Dueling .....

[pka]

Although I agree with your formula, I believe you have the wrong numbers
\(\displaystyle \begin{align*} \mathcal{P}(G_3) &=\mathcal{P}(G_3G_2)+\mathcal{P}(G_3R_2)\\ &=\mathcal{P}(G_3|G_2)\mathcal{P}(G_2) +\mathcal{P}(G_3|R_2) \mathcal{P}(R_2) \\&= \frac{1}{4} \cdot \frac{7}{8} +\frac{\boldsymbol{\color{red}{1}}}{8} \cdot\frac{7}{8}\\&=\dfrac{21}{64} \end{align*}\)

$\displaystyle \left\{\begin{array}{l}\mathcal{P}(G_2)=\frac{7}{8}\\\color{blue}{\mathcal{P}(R_2)=\frac{1}{8}}\\ \mathcal{P}(G_3|G_2)=\frac{1}{4}\\ \color{blue}{\mathcal{P}(G_3|R_2)=\frac{7}{8}} \end{array} \right.$

ref: http://www.freemathhelp.com/forum/t...e-red-light-is-on-what-is-the-prob-that/page2

As can seen that was a simple typo. The fact is the parts in blue above are correct values. So typo.
As I told you in replying to your PM, it clear in post #10.

Maybe I did not express it precisely enough but the question in this exercise is about that: you see a red light and then you press a button three times - what is a P to see a green at the end?

pkbgamma meant something different from my reading.

His is this: you go into a room, see a red light, then press the button three times. Then ask, what is the probability that the light is now green? Found by adding the probability of $\displaystyle RRG,~RGG,~GRG,\text{ or }GGG$
Given that I agree with pkbgamma, this is best done with Markov Chains.

 

[Ishuda]

As can seen that was a simple typo. The fact is the parts in blue above are correct values. So typo.
As I told you in replying to your PM, it clear in post #10. pkbgamma meant something different from my reading.

His is this: you go into a room, see a red light, then press the button three times. Then ask, what is the probability that the light is now green? Found by adding the probability of $\displaystyle RRG,~RGG,~GRG,\text{ or }GGG$
Given that I agree with pkbgamma, this is best done with Markov Chains.

Since you bring it up, as I told you an you replied

The machine works this way: if a red light is on, pressing of a button changes it to a green light with a probability of 7/8 and with a probability of 1/8 it stays red. But if a green light is on, pressing of a button changes it to a red light with a probability of 3/4 and with a probability of 1/4 it stays green. The question is: if red light is on, what is a probability too see green light after second and after third press?
RRG = 1/8*7/8
RRR = 1/8*1/8
RGG = 7/8*2/8
RGR = 7/8*6/8

According to the person's reply, I read incorrectly. Of course I disagree.
Only the BLUE cases above are relevant as I read it.

According to him, it is as if you walk into a room find the light red and then you press the bottom the times, starting with the light already red.
RRG, RGG, GGG, GRG are the only cases as he reads it that endup green after three pushes.

As you can see, I was talking about the typo you made. Even given your blue highlighting, you still ignored the fact that it was the numbers I was disagreeing with in you post, not the formula. I only posted here after you ignored that.

 

[pka]

you still ignored the fact that it was the numbers I was disagreeing with in you post, not the formula. I only posted here after you ignored that.

I really have no idea what you mean. I find no mistake in any numbers I posted.
\(\displaystyle \begin{array}{*{20}{c}}{}|& R&G\\R& {1/8}&{7/8}\\G& {3/4}&{1/4}\end{array}\) I hope that you will agree that is the correct transition matrix.
If so, see this. There you see that if we start with the light red (row I) after two pushes the light being green (column II) with probability $\displaystyle \frac{21}{64}$.

Here you can see what happens after pressing the button three times.

Please do point out my mistake.

 

[Ishuda]

I really have no idea what you mean. I find no mistake in any numbers I posted.
\(\displaystyle \begin{array}{*{20}{c}}{}|& R&G\\R& {1/8}&{7/8}\\G& {3/4}&{1/4}\end{array}\) I hope that you will agree that is the correct transition matrix.
If so, see this. There you see that if we start with the light red (row I) after two pushes the light being green (column II) with probability $\displaystyle \frac{21}{64}$.

Here you can see what happens after pressing the button three times.

Please do point out my mistake.

Let's see, you admit to a typo but can see nothing incorrect in your post. O.K.