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

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.

Compute the first two moments from the probability distribution:The variance is $E[X^2]-E[X{]}^2=9/5-36/25=9/25$, so the expected value and standard deviation are

The interval within one standard deviation is $[3/5,9/5]$. Of the possible integer outcomes $0,1,2$, only $1$ lies in this interval. Hence

Three draws cannot all be green because there are only two green balls. Thus the probability of at least one red ball is $1$. At least two red balls means $X\le 1$, so

Let $Y$ indicate whether the draw from the second bag is green. It is independent of $X$, with $P(Y=1)=1/3$ and $P(Y=0)=2/3$. Exactly one green overall arises from the disjoint alternatives $(X,Y)=(1,0)$ or $(0,1)$. Therefore

The sole green ball comes from the first bag exactly on the event $(X,Y)=(1,0)$. Divide this joint probability by the conditioning event's probability:This is the conditional probability form of Bayes' theorem.