Past exam of the mathematics course of the University of Cambridge / 2008 / ia / Paper 2 / 3F (Probability)

The sampling without replacement makes all $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ unordered pairs equally likely. The count of red-red pairs is $\left(\genfrac{}{}{0ex}{}{3}{2}\right)=3$, while the count of mixed pairs is $3(n-3)$. Thus their probabilities satisfyand cancellation gives $n-3=3$. Hence $\boxed{n=6}$, with three black socks. The count of black-black pairs is also $\left(\genfrac{}{}{0ex}{}{3}{2}\right)=3$, soThe mixed, red-red and black-black probabilities are respectively $3/5,1/5,1/5$, which add to one and verify the given ratio.