Let's assume we have $n$ token consisting of red($R$) and blue($B$) colors with $|R|=n_1$ and $|B|=n_2$ ($n=n_1+n_2$). If we draw one token each time without replacement and repeat this process $T$ times, how many $R$ and how many $B$ tokens are expected to be drawn?
I tried to solve this but the conditional probability tree grows exponentially so that in $T$'th round I end-up with $2^T$ events (while the probability of most of them are $0$ because $r>=n_1$ or $b>=n_2$). Is there another method to solve this in less computational complexity? Also please explain how to solve a similar problem with more than two tokens.
For $k=1,\dots,T$, let $X_K$ be a random variable whose value is $1$ if draw number $k$ is red and $0$ if it is blue. The expected number of red balls is $$E\left(\sum_{k=1}^T X_k\right)=\sum_{k=1}^TE(X_k)=\frac{Tn_1}{n_1+n_2}$$ because $E(X_k)=\frac{n_1}{n_1+n_2}$ for every $k$.
What you say in a comment about the changing probability as balls are drawn is a statement about the conditional probability, but it's easier to just use linearity of expectation, as above.
If this isn't clear, imagine that instead of colored balls, we have cards with red and blue backs. The cards are shuffled and $T$ of them are dealt face-up in a line. What is the probability that the $k$th card in the line has a red back?
If you want to do it more formally, compute the probability that $X_2=1.$ Remember, $$\begin{align}\Pr(X_2=1)=&\Pr(X_2=1|X_1=1)\Pr(X_1=1)+\\&\Pr(X_2=1|X_1=0)\Pr(X_1=0)\end{align}$$