Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 1 / 16

Assume the three standard dice are fair and independent. For one die $X$, a discrete uniform distribution on $1,\dots ,6$ givesThe variance is $\mathrm{Var}X=\mathbb{E}{X}^2-(\mathbb{E}X{)}^2=35/12$. For the sum $Q=X_1+X_2+X_3$, linearity of expectation and addition of variances for independent random variables give

The sample space has $6^3=216$ equally likely ordered outcomes. The repeated value can be any of the six faces, giving six outcomes $(j,j,j)$. Therefore

Choose the first face in six ways, the second in five ways avoiding the first, and the third in four ways avoiding both. ThusEach ordered outcome is equally likely in this sample space, so dividing the count by $216$ gives the probability.

To find the probability of exactly one pair, choose its repeated face in six ways, the distinct face in five ways, and which of the three dice has that distinct face in three ways. HenceAs a check, the disjoint events $S,D,T$ exhaust the sample space, and $1/36+5/9+5/12=1$.

The sample space outcomes with qualifying sums have totals three, four and five. Their ordered patterns are $(1,1,1)$, the three permutations of $(1,1,2)$, and the three permutations each of $(1,1,3)$ and $(1,2,2)$. Thus there are $1+3+3+3=10$ qualifying outcomes:

Of the ten outcomes in $F$, all except $(1,1,1)$ have exactly one repeated pair. Hence $P(T\mid F)=9/10$. Applying Bayes' theorem in the requested direction,Equivalently there are nine favorable outcomes among the ninety outcomes in $T$. This direct count confirms the conditional probability calculation.