Question

1. A Toyota dealer consumes four kinds of engine oil A, B, C and D. This dealer buys its consumption each week. The dealer never buys the same brand in successive weeks, except brand D. If the dealer buys engine oil D then with same probability it can buy all kinds of engine oils next week. If the dealer perches brand C then next week it will buy D. However, if the dealer buys engine oil B then the next week it is four times as likely to perches A as the other brands and finally if the dealer buy engine oil A it will buy D or C with same probabilities. What is the probability that mentioned dealer buys oil C when we know that oil B was purchased three weeks ago?

Answer 1

We can model the problem as Mrakov chain with states A, B, C and D.

The tratnsition probability from state A to state C or D is 1/2.  The tratnsition probability from state A to state A or B is 0.

The transition probability from state B to state A is 4 times other transition probabilities (k) to C or D.

Thus 4k + k + k = 1

=> k = 1/6

The transition probability from state B to state A is 4/6 = 2/3 times and the transition probabilities to C or D is 1/6.

The tratnsition probability from state C to state D is 1.  The tratnsition probability from state C to state A, B or C is 0.

The tratnsition probability from state D to state A, B, C or D is 1/4.

The transition probability matrix is,

The probability that mentioned dealer buys oil C when we know that oil B was purchased three weeks ago

= P(X3 = C | X0 = B)

= P(X0 = B, X1 = A, X2 = D, X3 = C) + P(X0 = B, X1 = C, X2 = D, X3 = C) + P(X0 = B, X1 = D, X2 = A, X3 = C) + P(X0 = B, X1 = D, X2 = B, X3 = C) + P(X0 = B, X1 = D, X2 = D, X3 = C)

= (2/3) * (1/2) * (1/4) + (1/6) * 1 * (1/4) + (1/6) * (1/4) * (1/2) +  (1/6) * (1/4) * (1/6) +  (1/6) * (1/4) * (1/4)

= 0.1631944