Problem with my answers:

Reasoning:

(a) [imath] P_{00} = P_{02} = P_{20} = 0 [/imath] are impossible events, since a swap must happen, and starting from state 0 black pens in the right pocket can only to state 1. Similarly, starting with 2 black pens in the right pocket cannot go to 0 (only 1 or stay at 2).

(b) [imath] P_{01} = 1 [/imath], per reasoning in part (a)

[imath] P_{10} = P(\text{taking black from right})P(\text{taking white from left}) = (1/2)(1/3) = 1/6 [/imath]

[imath] P_{11} = P(\text{taking white from right})P(\text{taking white from left}) + P(\text{taking black from right})P(\text{taking blackfrom left}) = (1/2)(1/3) + (1/2)(2/3) = 3/6 [/imath]

[imath] P_{12} = P(\text{taking white from right})P(\text{taking black from left}) = (1/2)(2/3) = 2/6 [/imath]

[imath] P_{21} = P(\text{taking black from right})P(\text{taking white from left}) = (1)(2/3) = 2/3 = 4/6 [/imath]

[imath] P_{22} = P(\text{taking black from right})P(\text{taking black from left}) = (1)(1/3) = 1/3 = 2/6 [/imath]

(c) Labelled above using the transition probabilities

(d) True. Since all states are are reachable from all others, the chain is one big communicating class [imath] C = {0, 1, 2} [/imath]

(e) True. If the chain is irreducible then every state is recurrent; there's always a way to get back to each state

(f) True. No state has a self-loop with [imath] P_{ii} = 1 [/imath]

(g) True. The presence of a self-loop ensures that the chain is aperiodic, i.e. the [imath] gcd(1, 2, 3, ...) = 1 [/imath]

(h) Solving the balance equation [imath] \pi = \pi P [/imath], where [imath] \pi = [\pi_0 \pi_1 \pi_2 ] [/imath] and [imath] P [/imath], is the transition matrix, I get a system of equations and adding the fact that [imath] \pi_0 + \pi_1 + \pi_2 = 1 [/imath] I get [imath] \pi_2 [/imath] = 0.3