Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2012 / Paper 2 / 18

There are eighteen equally likely first choices. Counting the relevant subsets givesFor conditional probabilities, restrict the denominator to the conditioning subset:The five counts come from the same finite sample space, so no independence assumption is involved.

Insert the values from (a) into Bayes' theorem:Thus the identity is satisfied. Equivalently, both $P(A\mid B)P(B)$ and $P(B\mid A)P(A)$ equal the joint probability $4/9$.

Condition on whether the first selection was a red apple. Its probability is $3/18$. If it was, two red apples remain among seventeen fruits; otherwise three remain. The law of total probability givesThe same result follows from exchangeability: before either selection is observed, each ordered draw position has the original red-apple fraction $3/18$, despite dependence between the draws.

Let $C$ denote a green pear on the first draw and $D$ a red apple on the second. The two categories are disjoint, so after $C$ there are still three red apples among seventeen remaining fruits. Thus $P(C\cap D)=(8/18)(3/17)$. Using $P(D)=1/6$ from (c), conditional probability givesEquivalently, conditioning on the second fruit being one particular category removes a red apple and leaves the other seventeen fruits symmetrically eligible for the first position.