Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 2 / 16 / a / i / Solution

Let $X$ count green balls. Sampling without replacement gives a hypergeometric distribution. The possible values are $0,1,2$, and counting equally likely three-element samples givesThereforeTheir sum is $(1+6+3)/10=1$, as required for a probability distribution.