Answer:
Step-by-step explanation:
On a given day , a particular raccoon will eat the trash from one of three different houses.
Let assume
be a random variable that illustrating the house raccoon will eat on an unknown given nth day.
If he eats from the trash of a particular house, he has a 50% chance to eat from the same house the next day, and a 25% chance each to eat from one of the other two houses.
There are three states given in the above statement.
So, we can have state 1, state 2 and state 3
Assuming that:
state 1 = house 1
state 2 = house 2
state 3 = house 2
If he eats from the trash of a particular house,
For state 1 : he has a 50% chance to eat from the same house the next day
i.e state 1 = 0.50
and a 25% chance each to eat from one of the other two houses.
For state 2 and state 3: = 0.25
i.e state 2 = 0.25
state 3 = 0.25
NOW:
![\mathtt{P[Y_{n+1 }= 0 \ | \ Y_n = 0] = 0.5}](https://tex.z-dn.net/?f=%5Cmathtt%7BP%5BY_%7Bn%2B1%20%7D%3D%200%20%20%5C%20%7C%20%20%5C%20%20Y_n%20%3D%200%5D%20%3D%200.5%7D)
![\mathtt{P[Y_{n+1 }= 1 \ | \ Y_n = 0] = 0.25}](https://tex.z-dn.net/?f=%5Cmathtt%7BP%5BY_%7Bn%2B1%20%7D%3D%201%20%20%5C%20%7C%20%20%5C%20%20Y_n%20%3D%200%5D%20%3D%200.25%7D)
![\mathtt{P[Y_{n+2}= 2 \ | \ Y_n = 0] = 0.25}](https://tex.z-dn.net/?f=%5Cmathtt%7BP%5BY_%7Bn%2B2%7D%3D%202%20%20%20%5C%20%7C%20%20%5C%20%20Y_n%20%3D%200%5D%20%3D%200.25%7D)
![\mathtt{P[Y_{n+1 }= 0 \ | \ Y_n = 1] = 0.25}](https://tex.z-dn.net/?f=%5Cmathtt%7BP%5BY_%7Bn%2B1%20%7D%3D%200%20%20%5C%20%7C%20%20%5C%20%20Y_n%20%3D%201%5D%20%3D%200.25%7D)
![\mathtt{P[Y_{n+1 }= 1 \ | \ Y_n = 1] = 0.5}](https://tex.z-dn.net/?f=%5Cmathtt%7BP%5BY_%7Bn%2B1%20%7D%3D%201%20%20%5C%20%7C%20%20%5C%20%20Y_n%20%3D%201%5D%20%3D%200.5%7D)
![\mathtt{P[Y_{n+1 }= 2 \ | \ Y_n = 1] = 0.25}](https://tex.z-dn.net/?f=%5Cmathtt%7BP%5BY_%7Bn%2B1%20%7D%3D%202%20%20%5C%20%7C%20%20%5C%20%20Y_n%20%3D%201%5D%20%3D%200.25%7D)
![\mathtt{P[Y_{n+1 }= 0 \ | \ Y_n = 2] = 0.25}](https://tex.z-dn.net/?f=%5Cmathtt%7BP%5BY_%7Bn%2B1%20%7D%3D%200%20%20%5C%20%7C%20%20%5C%20%20Y_n%20%3D%202%5D%20%3D%200.25%7D)
![\mathtt{P[Y_{n+1 }= 1 \ | \ Y_n = 2] = 0.25}](https://tex.z-dn.net/?f=%5Cmathtt%7BP%5BY_%7Bn%2B1%20%7D%3D%201%20%20%5C%20%7C%20%20%5C%20%20Y_n%20%3D%202%5D%20%3D%200.25%7D)
![\mathtt{P[Y_{n+1 }= 2 \ | \ Y_n = 2] = 0.5}](https://tex.z-dn.net/?f=%5Cmathtt%7BP%5BY_%7Bn%2B1%20%7D%3D%202%20%20%5C%20%7C%20%20%5C%20%20Y_n%20%3D%202%5D%20%3D%200.5%7D)
The stochastic matrix for this scenario can be computed as:
0 1 2
![P = \left\begin{array}{c}0\\1\\2\end{array}\right \left[\begin{array}{ccc}0.5&0.25&0.25\\0.25&0.5&0.25\\0.25&0.25&0.5\end{array}\right]](https://tex.z-dn.net/?f=P%20%3D%20%5Cleft%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C1%5C%5C2%5Cend%7Barray%7D%5Cright%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0.5%260.25%260.25%5C%5C0.25%260.5%260.25%5C%5C0.25%260.25%260.5%5Cend%7Barray%7D%5Cright%5D)
![\mathbf{ P =\left[\begin{array}{ccc}0.5&0.25&0.25\\0.25&0.5&0.25\\0.25&0.25&0.5\end{array}\right] }](https://tex.z-dn.net/?f=%5Cmathbf%7B%20P%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0.5%260.25%260.25%5C%5C0.25%260.5%260.25%5C%5C0.25%260.25%260.5%5Cend%7Barray%7D%5Cright%5D%20%7D)