certain colour ball being drawn second There is a bag with 410 balls in, there are 200 red balls, 200 green balls and 10 black balls. If a ball is taken out of the bag, all balls of that colour are taken out. What is the probability of the black ball being drawn second?, what formula would be used to generalise this solution too?
Thank you for the help (this is my first post)
I dont feel like i can presume the chance remains constant,
so far I have been going down the route of
1-P(drawing Black first) * ((1/40)/((1/40)+(1/2))) which I still think is way off.
Now i am thinking that I should have done the second half of this twice, so now i would be left with:
1-p(drawing black) * (((1/40)/((1/40)+(1/2)))+((1/40)/((1/40)+(1/2))))
This would leave me with
.975*(.0469*.0469)
.975*.0938
this leaves me with about 9.14%
 A: P(black after green)$=\frac{200}{410}\times \frac{10}{210}$=P(black after red)
P(second is black)=$2 \times \frac{200}{410}\times \frac{10}{210}$
If there are x-black, y-green and z- red balls
P(black after green)$=\frac{y}{x+y+z}\times \frac{x}{x+z}$
P(black after red)$=\frac{z}{x+y+z}\times \frac{x}{x+y}$
P(second is black)=$\frac{y}{x+y+z}\times \frac{x}{x+z}+\frac{z}{x+y+z}\times \frac{x}{x+y}$
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A: From comment/question of OP (i.e. original poster) following answer of Lion Heart.

So say if this same question went to 10 different coloured balls, with different probabilities, would there be any way to do this without having to loop through combination of another colour before my desired colour?

First, see the comment of lulu, which follows your original question.  This response is a generalization of her comment.
Suppose that you have $10$ different colors, $c_1, c_2, \cdots, c_{(10)},$ with respective probabilities of $p_1, p_2, \cdots, p_{(10)}$ so that $p_1 + p_2 + \cdots + p_{(10)} = 1.$  Suppose further that you want to know the probability that color $c_1$ is drawn on the 2nd ball.  Then, two things have to happen.  Color $c_1$ can not be drawn on the first ball.  Then, color $c_1$ must be drawn on the 2nd ball.
Certainly, the probability that color $c_1$ is not drawn on the 1st ball, is easily expressed as $(1 - p_1)$.  The difficulty is that the probability of color $c_1$ being drawn on the 2nd ball will depend on which color was drawn on the 1st ball.  Note that in your original posed problem, it was specified (in effect) that $p_2 = p_3$, because there were the same number of red and green balls.  So, in your original problem, the computation was simpler.
Let $k$ be any element in $\{2,3,\cdots,10\}.$ 
The probability of color $c_k$ being drawn on the 1st ball is $p_k$. 
Then, assuming that color $c_k$ is drawn on the 1st ball, the chance of color $c_1$ being drawn on the 2nd ball is 
$\displaystyle \frac{p_1}{1 - p_k}.$
Therefore, the overall probability of color $c_1$ being drawn on the 2nd ball is
$$p_1 \times \left[\sum_{k=2}^{10} \frac{p_k}{1 - p_k}\right]. \tag1 $$
So, the underlying question in your comment/question seems to be: can the RHS factor in (1) above be simplified?
$$\frac{p_k}{1 - p_k} = \frac{1}{1 - p_k} - 1. \tag2 $$
Substituting (2) above back in to (1) above gives
$$p_1 \times \left[\sum_{k=2}^{10} \left(\frac{1}{1 - p_k} - 1\right)\right] = 
p_1 \times \left\{ ~\left[\sum_{k=2}^{10} \frac{1}{1 - p_k} \right] - 9 ~\right\}. \tag3 $$
Personally, I don't see how the (final) RHS factor in (3) above can be simplified further.  That is, I don't see how the reciprocals of $(1 - p_k)$ can be simplified.
As a concrete example, suppose that there are exactly $4$ colors, instead of $10$ colors.  Further suppose that $p_1 = (1/4)$, and consider the following two cases:
$\underline{\text{Case 1}}$
$p_2 = p_3 = p_4 = (1/4).$ 
Then $~\displaystyle \sum_{k=2}^4 \frac{1}{1 - p_k} = \frac{4}{3} \times 3 = 4.$
$\underline{\text{Case 2}}$
$p_2 = p_3 = (1/3), p_4 = (1/12).$ 
Then $~\displaystyle \sum_{k=2}^4 \frac{1}{1 - p_k} = \frac{3}{2} + \frac{3}{2} + \frac{12}{11} = \frac{45}{11}.$
What the Case 1 / Case 2 contrast indicates is that the sum of the corresponding reciprocals is not dependent on $p_1$ alone.  Therefore, the reciprocals must be looped through.
