# Bayesian formula for weather exercise

If it is nice weather on one day, the probability that it is going to be nice again the next is $13/15$.

If it is raining on one day, the prob. that it is going to be raining again the next day is $4/5$

Now $S_k$ is the event: "k-th day is nice" and we assume $\mathbb P(S_k)=\mathbb P(S_{k+1})\forall k\in\mathbb Z$ and $\mathbb P(S_{k+1}\cap S_{k-1}| S_k)=\mathbb P(S_{k+1}| S_k)\mathbb P(S_{k-1}| S_k)$

(a) What is $\mathbb P(S_{k-1}| S_k)$ ? I think we can use bayesian formula $\mathbb P(S_{k-1}| S_k)=\frac{\mathbb P(S_{k}| S_{k-1})\mathbb P(S_{k-1})}{\mathbb P(S_k)}$ and $\mathbb P(S_{k}| S_{k-1})=13/15$, but what about the other terms? Here the next point

(b) What is the probability that there is nice weather on a specific day?

(c) Probability that after a nice day we have two more nice days?

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Would perhaps $P(S_k)=P(S_{k+1})$ imply $P(S_{k-1})=P(S_k)$? If so, then $P(S_{k-1}|S_k)$ reduces to $P(S_k|S_{k-1})=13/15$. But that doesn't help with question b)...
In c) you have: $$P(S_{k+2}\cap S_{k+1}|S_k)=P(S_{k+1}|S_k)P(S_{k+2}|S_k, S_{k+1})=P(S_{k+1}|S_k)P(S_{k+2}|S_{k+1})=[P(S_{k+1}|S_k)]^2=\left(\frac{13}{15}\right)^2$$ which follows from your statement $$P(S_{k+1}\cap S_{k-1}|S_k)=\left[P(S_{k-1}|S_k)P(S_{k-1}|S_k, S_{k+1})\right]=P(S_{k-1}|S_k)P(S_{k-1}|S_k)$$ where the middle part in brackets is the implied step. It tells us that, conditional on $S_k$, $S_{k-1}$ and $S_{k+1}$ are independent. I'm not 100 % sure, but maybe this can inspire you at least. Good luck!