1. **State the problem:** We have a Markov chain with three states. The transition probabilities depend on the current state. 2. **Given information for transitions from State 3:** - The particle is equally likely to go to State 1 or State 2. - The particle cannot stay in the same state (transitions are always to a different state). 3. **Define the transition probabilities from State 3:** Let $a_{31}$ be the probability of going from State 3 to State 1. Let $a_{32}$ be the probability of going from State 3 to State 2. Let $a_{33}$ be the probability of staying in State 3. 4. **Use the given conditions:** - Since the particle cannot stay in the same state, $a_{33} = 0$. - The particle is equally likely to go to State 1 or State 2, so $a_{31} = a_{32}$. 5. **Sum of probabilities from State 3 must be 1:** $$a_{31} + a_{32} + a_{33} = 1$$ Substitute $a_{33} = 0$ and $a_{31} = a_{32} = p$: $$p + p + 0 = 1$$ $$2p = 1$$ $$p = \frac{1}{2}$$ 6. **Final values:** $$a_{31} = \frac{1}{2}, \quad a_{32} = \frac{1}{2}, \quad a_{33} = 0$$