Thread: How would I go about solving this Problem (Matrices)

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.

Hello, WarZebra!

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

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


(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.

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

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


If it is Cloudy today:

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

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

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

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

Ridiculous question . . .

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

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