Past exam of the mathematics course of the University of Cambridge / 2002 / ia / Paper 2 / 9F (Probability)

For $\mathbb{P}(B)>0$, the conditional probability is $\mathbb{P}(A\mid B)=\mathbb{P}(A\cap B)/\mathbb{P}(B)$. Since the events $B_i$ partition the sample space, their intersections with $A$ are disjoint and exhaust $A$. The law of total probability therefore gives $\mathbb{P}(A)={\sum }_j\mathbb{P}(A\mid B_j)\mathbb{P}(B_j)$. Substituting this and $\mathbb{P}(A\cap B_i)=\mathbb{P}(A\mid B_i)\mathbb{P}(B_i)$ into the definition proves Bayes' theorem:The denominator is positive by the given hypothesis on $A$.

Every urn contains $n-1$ balls. Let $k=n-r$ be its number of blue balls; choosing an urn uniformly makes $k$ uniform on $0,\dots ,n-1$. Conditional on that urn, the probability of a blue first draw is $k/(n-1)$, so the law of total probability givesDrawing twice without replacement requires $n\ge 3$. Given $k$, the probability that both draws are blue is $k(k-1)/((n-1)(n-2))$, and henceTaking the ratio defining conditional probability gives the second requested probabilityAlthough a blue first draw removes a blue ball from its urn, it also makes an urn rich in blue balls more likely. Averaging with this changed conditional distribution is essential.

Two independent events satisfy $\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)$. If $\mathbb{P}(B)>0$, this is equivalent to $\mathbb{P}(A\mid B)=\mathbb{P}(A)$: learning that $B$ occurred does not change the probability of $A$. The product definition also covers zero-probability events without dividing by zero.

The 36 ordered outcomes are equiprobable. For $2\le s\le 12$, let $c_s=6-|s-7|$ count outcomes with sum $s$. Then $\mathbb{P}(A_s)=c_s/36$ and $\mathbb{P}(B_i)=1/6$ for $1\le i\le 6$. Their joint probability is $1/36$ if $1\le s-i\le 6$, and zero otherwise.

If the joint probability is zero, independence is impossible in these ranges because both marginal probabilities are positive. If it is $1/36$, the independent events condition requires $1/36=c_s/216$, or $c_s=6$. This happens only for $s=7$, and then all six values of $i$ allow the partner $7-i$. Therefore within the possible sums and facesIf arbitrary integer labels outside these ranges are included, an impossible sum or face defines an empty event, which is trivially independent of every event.