Homework Help on Simple Markov Chain I have the following problem and questions provided:

Assume the state of the weather in Sydney on any particular day can be modelled using a state space S={1≡sunny,2≡overcast,3≡wet}. Also assume that the likelihood of any state depends only on the state on the previous day. A sunny day is followed by a sunny day 60% of the time and otherwise it is overcast. An overcast day is equally likely to be followed by either a sunny or a wet day. 20% of days after a wet day are also wet, but 20% are sunny.

a) Write down the transition probability matrix for this Markov chain.
My answer:
#            Sunny   Overcast  Wet
# Sunny       .6      .4        0
# Overcast    .5      0         .5
# Wet         .2      .6        .2

P <- matrix(c(.6, .5, .2, .4, 0, .6, 0, .5, .2), nrow = 3)

b) Draw the state transition diagram for this chain and check if it is irre-ducible or reducible. If reducible, identify the equivalence classes.
My answer:
So when I draw out the diagram it looks like it is an irreducible chain insofar as you can get to any state from any other state. But I'm also confused about the meaning of irreducible because when I take P to the nth power, it does reduce like so:
#make the power function for matrix powers first

'%^%' <- function(P,n){ 
  if(n > 1){
    M <- P
    for(i in 2:n){
      P <- P%*%M
    }
  }
  return(P)
}

(reducedP <- (P%^%100))

c) If it is sunny all weekend, calculate the probability it is also sunny on Monday.
My answer:
pi0 <- matrix(c(1, 0, 0), nrow = 3)
(t(pi0)%*%(P%^%2))[1]

But, because the conditional probability of being sunny given the previous day being sunny is necessarily .6, shouldn't this basically be (.6^2)*1 = .36?
d) Calculate the probability that a Monday is wet if the previous Monday was wet.
My answer:
pi0 <- matrix(c(0, 0, 1), nrow = 3)
(t(pi0)%*%(P%^%7))[3]

e) Calculate the probability that Wednesday through Friday are all sunny,if the previous Sunday was sunny.
For this problem, it seems like i should see what probability the chain gives me for it being sunny on wednesday and then calculate two conditional probabilities, but I don't understand how to get the intersection. Disclaimer: I haven't had a proper probability course before.
I'm hoping someone can help me and let me know if I'm on the right track/right or what I'm missing/need to be doing because at this point I'm totally lost (my professor won't help much but to say I'm on the right track and he doesn't know why the book example does what it does as it's seemingly incorrect.

edit: so here is basically where I'm at, what my general problem is using the easy question from part c):
If any state pi(n) does not depend on the initial state, then, for example in part c, isn't the probability that it will be sunny on monday given it was sunny all weekend only actually depend on what the state was on sunday? So we don't even care about saturday... That means the probability should be .6 because you can only get to a sunny monday in one transition step if sunday was sunny, and going from sunny to sunny is a .6 probability. Otherwise, wouldn't the answer be 1*.6*.6, that is, the initial state which is known, the probability of transitioning to a sunny day at t + 1, and then the probability of transitioning to a sunny day again at t + 2 | t + 1? That gives different answers. And the very confusing bit is that if I use the P matrix, I get .56 (as in my original email), but if I use the P* matrix, I get 0.377 which is close to the 0.36 when you do 1*.6*.6 .
So I have two different paths to take, but no idea which one is the right path and why. 
 A: For your part (b), I don't see a contradiction. 
A Markov chain is said to be irreducible if it is possible to get to any state from any state. In fact, when you compute $P^{100}$, you have shown explicitly that each entries are positive. It is indeed irreducible.
For part (c), you should not compute $P^2$, in fact, whether Monday rains just depends on whether it rains on Sunday. 
For part $(e)$, let $Q$ be the event that it rains on previous Sunday, let $W$ be the event that it is sunny on Wednesday, $T$ be the event that it rains on Thursday and $F$ be the event that it rains on $Friday$.
\begin{align}
P(WTF|Q) &= P(W|Q)P(T|W)P(F|T)
\end{align}
Try to compute this quantity.
A: By taking $\mathbf{P}^{100}$ you get (correct to to seven places) the limiting matirix $\lim_{n\rightarrow\infty}\mathbf{P}^{n},$ of which
all three rows are $\lambda =  (0.4901961,\, 0.3137255,\, 0.1960784).$
In an ergodic chain such a limiting distribution always exists. For some finite power $m$ the matrix $\mathbf{P}^m$ has all positive elements. Your
chain is ergodic because $\mathbf{P}^2$ has all positive elements.
P %*% P
     [,1] [,2] [,3]
[1,] 0.56 0.24 0.20
[2,] 0.40 0.50 0.10
[3,] 0.46 0.20 0.34

The limiting distribution $\lambda$ is also the steady-state distribution vector $\sigma,$ such that $\sigma\mathbf{P}= \sigma.$ 
For ergodic $\mathbf{P},$ you can use the R function eigen to find $\sigma,$ as shown below. It is necessary
to use the transpose t(P) because eigen finds right eigenvectors and
we want a left eigenvector. It is convenient to use as.numeric to
avoid complex number notation in case some eigenvectors are not real.
For an ergodic chain, the steady-state vector is the eigenvector listed first (having smallest modulus) and it is always a real vector proportional to $\sigma.$
The last step ensures that the elements of $\sigma$ sum to unity.
eigen(t(P))$vectors[,1]
[1] -0.7981886 -0.5108407 -0.3192754

g = eigen(t(P))$vectors[,1]
sgm = as.numeric(g)/sum(g);  sgm
[1] 0.4901961 0.3137255 0.1960784

