I have worked through this problem based on the probability Matrixes provided. I would like to review what I have done to make sure my process to solve this problem was correct. Consider the “Weather Forecasting Problem” discussed in class and assume that this model has been used for quite a while to forecast the weather of an old town: Xt = The weather condition on day t Xt E {R,N} = {1,2} R N 1 2 P= R 1 | 0.40 0.60 | N 2 | 0.45 0.55 | R N 1 2 P2= R 1 | 0.43 0.57 | N 2 | 0.43 0.57 | a) Given that today is rainy, what is the probability that it will be rainy two weeks from now? Well, using a quick test to verify that we are at a steady state at P 2 Π * P = [ Π0 Π1 ] | 0.40 0.60 | | 0.45 0.55 | Π0 = 0.40 Π0 + 0.45 Π1 à Π0 = 0.75 Π1 Π1 = 0.60 Π0 + 0.55 Π1 Π0 + Π1 = 1 à 0.75 Π1 + Π1 = 1 à 1.75 Π1 = 1 à Π1 = 0.57 Π0 = 0.75 Π1 à Π0 = (0.75) * (0.57) à Π0 = 0.43 R N Π = [ 0.43 0.57 ] Therefore, R is a periodic state with a and m = 7 (14 days) Pr( x14 = R | x1 = R) = .43 b) Suppose it is a typical day in the life of this old town and you are a person from out of town who happens to be getting off the plane at the airport of this town. What is the probability that the weather on that day is not rainy? From a) R N On any given day, the probability of stepping off a plane and it is not rainy is .57 Π = [ 0.43 0.57 ] c)Referring to part (b), if you were to rent a convertible and drive around town until it rains again, how many days on average will you be expecting to keep the car? R N 1 2 P = R 1 | 0.43 0.57 | N 2 | 0.43 0.57 | mij = mNR = MFPT to R starting at N Therefore, mNR = (1)*(pNR) + (1 + mNR)*(pnn) mNR = 1 / (1 – pNR) mNR = 1 / (1 – 0.43) mNR = 1 / 0.57 mNR = 1.75 days