I have a sack of $b$ balls, $w$ of which are white. I uniformly and randomly throw exactly $t$ $(0 \leq t < b)$ of the balls away. I then randomly select one of the remaining $b-t$ balls.
What is the probability that I have selected a white ball?
Based off the information I am given, since there is an equal chance of $b$, $t$, $w$ being any number, I have a sense that the answer is $\frac{w}{b}$, but I am lacking the intuition behind this answer. Any guidance would be greatly appreciated.
Your intuition is correct, as your method is equivalent to pick a single ball from the whole sack. If you drew all the balls in order, each (such as the $t+1$th ball) would have probability $\frac{w}{b}$ of being white, even though these events would not be independent
The complicated calculation would be to consider the possibility that $v$ of the first $t$ were white and $t-v$ were black. The probability of this would be $\dfrac{{w \choose v}{b-w \choose t-v}}{b \choose t}$. The conditional probability that the $t+1$th ball would then be white would be $\frac{w-v}{b-t}$. So the overall probability that the $t+1$th ball is white is $$\sum_v\frac{{w \choose v}{b-w \choose t-v}}{b \choose t} \frac{w-v}{b-t}$$ and this can be rewritten as $$=\sum_v\frac{w{w-1 \choose v}{b-w \choose t-v}}{b{b-1 \choose t}}=\frac{w}{b}\sum_v\frac{{w-1 \choose v}{b-w \choose t-v}}{{b-1 \choose t}} = \frac{w}{b} $$