Modeling the problem

[kan]

I'm having a problem while solving this problem, when i implement the solution it happens a loop and it doesn't go to the answer

Consider a model for the long-term eating behavior of students at a University. It was found that 25% of students who ate at canteen A would come back to eat again the next day, while those who ate at canteen B had a 93% chance of going back to eating again the next day. Assume that this university has only 2 canteen on campus and further assumes that all students eat at either canteen. Then, in the long run, what is the percentage of students going to eat at canteen A and B, respectively?

A. 8% and 92%
B. 8,5% and 91,5%
C. 9,5% and 90,5%
D. 10% and 90%

Hope everyone help, thank you everyone!

 

[Deleted member 4993]

I'm having a problem while solving this problem, when i implement the solution it happens a loop and it doesn't go to the answer

Consider a model for the long-term eating behavior of students at a University. It was found that 25% of students who ate at canteen A would come back to eat again the next day, while those who ate at canteen B had a 93% chance of going back to eating again the next day. Assume that this university has only 2 canteen on campus and further assumes that all students eat at either canteen. Then, in the long run, what is the percentage of students going to eat at canteen A and B, respectively?

A. 8% and 92%
B. 8,5% and 91,5%
C. 9,5% and 90,5%
D. 10% and 90%

Hope everyone help, thank you everyone!

".....when i implement the solution..."

Please share your solution along with detailed steps to arrive at the solution.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520

 

[HallsofIvy]

Let a be the probability a student eats in canteen A on a given day and b be the probability a student eats in canteen B on that day. We can represent this as [math]\begin{bmatrix}x \\ y\end{bmatrix}[/math]. If the "transition" matrix, from one day to the next, is [math]\begin{bmatrix}p & q \\ r & s \end{bmatrix}[/math] then [math]\begin{bmatrix}p & q \\ r & s\end{bmatrix}[/math][math]\begin{bmatrix}1 \\ 0 \end{bmatrix}[/math][math]= \begin{bmatrix} p \\ r \end{bmatrix}=[/math][math]\begin{bmatrix} 0.25 \\ 0.75\end{bmatrix}[/math] so p= 0.25 and r is 0.75. Also [math]\begin{bmatrix} p & q \\ r & s \end{bmatrix}\begin{bmatrix}0 \\ 1 \end{bmatrix}[/math][math]= \begin{bmatrix}q \\ s \end{bmatrix}=[/math][math]\begin{bmatrix} 0.07 \\ 0.93\end{bmatrix}[/math].

So the transition matrix is [math]\begin{bmatrix}0.25 & 0.07 \\ 0.75 & 0.93 \end{bmatrix}[/math]. The columns are the two vectors.

Now, the question is about the long term result. Applying this "n" times is the same as [math]\begin{bmatrix}0.25 & 0.07 \\ 0.75 & 0.93 \end{bmatrix}^n\begin{bmatrix} x \\ y\end{bmatrix}[/math]. Find the limit as n goes to infinity. You might need to "diagonalize" the matrix in order to do that.

 

[kan]

Let a be the probability a student eats in canteen A on a given day and b be the probability a student eats in canteen B on that day. We can represent this as [te]\begin{bmatrix}x \\ y\end{bmatrix}\(\displaystyle . If the "transition" matrix, from one day to the next, is \(\displaystyle \begin{bmatrix}p & q \\ r & s \end{bmatrix}\). Then \(\displaystyle \begin{bmatrix}p & q \\ r & s\end{bmatrix}\begin{bmatrix}1 \\ 0 \end\bmatrix}= \begin{bmatrix} p & r \end{pmatrix}= \begin{0.25 \\ 0.75\end{\bmatrix}\) so p= 0.25 and r is 0.75. Also \(\displaystyle \begin{bmatrix} p &\)\)

Can you modify it? Thanks you for much!