pka
Elite Member
- Joined
- Jan 29, 2005
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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.
pkbgamma meant something different from my reading.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?
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.
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