Answer:
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 \ Y_n = 0] = 0.5](https://img.qammunity.org/2021/formulas/mathematics/college/vow58hqh6q0b0l1vp7r5pjkm7m2sdcxgac.png)


![\mathtt \ Y_n = 1] = 0.25](https://img.qammunity.org/2021/formulas/mathematics/college/p0ielks6canr3nbu8qpjfgt6dhm3k1xjbq.png)

![\mathtt \ Y_n = 1] = 0.25](https://img.qammunity.org/2021/formulas/mathematics/college/az1rpr5si6w7x4zqjmf4zouw9oao9p37nn.png)


![\mathtt \ Y_n = 2] = 0.5](https://img.qammunity.org/2021/formulas/mathematics/college/ige60ei2890jr7z6jvdu8ig0z0rriyxlbh.png)
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://img.qammunity.org/2021/formulas/mathematics/college/3mmobqgjtnpxpio7cx6cv4l39byzrnjch1.png)
![\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://img.qammunity.org/2021/formulas/mathematics/college/ld4wngv4q3tgc5do8a27l4l5r0jtijbu9e.png)