For this, I'm going to introduce a notation you're probably not aware of, which comes from the methods curriculum -
)
is the probability of A if you already know that B has happened. We can then put this into a matrix equation like so:
\\ Pr(B_1)\end{array}<br />=\begin{array}{|cc|}Pr(A_1|A_0)&Pr(A_1|B_0)\\ Pr(B_1|A_0)&Pr(B_1|B_0)\end{array}\cdot<br />\begin{array}{|c|}Pr(A_0)\\ Pr(B_0)\end{array})
Remembering this whole set up can be a little annoying, so here's some tips - first, for the two column matrices, the one on the far right will always be the state before the one on the initial left, and the matrix in the middle is the transition matrix. To remember how we laid out the transition matrix, for
)
, X will be whatever is in the left most matrix in the same row, and Y will be whatever you'd get if you multiplied the matrices out. If you can find another way which works for you, use that - the best way to remember this is whatever way works for you!
Now, to interpret this question. It says that 10% of the wagons that were at point A ended at point B - this means that
 = 10\%)
. It also says that 8% of the wagons that started at point B ended at point A, this means that
 = 8\%)
. So, this gives us the following matrix equation:
\\ Pr(B_1)\end{array}<br />=\begin{array}{|cc|}Pr(A_1|A_0)&0.08\\ 0.10&Pr(B_1|B_0)\end{array}\cdot<br />\begin{array}{|c|}Pr(A_0)\\ Pr(B_0)\end{array})
Now, here's something cool - the columns of any transition matrix MUST add up to one. Using this information, we can now complete the transition matrix, giving us:
\\ Pr(B_1)\end{array}<br />=\begin{array}{|cc|}0.90&0.08\\ 0.10&0.92\end{array}\cdot<br />\begin{array}{|c|}Pr(A_0)\\ Pr(B_0)\end{array})
Now, for part b, we're told that initially 125 wagons were at each starting point. So, we can now plug this into the far-right matrix (also called the initial state matrix, for this reason):
\\ n(B_1)\end{array}<br />=\begin{array}{|cc|}0.90&0.08\\ 0.10&0.92\end{array}\cdot<br />\begin{array}{|c|}125\\ 125\end{array})
(note: if you know anything about probabilities, you'll know that they must be between 0 and 1. Don't worry, though - what I'm doing here is perfectly fine if you change those "Pr"s to "n"s like I did above - and it still works!
)
However, using this transition matrix only counts for one week, so in truth we actually have to apply it twice to get two weeks like the question wants us to. Using a CAS, this gives us:
\\ n(B_1)\end{array}<br />=\begin{array}{|cc|}0.90&0.08\\ 0.10&0.92\end{array}^2\cdot<br />\begin{array}{|c|}125\\ 125\end{array}<br />=\begin{array}{|c|}130\\ 120\end{array})
So, 130 carts are at point A and 120 carts are at point B.
Part c is the same principle, but we don't have whole numbers to work with. So, we know that 40% start at point A, and that columns must add up to 100%, so we get:
\\ Pr(B_1)\end{array}<br />=\begin{array}{|cc|}0.90&0.08\\ 0.10&0.92\end{array}\cdot<br />\begin{array}{|c|}0.40\\ 0.60\end{array})
But once again - the use of one transition matrix only counts for one week, and this questions wants to know about 6 weeks, so we have to use it 6 times, giving us:
\\ Pr(B_1)\end{array}<br />=\begin{array}{|cc|}0.90&0.08\\ 0.10&0.92\end{array}^6\cdot<br />\begin{array}{|c|}0.40\\ 0.60\end{array})
Plugging this into the CAS, we get:
\\ Pr(B_1)\end{array}<br />=\begin{array}{|cc|}0.90&0.08\\ 0.10&0.92\end{array}^6\cdot<br />\begin{array}{|c|}0.40\\ 0.60\end{array}<br />=\begin{array}{|c|}0.51\\ 0.49\end{array})
So, 51% of the carriages will be at point A, and 49% of the carriages will be at point B.
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