There are $n$ white balls in an urn. We draw one ball with replacement. If it is white, we paint it red before putting it back. Let $X$ be the random variable representing the number of red balls in the urn after $k$ draws. Find the expected value $\mathbb{E}[X]$.
Let $X_i$ be a random variable such that:$$X_i = \begin{cases} 1 & \text{if a white ball is drawn in the } i\text{-th draw}, \\0 & \text{if a red ball is drawn in the } i\text{-th draw}.\end{cases}$$Then $X = X_1 + X_2 + \ldots + X_{k}$.
By the linearity of expectation, we have:$$ \mathbb{E}[X] = \mathbb{E}[X_1] + \mathbb{E}[X_2] + \ldots + \mathbb{E}[X_{k}]. $$
For each $i \leq k$, we have:$$ \mathbb{E}[X_i] = 1 \cdot P(X_i = 1) + 0 \cdot P(X_i = 0) = P(X_i = 1), $$so$$ \mathbb{E}[X] = \sum_{i=1}^{k} P(X_i = 1). $$
Let $R_i$ denote the number of red balls in the urn after the $i$-th draw. Note that $R_i = X_1 + X_2 + \ldots + X_{i-1}$ and $R_1 = 0$, hence:$$ P(X_i = 1) = \frac{n - R_{i-1}}{n} = 1 - \frac{ \sum_{j=1}^{i-1} X_j}{n}. $$
However, looking for $R_i$ brings us back to the original problem. On the other hand, it is the only approach I have come up with. My main issue is finding $P(X_i = 1)$. Thank you in advance for your help.
