Urn and dependent events with same probability Say I have a urn with $4$ green balls, $5$ red balls and $6$ yellow balls. Now I extract $5$ balls, one after another. Which is the probability of having drawn a green ball for every different draw?
I know that all probabilities are the same : $\frac{4}{15}$. I'm struggling to show why though.
My attempt :
Define the event $E_k = \{\text{extract a green ball at k-draw\}}$ with $k \in \{1,2,3,4,5\}$.
Clearly $E_1 = \frac{4}{15}$ and $E_2=\frac{4}{15}\cdot\frac{3}{14}+\frac{11}{15}\cdot\frac{4}{14}=\frac{4}{15}$.
Now I can express the event $E_k$ by saying that we must have less than $4$ green balls drawn in the first $k-1$ extractions. Then I can say I have a hypergeometric distribution for every value $p$ such that $0\leq p \leq \min(3,k-1)$ , so :
$$P(\text{drawing } p \text{ green balls in } k-1 \text{ draws}) = \frac{\binom{4}{p} \binom{15 - 4}{k-p-1}}{\binom{11}{k-1}}$$
Then I have to draw a green ball at $k$ extraction, which happens with probability $\frac{4-p}{15-k+1}$.
So putting all together :
$$P(E_K)=\sum_{p=0}^{\min(3,k-1)}\frac{\binom{4}{p} \binom{11}{k-p-1}(4-p)}{\binom{11}{k-1}(16-k)}$$
But now, I'm stuck , I don't know how to simplify this sum.
Is my reasoning correct? Is there a way to simplify the sum ? How can it be solved in other better ways?
 A: Draw all the balls out.  What is the probability that the $k$th ball was green?  There are $\binom{15}4$ equally likely ways to choose the positions of the green balls, and in $\binom{14}3$ of them, ball $k$ is green.
We have $$\frac{\binom{14}3}{\binom{15}4}=\frac4{15}$$
EDIT
In response to the OP's comment.
Actually, my initial solution was something like this: "The $k$th ball is equally likely to be any of the $15$ balls, so the question is, 'What is the probability that a randomly-selected ball is green,' and the answer is clearly $\frac4{15}$." That is, choosing the $k$th ball is just a mechanism for picking a ball at uniformly at random.
However from your question, I thought you wanted something more computational, and I thought somewhat along the following lines.  First, we only care about the green balls, so concentrate on them.  Second, since the answer should be the same for any $k$, look for a solution that doesn't really depend on $k$.  Frankly, having read your question, I started for a moment along the same lines, but then I thought, "This isn't symmetric enough."  Once I thought that, I immediately saw the argument above.
