Question

The pattern of sunny and rainy days on the planet rainbow is a homogeneous markov chain with two states. every sunny day is followed by another sunny day with probability 0.8. every rainy day is followed by another rainy day with probability 0.6.

a.today is sunny on rainbow. what is the chance of rain the day after tomorrow?

b.compute the probability that april 1 next year is rainy on rainbow.

Answer 1

The attatchment shows the solution. PFA.

Answer 2

Answer:

Concept:

Simply put, the probability is the likelihood that something will occur. When we don't know how an event will turn out, we can discuss the likelihood or likelihood of several outcomes. Statistics is the study of events that follow a probability distribution.

Explanation:

Given:

The pattern of sunny and rainy days on the planet rainbow is a homogeneous Markov chain with two states.

Every sunny day is followed by another sunny day with a  probability of 0.8

Every rainy day is followed by another rainy day with a probability of 0.6.

Find:

(a) Today is sunny on the rainbow. what is the chance of rain the day after tomorrow?

(b) Compute the probability that April 1 next year is rainy on a rainbow.

Solution:

(a)  Let S and R stand for the planet's sunny and rainy days. Given by is the transition probability matrix.

                              $T=\underset{R}{S}\left(\begin{array}{cc}S & R \\0.8 & 0.2 \\0.4 & 0.6\end{array}\right)$

We have to find the probability of rain after two days. We have to find $\mathrm{T}^{2}$

     $\left(\begin{array}{ll}0.8 & 0.2 \\ 0.4 & 0.6\end{array}\right)\left(\begin{array}{ll}0.8 & 0.2 \\ 0.4 & 0.6\end{array}\right)$         =   $\underset{R}{S}\left(\begin{array}{cc}S & R \\0.06 +0.08 & 0.16+0.12 \\0.32+0.24 & 0.08+0.36\end{array}\right)$

                                  $=\underset{R}{S}\left(\begin{array}{cc}S & R \\0.72 & 0.28 \\0.56 & 0.44\end{array}\right)$

Hence if today is sunny on the planet, the chance of rain the day after tomorrow is 0.28.

(b)  We can use the steady-state distribution because April 1 in 2019 is so far away from many changes. We resolve the set of equations to determine it.

                             $\pi P=\pi, \pi_{1}+\pi_{2}=1 .$

These equations are:

                           $\begin{aligned}0.8 \pi_{1}+0.4 \pi_{2} &=\pi_{1} \\0.2 \pi_{1}+0.6 \pi_{2} &=\pi_{2} \\\pi_{1}+\pi_{2} &=1\end{aligned}$

From the first equation, $\pi_{1}=2 \pi_{2}$

From the second equation again, $\pi_{1}=2 \pi_{2}$

We know that one equation will follow the others. Substituting $\pi_{1}=2 \pi_{2}$ into the last equation, we get $2 \pi_{2}+\pi_{2}=1$.

From here, $\pi_{2}=1 / 3$ and $\pi_{1}=2 / 3$.

Hence, the probability that April 1 next year is rainy is

$\pi_{2}=1 / 3 .$

#SPJ3