there are two states:


the car is where he is: CWH

the car is not where he is: CNWH


if at CWH, the next state is CWH with probability p (since if it rains, he will drive the car and the car will be with him in the next state), and the next state is CNWH with probability 1-p (if it doesn't rain, he won't drive, so the car won't be with him in the next state)


if at CNWH, the next state is always CWH with probability 1 (whether it rains or not, he will walk back to where the car is)

writing the input probability flow equations for CWH, we get

input: CNWH*1 + CWH*p

this should be equal to the state probability CWH

CWH = CNWH + p * CWH

we also know that the sum of the probabilities of the two states is 1

CWH = 1- CWH + p*CWH

CWH (2-p) = 1

CWH = 1/(2-p)


on any given day, he will walk in the rain if he doesn't have the car in the morning and it rains, or if he has the car and it doesn't rain and he leaves the car at home, but then on his way back home it rains

CNWH*p + CWH*(1-p)*p

=p * (CNWH + CWH - p * CWH)

= p * (1 - p*CWH)

=p (1- p/(2-p))

=p- p^2 / (2-p)

=2p (1- p)/(2-p)