Markov chain - is my diagram/matrix correct? A boy goes to school on a bike or on foot.
If one day he goes on foot, then on the second day he takes a bike with probability $0.8$.
If he goes on a bike one day, then he falls off the bike with probability $0.3$ and goes on foot the next day.
What is the probability that one the last day of school he goes to school on a bike?
We assume that the school year has just started.
I can see that we need to define a Markov chain.
I've tried coming up with a directed graph which I cannot draw here, but I've pasted it here 
This is what the matrix would look like:
$$A= \begin{array}{l}
\mbox{bike} \\
\mbox{falls}  \\
\mbox{on foot}  \\
\mbox{doesn't fall}
\end{array} \left( \begin{array}{cccc}
0 & 0.3 & 0 & 0.7 \\
0 & 0 & 1 & 0\\
0.8 & 0 & 0.2 & 0 \\
1 & 0 & 0 & 0 \end{array} \right) $$
This matrix is ergodic and solving the equation $[x_1, x_2, x_3, x_4]A = [x_1, x_2, x_3, x_4], \ \ x_1 + x_2  + x_3 +x_4=1$ I get that $x_1=8/19$.
So it would seem that the probability that on the last day of school the boy travels on a bike is $8/19$.
Is that true? Is my solution correct?

I think you can simplify things a lot by just using two states, foot (state $1$) and bicycle (state $2$). Then you have the transition matrix: $$\begin{pmatrix} 0.2 & 0.8 \\ 0.3 & 0.7 \end{pmatrix}$$
In fact, the way you've done it is a bit confusing because falling off a bicycle is not a state in the same sense that going to school by foot/bicycle is. For one time step (a day) his state is the mode of transport he uses. But he doesn't spend a full day falling off a bike, so it doesn't make sense to use this as a state.
