Transition matrices

Question

Question is as follows :

Write the $3 \times 3$ transition matrix for a chemistry course that is taught in two sections, if every week $\frac{1}{4}$ of those in section A and $\frac{1}{3}$ of those in section B drop the course, and $\frac{1}{6}$ of each section transfer to the other section.

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My approach

I wrote the formula

$$ x=\frac{3}{4}x_0+\frac{1}{6}y_0\\ y=\frac{1}{6}x_0+\frac{2}{3}y_0\\ $$

And from here I derived the matrix

$$ \begin{bmatrix} \frac{3}{4} & \frac{1}{6} \\ \frac{1}{6} & \frac{2}{3} \end{bmatrix} $$

But seems like my matrix is wrong according to the answer sheet. I mean, seriously what is wrong here? Answer sheet provides a 3 by 3 matrix instead. Okay I understand it could be wrong but the correct answer to the question can not be a 3 by 3 matrix, how come?

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EDIT

According to the answer sheet the correct answer is :

$$\begin{bmatrix} \frac{7}{12} & \frac{1}{6} & 0 \\ \frac{1}{6} & \frac{1}{2} & 0 \\ \frac{1}{4} & \frac{1}{3} & 1 \\ \end{bmatrix}$$

Answer 1

There are three states: $A$, $B$ and $O$ where the last state is being in neither section. The answer sheet is correct:

Let $p_{ij}$ denote the $ij$th entry of the transition matrix (the probability of going from state $i$ to state $j$).

• $p_{AO}=1/4$ because $1/4$ of those in $A$ drop the course
• $p_{BO}=1/3$ because $1/3$ of those in $B$ drop the course
• $p_{OO}=1$ assuming a course cannot "undropped"
• $p_{AB}=p_{BA}=1/6$ because $1/6$ switch sections

The other probabilities can be found using the fact that columns have to sum to one, i.e. $$p_{Ai}+p_{Bi}+p_{Oi}=1.$$

Answer 2

I think the answer sheet is wrong in stating the matrix is 3x3, unless they consider the sum of enrolled students as part of the state.

As for the actual numbers, notice that you may get two difference answers, based on how you interpret the $1/6$ that drops: is it $1/6$ of the initial number $x_0$ or is it $1/6$ of those that do not drop?