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The weather on a given planet can be either cloudy or clear, with a constant probability. In $60\%$ of the cloudy days, the next day was clear and in $30\%$ of the clear days, the next day was cloudy. What percent of the days are cloudy?

I defined the following:

$\Omega = \{((i,weather),(i+1,weather))\}, 1 \leq i \leq 364$, weather is cloudy or clear.

$ Cloudy_1 - \{((i,cloudy),(i+1,weather))\} $

$ Cloudy_2 - \{((i,weather),(i+1,cloudy))\} $

$ Clear_1 - \{((i,clear),(i+1,weather))\} $

$ Clear_2 - \{((i,weather),(i+1,clear))\} $

I need to calculate: $P(Cloudy_1 \cup Cloudy_2) $

My conclusions from the questions are:

  1. $P(Clear_2 | Cloudy_1)=\frac{6}{10}$, $P(Cloudy_2 | Cloudy_1)=\frac{4}{10}$

  2. $P(Cloudy_2 | Clear_1)=\frac{3}{10}$, $P(Clear_2 | Clear_1)=\frac{7}{10}$

I played with the equations but looks like a dead end. Any clue?


Don't give me the answer

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  • $\begingroup$ Use Bayes' theorem $\endgroup$ – john doe Apr 15 '18 at 20:35
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    $\begingroup$ @johndoe this has nothing to do with bayes' theorem. $\endgroup$ – JMoravitz Apr 15 '18 at 20:45
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    $\begingroup$ The correct search terminology here is "regular markov chain" and "steady state" $\endgroup$ – JMoravitz Apr 15 '18 at 20:46
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Hint: if $A_i$, $B_i$ are the events that day $i$ is clear and cloudy, respectively, then $$P(A_{i+1})=P(A_{i+1}|A_i)P(A_i)+P(A_{i+1}|B_i)P(B_i)$$ using the law of total probability. Now if $a$ is the probability that a given day is cloudy, we have $P(A_{i+1})=P(A_i)=a$ and $P(B_i)=1-a$. Solve for $a$ using the equations you have already written down.

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  • $\begingroup$ Did you conclude from: with a constant probability, that: $P(A_{i+1})=P(A_i)$ ? $\endgroup$ – Stav Alfi Apr 17 '18 at 18:59
  • $\begingroup$ If yes, does it means that $P(A_{i+1}|A_i)=P(A_i)$ ? $\endgroup$ – Stav Alfi Apr 17 '18 at 19:42

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