# math (Probability)

#### romsek

Math Team
There is another very elegant way to solve this if you are interested.

We treat the whole thing as an absorbing Markov chain.

The state transition matrix is

$$\displaystyle T = \begin{pmatrix} 0 &\dfrac 1 2 &\dfrac 1 2 &0\\ \dfrac 5 6 &0 &0 &\dfrac 1 6\\ 0 &0 &1 &0\\ 0 &0 &0 &1 \end{pmatrix}$$

This breaks up into the $Q$ and $R$ matrices as

$Q=\begin{pmatrix}0 &\dfrac 1 2 \\\dfrac 5 6 &0\end{pmatrix}\\ R = \begin{pmatrix}\dfrac 1 2 &0 \\0 &\dfrac 1 6\end{pmatrix}\\ N = \left(I_2 - Q\right)^{-1} = \begin{pmatrix} \dfrac{12}{7} & \dfrac{6}{7} \\ \dfrac{10}{7} & \dfrac{12}{7} \\ \end{pmatrix}\\ B = NR = \begin{pmatrix} \dfrac{6}{7} & \dfrac{1}{7} \\ \dfrac{5}{7} & \dfrac{2}{7} \\ \end{pmatrix}$

From $B$ we can directly read off that $P[\text{MW|start with MP}] = \dfrac 6 7$ and $P[\text{MW|start with SP}] = \dfrac 5 7$

#### annasilva3

There is another very elegant way to solve this if you are interested.

We treat the whole thing as an absorbing Markov chain.

The state transition matrix is

$$\displaystyle T = \begin{pmatrix} 0 &\dfrac 1 2 &\dfrac 1 2 &0\\ \dfrac 5 6 &0 &0 &\dfrac 1 6\\ 0 &0 &1 &0\\ 0 &0 &0 &1 \end{pmatrix}$$

This breaks up into the $Q$ and $R$ matrices as

$Q=\begin{pmatrix}0 &\dfrac 1 2 \\\dfrac 5 6 &0\end{pmatrix}\\ R = \begin{pmatrix}\dfrac 1 2 &0 \\0 &\dfrac 1 6\end{pmatrix}\\ N = \left(I_2 - Q\right)^{-1} = \begin{pmatrix} \dfrac{12}{7} & \dfrac{6}{7} \\ \dfrac{10}{7} & \dfrac{12}{7} \\ \end{pmatrix}\\ B = NR = \begin{pmatrix} \dfrac{6}{7} & \dfrac{1}{7} \\ \dfrac{5}{7} & \dfrac{2}{7} \\ \end{pmatrix}$

From $B$ we can directly read off that $P[\text{MW|start with MP}] = \dfrac 6 7$ and $P[\text{MW|start with SP}] = \dfrac 5 7$
OH thank you so much