5 An inspector has three factories, A, B, C, to check. He spends each day in one of the factories. He chooses the factory to visit on a particular day according to the following rules.
- If he is in A one day, then the next day he will never choose A but he is equally likely to choose B or C .
- If he is in B one day, then the next day he is equally likely to choose \(\mathrm { A } , \mathrm { B }\) or C .
- If he is in C one day, then the next day he will never choose A but he is equally likely to choose B or C .
- Write down the transition matrix, \(\mathbf { P }\).
- On Day 1 the inspector chooses A.
(A) Find the probability that he will choose A on Day 4.
(B) Find the probability that the factory he chooses on Day 7 is the same factory that he chose on Day 2. - Find the equilibrium probabilities and explain what they mean.
The inspector is not satisfied with the number of times he visits A so he changes the rules as follows.
- If he is in A one day, then the next day he will choose \(\mathrm { A } , \mathrm { B } , \mathrm { C }\), with probabilities \(0.8,0.1,0.1\), respectively.
- If he is in B or C one day, then the probabilities for choosing the factory the next day remain as before.
- Write down the new transition matrix, \(\mathbf { Q }\), and find the new equilibrium probabilities.
- On a particular day, the inspector visits factory A. Find the expected number of consecutive further days on which he will visit factory A.
Still not satisfied, the inspector changes the rules as follows.
- If he is in A one day, then the next day he will choose \(\mathrm { A } , \mathrm { B } , \mathrm { C }\), with probabilities \(1,0,0\), respectively.
- If he is in B or C one day, then the probabilities for choosing the factory the next day remain as before.
The new transition matrix is \(\mathbf { R }\).
On Day 15 he visits C . Find the first subsequent day for which the probability that he visits B is less than 0.1.Show that in this situation there is an absorbing state, explaining what this means.
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