Probability minimum value question A bag contains some coloured counters, some of these counters are black. You remove two counters from the bag and they are both black.  The probability of you doing this was 132/380.  What is the minimum number of counters in the bag and how many of these are black?

Let us assume there are $n$ counters, $k$ of which are black. The probability of removing a black counter on the first turn equals $\frac{k}{n}$, and the probability of removing a black counter on the second turn (given a successful first turn) equals $\frac{k-1}{n-1}$. In order for $k (k-1)$ to equal 132 and $n (n-1)$ to equal 380, $k$ must be 12 and $n$ must be 20. Indeed:
$$\frac{12}{20} \cdot \frac{11}{19} = \frac{132}{380}$$
