Hello The graph is showing the functions PAA(t), PAB(t) and PAC(t) where, for example: PAA(0) = 1 PAA(1) = 0.5 (from the transition probability matrix) PAA(2) = 0.375 (from part (i)) as k -> inf PAA(k) -> 1/3 (pi_1 from part (ii)) Similarly: PAB(0) = 0 PAB(1) = 0.25 (from the transition probability matrix) PAB(2) = 0.375 (from part (i)) as k -> inf PAA(k) -> 1/2 (pi_2 from part (ii)) The Examiners' Report shows more calculated probabilities (eg PAA(3), PAA(4) etc) but given the question says 'using your answers to part (i) and (ii)' I don't see how they would have expected you to do any more calculations. So I would have drawn a graph showing PAX(0), PAX(1), PAX(2) for X = A, B and C and then shown a dotted line tending to the stationary probabilities and put labels on the x-axis to the effect of 0, 1, 2, ...., inf. Hope this helps! Andy
