# How many x-coloured balls are there in the box "most probably"?

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?

• A technical proof of that result would indeed seem to be too much to ask of an introductory student, but if you could add some conceptual explanation, that would probably really help. Commented Oct 9, 2017 at 18:36

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.
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 $n$.