# Markov Chains (probability of event not occurring)

Given a transition matrix $$P(3X3) = \begin{pmatrix} 0.3& 0.7& 0\\ 0.4 &0 & 0.6\\ 0 &0.5& 0.5 \end{pmatrix}$$

Starting with level $$0$$ on the top left and moving down to level $$2$$ towards the bottom.

Question: If the person starts on level $$0$$ what is the probability they will not reach level $$2$$ once in the next four years?

I have tried multiple things such as calculating the probability of reaching level 2 at each year then multiplying those together and subtracting them from one but I don't get the right answer for any. Please help.

• What is the right answer? – callculus Feb 14 at 16:22
• I don't have what the answer is – KombatWombat Feb 14 at 16:34
• make level 2 an absorbing state.... – user8675309 Feb 14 at 19:48
• By “not reach level 2 once in the next four years” do you mean that the process doesn’t enter state 2 at any time during that period? – amd Feb 14 at 19:55
• Yes that is what I mean – KombatWombat Feb 15 at 5:24

A person starts on level zero among levels $$0,1,2$$, and can transition between levels every year. The probabilities of each transition are given in the transition matrix $$P_{3 \times 3} = \begin{pmatrix} 0.3& 0.7& 0\\ 0.4 &0 & 0.6\\ 0 &0.5& 0.5 \end{pmatrix}$$
Where $$P_{ij}$$ denotes the probability of moving from level $$i-1$$ to level $$j-1$$.
Question: If the person starts on level $$0$$ what is the probability they will not reach level $$2$$ once in the next four years?
Following an idea from the comments: if we make level 2 an absorbing state, i.e. if we replace the bottom row with $$(0,0,1)$$ to form the matrix $$Q$$, then the probability of reaching level 2 at some point is the $$1,3$$ entry of the matrix $$Q^4$$. That is, it is the $$1,3$$ entry of the matrix $$Q^4$$, where $$Q = \pmatrix{0.3& 0.7& 0\\ 0.4 &0 & 0.6\\ 0 & 0 & 0.5}.$$ With a direct computation, we see that this probability is $$0.7014$$. It follows that the desired probability is equal to $$1 - 0.7014 = 0.2986$$.