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Suppose each day there is a $0.2$ probability will rain in the morning. $P(\text{rains afternoon}|\text{rain morning})=0.6$ and $P(\text{rains afternoon}|\text{not rain morning})=0.3$. Suppose John would go to the office in the morning and leave in the afternoon and he totally get 3 umbrellas. Let $X_n$ be the number of umbrellas storing in his office at $nth$ night. Whats the proportion of time he has $0$ umbrella in his office at night?

I tried to make the transition probability matrix and find the stationary distribution at state $0$ and found that $\pi_0=0.19$ but seems not consistent with the answer but i don't know what is the mistake. Any hints or solution are welcome.

$P_{transition} =\begin{pmatrix}0.92&0.08&0&0\\0.24&0.68&0.08&0\\0&0.24&0.68&0.08\\0&0&0.24&0.76\end{pmatrix}$ so $P-I=\begin{pmatrix}-0.08&0.08&0&0\\0.24&-0.32&0.08&0\\0&0.24&-0.32&0.08\\0&0&0.24&-0.24\end{pmatrix}$

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    In a transition matrix, the entries in each column must add up to $1$ (or, if you are using the other convention, the entries in each row must add up to $1$). So have another think about how to set up the transition matrix.2012-12-17
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    O i made some mistake,2012-12-17
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    @GerryMyerson Ya i think you are right, i needa compute again2012-12-17
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    @GerryMyerson have write in the answer space, is it correct?2012-12-17

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