Complicated Problem (help?)

[PhilA]

I feel a little silly posting this here, as this is a problem I need to figure out for work, not school... my math skills are rusty, to say the least.
The problem I'm presenting is an abstraction; it's not literally the problem I'm facing (I don't work in a block factory), but the numbers and relationships to to each other are. I've gotten the math up to this point myself, but I have no idea how to figure out this last step and get the final ratio.

A factory produces blue blocks and red blocks.
Each time the factory produces a blue block, the next block it produces has a 38% chance to be red, otherwise it will be blue.
Each time the factory produces a red block, the next block it produces has a 45% chance to be red, otherwise it will be blue.
What percentage of the blocks will be red?

Thanks in advance for the help guys.

 

[PhilA]

Same problem, different numbers:

A second produces blue blocks and red blocks.
Each time this factory produces a blue block, the next block it produces has a 47% chance to be red, otherwise it will be blue.
Each time this factory produces a red block, the next block it produces has a 54% chance to be red, otherwise it will be blue.
What percentage of the blocks will be red?

 

[Deleted member 4993]

Same problem, different numbers:

A second produces blue blocks and red blocks.
Each time this factory produces a blue block, the next block it produces has a 47% chance to be red, otherwise it will be blue.
Each time this factory produces a red block, the next block it produces has a 54% chance to be red, otherwise it will be blue.
What percentage of the blocks will be red?

Please let us know what the original problem was.

 

[Deleted member 4993]

I feel a little silly posting this here, as this is a problem I need to figure out for work, not school... my math skills are rusty, to say the least.
The problem I'm presenting is an abstraction; it's not literally the problem I'm facing (I don't work in a block factory), but the numbers and relationships to to each other are. I've gotten the math up to this point myself, but I have no idea how to figure out this last step and get the final ratio.

A factory produces blue blocks and red blocks.
Each time the factory produces a blue block, the next block it produces has a 38% chance to be red, otherwise it will be blue.
Each time the factory produces a red block, the next block it produces has a 45% chance to be red, otherwise it will be blue.
What percentage of the blocks will be red?

Thanks in advance for the help guys.

Please let us know what the original problem was.

 

[soroban]

Hello, PhilA!

I believe this is a "Markov process".
If you're not familiar with it, here is a mini-course . . .

A factory produces blue blocks and red blocks.
Each time the factory produces a blue block, the next block it produces has a 38% chance to be red, otherwise it will be blue.
Each time the factory produces a red block, the next block it produces has a 45% chance to be red, otherwise it will be blue.
What percentage of the blocks will be red?

\(\displaystyle \text{We have: }\;\begin{array}{cccccccccc}P(B\to B) \:=\:0.62 && P(B\to R) \:=\:0.38 \\ P(R\to B) \:=\:0.55 && P(R\to R) \:=\:0.45 \end{array}\)

\(\displaystyle \text{We have a "transition matrix": }\;A \;=\;\begin{bmatrix}0.62 & 0.38 \\ 0.55 & 0.45 \end{bmatrix}\)

\(\displaystyle \text{The matrix after }n\text{ blocks are produced is: }\:\begin{bmatrix}0.62 & 0.38 \\ 0.55 & 0.45\end{bmatrix}^n\)
\(\displaystyle \text{As more and more blocks are produced, the matrix approaches: }\:\lim_{n\to\infty}\begin{bmatrix}0.62 & 0.38 \\ 0.55 & 0.45\end{bmatrix}^n\)
\(\displaystyle \text{This limit is called the "steady state matrix": }\;\begin{bmatrix} p & q \\ p & q\end{bmatrix}\)



\(\displaystyle \text{This is the procedure for finding the steady-state matrix . . .}\)

\(\displaystyle \text{Set up two equations: }\;\begin{array}{cccc} (1) & (p,q)\cdot A &=& (p,q) \\ (2) & p + q &=& 1 \end{array} \quad\text{ and solve for }p\text{ and }q.\)


\(\displaystyle \text{Here we go!}\)


\(\displaystyle \text{We have: }\;(1)\;\;(p,q)\cdot\begin{bmatrix}0.62 & 0.38 \\ 0.55 & 0.45\end{bmatrix} \;=\;(p,q)\)

\(\displaystyle \text{This gives us two equations: }\;\begin{array}{ccc} 0.62p + 0.55q &=& p \\ 0.38p + 0.45q &=& q \end{array} \quad \hdots\text{ but they are }equivalent.\)

\(\displaystyle \text{Select one equation, say, the second one: }\;0.38p \,+\, 0.45q \:=\:q \quad\Rightarrow\quad 0.38p \,-\, 0.55q \:=\:0\)


\(\displaystyle \text{Together with (2), we have a system of equations: }\;\begin{Bmatrix} 0.38p - 0.55q &=& 1 & (a) \\ p \;+\; q &=& 1 & (b)\end{Bmatrix}\)

\(\displaystyle \begin{array}{ccccccc}\text{Multiply (b) by 0.55:} & 0.55p + 0.55q &=& 0.55 \\ \text{Add (a):} & 0.38p - 0.55q &=& 0 \end{array}\)

\(\displaystyle \text{And we have: }\;0.93p \:=\:0.55 \quad\Rightarrow\quad p \:=\:\frac{0.55}{0.93} \quad\Rightarrow\quad \boxed{p \:=\:\frac{55}{93}}\)

\(\displaystyle \text{Substitute into (2): }\;\frac{55}{93} + q \:=\:1 \quad\Rightarrow\quad \boxed{q \:=\:\frac{38}{93}}\)


\(\displaystyle \text{The steady-state matrix is: }\;\begin{bmatrix}\frac{55}{93} & \frac{38}{93} \\ \\[-3mm] \frac{55}{93} & \frac{38}{93} \end{bmatrix}\)


\(\displaystyle \text{This means that }eventually: \;\begin{array}{ccc}\frac{55}{93} &\approx& 59\%\text{ will be blue} \\ \\[-3mm] \frac{38}{93} &\approx& 41\%\text{ will be red.} \end{array}\)

 

[PhilA]

Please let us know what the original problem was.

It's actually an impulse purchase calculation. A when a customer sees the customer in line before him pick up a small item displayed by the register (gum, candy, magazine, etc) they are slightly more likely to do the same.

Thanks for the help guys!

 

[Edith]

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