I got the following question:
In a house with 9 rooms. There is 1 mouse that is looking for some food. This can be found in 2 rooms, but there are also 2 cats, these are in different rooms. When the mouse enters a room with a cat, he will be eaten right away, but they are quite lazy and always stay in the same room. The mouse is very forgetful, so he immediately forgets that he has entered a room. And the probability to enter a room will always have the same probability of the time before.
The following picture shows how everything is arranged:
I have numbered the rooms to calculate the expected value.
A) What is the probability that the mouse takes one of the food before it is eaten by a cat?
B) What is the probability that the mouse eats both food before it is eaten by a cat?
C) What is the expected value? The expected value means in this case the total amount steps the mouse walks before he will be in the room with the cat.
A) $$ {1\over2} * {2\over3} = {1\over3}$$ B) $$ {1\over2} * {1\over3} = {1\over6}$$
My answer of A & B are not correct, because my teacher told me if I used this on my exam I will not get all the points. And I am really sure how to do it the right way.
I tried the following to calculate the expected value:
Define X= #Total steps of mouse
Pi(X=k) = Chance that X=k if you start in room i.
I know that if the mouse is in the room with the cat I will not walk anymore. So I gave the rooms with the cat 1.
$$ P1 = P5= 1 $$
$$ P4(X=k) = {1\over2}P5(X=k-1) + {1\over2}P3(X=k-1) $$ $$ P5(X=k) = {1\over4}P8(X=k-1) + {1\over4}P2(X=k-1) + {1\over4}P7(X=k-1) + {1\over4}P4(X=k-1) $$ $$ P2(X=k) = {1\over2}P1(X=k-1) + {1\over2}P3(X=k-1) $$ $$ P6(X=k) = {1\over2}P7(X=k-1) $$ $$ P7(X=k) = {1\over2}P6(X=k-1) + {1\over2}P3(X=k-1) $$ $$ P8(X=k) = {1\over2}P7(X=k-1) + {1\over2}P8(X=k-1) $$ $$ P9(X=k) = {1\over2}P8(X=k-1) $$
I'm not sure if this is correct and how to continue from here. I hope that someone could show me how to do this the correct way or give me some tips.