How would I go about solving this Problem (Matrices)

**WarZebra** · Oct 23rd 2010, 02:31 PM

http://img257.imageshack.us/img257/1311/58856961.png

I can't seem to figure out how to do it, all I know is I need to use matrices.

**Soroban** · Oct 23rd 2010, 07:42 PM

Hello, WarZebra!

This is a Markov process.
Are your familiar with it?

. . . . . . . . $\begin{array}{cccccccc}C && R && \;S \end{array}$
. . $\begin{array}{c}C \\ R \\ S\end{array}\begin{pmatrix}<br /> 0.60 & 0.10 & 0.30 \\ 0.20 & 0.55 & 0.25 \\ 0.10 & 0.15 & 0.75 \end{pmatrix}$

(a) If today is cloudy, find the probability that it will be sunny two days from now.

(b) What about cloudy?

(c) What about rainy?

For the two-day probabilities, square the matrix.

$A^2 \;=\;\begin{pmatrix}0.60 & 0.10 & 0.30 \\ 0.20 & 0.55 & 0.25 \\ 0.10 & 0.15 & 0.75\end{pmatrix}\begin{pmatrix}0.60 & 0.10 & 0.30 \\ 0.20 & 0.55 & 0.25 \\ 0.10 & 0.15 & 0.75\end{pmatrix}$

. . . . . . . . . . . $\begin{array}{ccccccccccc}C && \;\;R && \quad S \end{array}$
. . . $=\; \begin{array}{c}C \\ R \\ S\end{array}\begin{pmatrix}<br /> 0.410 & 0.160 & 0.430 \\ 0.255 & 0.3605 & 0.385 \\ 0.165 & 0.205 & 0.630 \end{pmatrix}$

If it is Cloudy today:

$(a)\;P(\text{S in 2 days}) \:=\:0.430$

$(b)\;P(\text{C in 2 days}) \:=\:0.410$

$(c)\;P(\text{R in 2 days}) \:=\:0.160$

Now do the same calculations for 8 days from now, 10 days from now.
What do you notice?

Ridiculous question . . .

We need to determine $\,A^8$ and $A^{10}$ .

And I don't know anyone willing to do that.

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