For the rat in the maze example discussed in the handouts su

For the rat in the maze example discussed in the handouts, suppose the rat has a probability of 1/4 of starting in any compartment a:Find the probability that it is in compartment 1 after one move. b:Find the probability that it is in compartment 3 after two moves. Example: A rat moves through the maze shown. Each time the rat makes a move from one compartment to another, it chooses a passageway at random. Let X(n) be the compartment occupied by the rat after n moves. Time in this case is measured by the moves of the rat rather than by the clock. If the rat begins in compartment 1, what is the probability that it first goes to 2, and then to 4, and then back to 2

Solution

Let P(ij) denote the probablity moving from compartment i to j

So, the initial state probabilites are --> P(01) = P(02) = P(03) = P(04) = 1/4

After first move,

P(11) = 0 (as the rat is moving one compartment to another, it\'s probability of remaining in the same compartment is 0) , similarly P(22) = P(33) = P(44) = 0, so the probability of moving from one compartment to rest three will be 1/3 each.

P(12) = P(13) = P(14) = 1/3 , similarly, P(21) = P(23) = P(24) = 0 ... and so on ...

So probability of being in compartment 1 after first move = P(01)*P(11) + P(02)*P(21) + P(03)*P(31) + P(04)*P(41) = 1/4*0 + 1/4*1/3 + 1/4*1/3 + 1/4*1/3 = 1/4

The probablities are multiplied, because the movement of the rate from one compartment to the next move after move are disjoint set of events, as the rat does not follow any pattern and chooses a passageway at random.

Similarly, calculate the Probability of rat being in compartment 3 after 2 moves, as below --

P(01)*P(12)*P(23) + P(01)*P(14)*P(43) +

P(02)*P(21)*P(13) + P(02)*P(24)*P(43) +

P(03)*P(31)*P(13) + P(03)*P(32)*P(23) + P(03)*P(34)*P(43) +

P(04)*P(41)*P(13) + P(04)*P(42)*P(23)