What independent events actually means? A bin has $2$ balls, one is black and one is white. Every round a uniformly chosen ball is drawn from the bin. If the color of the ball is white, then the ball is returned to the bin with an additional white ball. If the ball is black the experiment is over. Let $X$ be the number of rounds in the above experiment.
  
  
  Compute the distribution of $X$.
  


$$P(X=k)= ?$$

I have 3 questions regarding my (partial) solution.

My (partial) solution:

For $1&lt;=i&lt;=k$, I defined $X_i$ to be the number of white balls drawn in the $i$th round.


Is it valid to conclude without any explanation or proof that (and everything I did so far):
$$ P(X=k)=P(X_1=1,X_2=1,....X_{k-1}=1,X_k=0) $$
for every $1&lt;=i&lt;j&lt;=k$, $X_i$ and $X_j$ seems to be dependent because if I draw a white ball then in the next round, the probability to draw a white ball is smaller. Also, the only way to go to the next round is if we drew a white ball. Then, logically, why they are independent? (in the following section I'm trying to prove that they are independent mathematically).


I was trying to prove that for every $1&lt;=i&lt;j&lt;=k-1$, $X_i$ and $X_j$ are independent (Eventually, I will prove for $1&lt;=i&lt;j&lt;=k$).

$$ P(X_i=1,X_j=1)=P(X_j=1 | X_i=1)*P(X_i=1)=\frac{j}{j+1} * \frac{i}{i+1} $$
The above is true because $i&lt;j$, so we can conclude that because $P(X_i=1)$ then the experiment isn't over so the event $X_i=1$ doesn't have any infloence over $P(X_j=1)$.

But the opposite seems to me not true:
$$ P(X_i=1,X_j=1)=P(X_i=1 | X_j=1)*P(X_j=1)= 1 * \frac{j}{j+1} $$

The above is true because $i&lt;j$ so the experiment wasn't over at the round number $i$, so at the round number $i$ we must have drawn a white ball - and there is no other option.


$ P(X_i=1,X_j=1)=P(X_j=1 | X_i=1)*P(X_i=1)=P(X_i=1 | X_j=1)*P(X_j=1)$. Why am I wrong?
A difficulty encountered by your argument is that you do not have a clear definition of $X_j$ from the problem statement.

You seem to be tempted to say $X_j = 0$ if the experiment ends before the $j$th drawing.
But if there is no $j$th drawing, is "the number of white balls in the $j$th drawing" even defined?

One way to resolve this is to make it part of the definition of $X_j$ that $X_j = 0$ if the experiment ends before $j$ balls are drawn.
Then $X_i$ and $X_j$ clearly are not independent.
In particular, because $X_2 = 1 \implies X_1 = 1,$
$$P(X_1 = 1 \cap X_2 = 1) = P(X_2 = 1) \neq P(X_1 = 1)P(X_2 = 1).$$

Both of your attempts to evaluate $P(X_1 = 1 \cap X_2 = 1)$
by means of conditional probability were incorrect,
because you assumed that in general $P(X_i = 1) = \frac{i}{i+1}$
for every positive integer $i.$
In fact, according to the way you defined $X_i,$
$P(X_i = 1) &lt; \frac{i}{i+1}$ whenever $i &gt; 1.$

Here is a way out of this difficulty. The value of $X_j$ is irrelevant to the outcome of the experiment when there is no $j$th drawing.
So you can define $X_j$ any way you like in that case.
Define $X_j$ as follows for any positive integer $j$:
$$
X_j = \begin{cases}
1 &amp; \text{at least $j$ balls are drawn and the $j$th ball is white,} \\
0 &amp; \text{at least $j$ balls are drawn and the $j$th ball is black,} \\
1 &amp; \text{with probability $\frac{j}{j+1}$ if fewer than $j$ balls are drawn,}\\
0 &amp; \text{otherwise.}\\
\end{cases}
$$

When you define $X_j$ this way, it turns out that indeed 
$P(X_j = 1) = \frac{j}{j+1},$ 
and $X_j$ is independent of $X_i$ whenever $i \neq j.$

Now you can proceed to compute $P(X = k)$ without getting hung up on the dependence of $X_j$ on $X_i$ or on whether $X_j$ is even defined.
Yes, $P(X=k)=P(1, 1,\cdots1, 0)$, by definition.
Of course they are dependent, because $X_i=0\implies P(X_j=0)=P(X_j=1)=0$. And conversely $X_j=0\implies P(X_i=1)=1$, $X_j=1\implies P(X_i=1)=1$.


In this exercise, we have an unusual situation, as $X_j=0,X_j=1$ are not the only cases. $X_j\text{ not drawn}$ is also possible, with some probability.