{n-1}^1=0\cap X_{n-1}^2=1)=$$
$$=\frac{P(X_{n}^1=0\cap X_{n}^2=0\cap X_{n-1}^1=1\cap X_{n-1}^2=0)}{P(X_{n-1}^1=1\cap X_{n-1}^2=0)+ P(X_{n-1}^1=0\cap X_{n-1}^2=1)}+$$
$$+\frac{P(X_{n}^1=0\cap X_{n}^2=0\cap X_{n-1}^1=0\cap X_{n-1}^2=1)}{P(X_{n-1}^1=1\cap X_{n-1}^2=0)+P(X_{n-1}^1=0\cap X_{n-1}^2=1)}.$$
If $\{X_n^1\}$ and $\{X_n^2\}$ are independent and stationary Markov processes then we can do the calculation. (Without referring to any combined Markov process.)
So,
$$P(X_{n}^1=0\cap X_{n}^2=0\cap X_{n-1}^1=1\cap X_{n-1}^2=0)=$$
$$=P(X_{n}^1=0\cap X_{n-1}^1=1)P(X_{n}^2=0\cap X_{n-1}^2=0)=$$
$$=P(X_{n}^1=0\mid X_{n-1}^1=1)\pi_1^1P(X_{n}^2=0\mid X_{n-1}^2=0)\pi_2^0$$
and 
$$P(X_{n-1}^1=1\cap X_{n-1}^2=0)=P(X_{n-1}^1=1)P( X_{n-1}^2=0)=\pi_1^1\pi_2^0,$$
$$P(X_{n-1}^1=0\cap X_{n-1}^2=1)=P(X_{n-1}^1=0)P( X_{n-1}^2=1)=\pi_1^1\pi_2^1.$$
The conditional probabilities are given by the OP and the stationary probabilities can be calculated.

NOTE
The title of the OP is misleading. It asks for a probability matrix as if the number of switches ON formed a Markov process. This is not the case. Even so, the indexed probabilities $P_{i,j}$ can be calculated. (Here $i,j \in \{0,1,2\}.$)
ALSO
Take a look at Did's notes below. 
There: "if they are at stationarity" means that my calculations are exact if we set the switches randomly -- at the beginning -- with probabilities $\pi_1^1$, $\pi_1^0$,$\pi_2^1$,$\pi_2^0$.
