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Here is another question from the book of V. Rohatgi and A. Saleh. I would like to ask help again. Here it goes:

An urn contains $r$ red and $g$ green marbles. A marble is drawn at random and its color noted. Then the marble drawn, together with $c > 0$ marbles of the same color, are returned to the urn. Suppose that $n$ such draws are made from the urn. Find the probability of selecting a red marble at any draw.

So here is what I have so far:

$$\text{P{a red ball in at least one of the $n$ draws}}=1-\text{P{green balls in all $n$ draws}}$$ I noted that $$\text{P{all green balls}}=\text{P{X$_1$$\;$=$\;$green}}\times{...}\times\text{P{X$_n$$\;$=$\;$green$\;$|$\;$X$_1$$\;$=$\;$${...}$$\;$=$\;$X$_{n-1}$=$\;$green}}$$ Thus, I calculated $\text{P{a red ball in at least one of the $n$ draws}}$ as $$1-\left(\frac{g}{r+g}\right)\left(\frac{g+c}{r+g+c}\right){...}\left(\frac{g+(n-1)c}{r+g+(n-1)c}\right)$$

I do not know if my way is correct or not. However, I feel like this is wrong since the text gave an answer of $$\frac{r}{r+g}$$ If that is the actual answer, can anyone help explain why? Thanks.

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The question is to get $$\text {The probability of selecting a red marble at any draw}$$ Not $$\text {The probability of selecting a a red marble in at least one of the $n$ draws}$$ The First time you draw a marble the Probability is: $${\text{Red Marbles}\over\text{Total marbles}}={\frac{r}{r+g}}$$
The Second time you draw a marble the Probability is: $${\frac{r}{r+g}}*{\frac{r+c}{r+g+c}}+{\frac{g}{r+g}}*{\frac{r}{r+g+c}}={\frac{r}{r+g}}$$
And so on...
As you can see the Probability does not depend on $\text{ n or c}$.

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    $\begingroup$ Oh so that was a trick question. I should've noticed. I was tricked by the $c<0$. Thanks a lot. $\endgroup$ – math_stat_enthusiast Oct 20 '13 at 7:06

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