So, there are n balls in a box, with some of them coloured white. If the first two you choose are white, how many white balls are there in the box, "most probably"?
Intuition tells me it's just n... for an introductory course in probability theory (for statistics students, i.e probability for the sake of probability), is that the simplest answer?
Knowledge of prior probability is required to solve this problem. Consider the proposition: \begin{align} N_k\equiv\textrm{There are $k$ number of white balls; }0\leq k\leq n \end{align} I shall assume uniform probability for number of white balls i.e. \begin{align} P(N_0)=P(N_1)=P(N_2)=...=P(N_n)=\frac{1}{n+1} \end{align} Now consider the proposition: \begin{align} W_j\equiv\textrm{$j$-th pick (without replacement) is a white ball; }1\leq j\leq n \end{align} In what follows product of propositions stands for logical AND between them. What you need is $P(N_k|W_1W_2)$ which is the probability that $i$ number of balls are white given that first two picks without replacement were both white.
It is easier to calculate the reverse probabilities: \begin{align} P(W_1|N_k)& =\frac{k}{n},\quad 0\leq k\leq n\\ P(W_2|W_1N_k)& =\frac{k-1}{n-1},\quad 1\leq k\leq n\\ P(W_2W_1|N_k) & = P(W_2|W_1N_k)P(W_1|N_k)=\frac{k(k-1)}{n(n-1)},\quad 0\leq k\leq n\\ P(W_1W_2)& =\sum_{k=0}^nP(N_k)P(W_2W_1|N_k)=\frac{1}{n+1}\sum_{k=2}^n\frac{k(k-1)}{n(n-1)}=\frac{1}{3} \end{align} where in the third and fourth equations I have used Bayes theorem to split up the probabilities.
Thus: \begin{align} P(N_kW_1W_2) & =P(W_1W_2)P(N_k|W_1W_2)=P(N_k)P(W_1W_2|N_k)\\ P(N_k|W_1W_2)&=\frac{P(N_k)P(W_1W_2|N_k)}{P(W_1W_2)}\\ &=\frac{3k(k-1)}{n(n-1)(n+1)} \end{align} Above is a monotonic increasing function of $k$. Therefore $k=n$ has the highest probability, equal to $3/(n+1)$. I don't know what your definition of "most probably" is, but in this problem most probable number of white balls is equal to the total number of balls $n$.
Let X equal the total number of balls in the box.
"Most probably" means more than 50%, and the question does not state that they are replaced, so you can say that n/X * (n-1)/(X-1) > 50, or (n^2-n)/(X^2-X) > 50, meaning that there are more white balls in the box than non white balls, and that n has to be greater than 1. Due to the lack of numbers that is as precise as you can be.
