Answer:
Step-by-step explanation:
Answer:
The number of umbrellas at her current location = Xn
The state space is X = {0,1,2,3,4,…r}
Probability of rain = p = 0.2
Probability it does not rain q = 1-p = 1-0.2 = 0.8
r = number of umbrellas she possess
let umbrellas at current place = i
Therefore umbrellas at other place = r-i
When she gets to the other place, there will still be r-i umbrellas if it does not rain
Pi,r-I = q = 0.8 (it does not rain)
Pi, r-i+1 = p = 0.2 (it rains) for i = 1,2,3…r
P0r = 1
Pij = 0, if i+j > r+2 or i+j<r
The markov chain in this case cannot be reduced and is finite and is therefore a positive recurrent. The only stationary distribution in this case is the limiting distribution.
m
∑ πiPi0= πrPr0 =q / (r+q) =π₀
i=0
m
∑ πi Pij=πr-j Pr-j,j +πr -j +1 Pr - j=1,j= q/(r + q) +p/(r+q) =1/(r+q)=πr
i=0
When i =0, no umbrella at current place, then \pi = q/(r+q) = 0.8/(3+0.8) = 0.210
When i = 1,2....r, carries umbrella at current place, then \pi = 1/(r+q) = 1/(3+0.8) = 0.263
Therefore the π's given is the stationary distribution and is a limiting distribution
The Professor can get wet only when she has no umbrella at her current location and it rains. Therefore the probability is
π0p=pq/(r+q) = (0.2*0.8)/(3+0.8) = 0.042
