Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / 16

Uniform sampling without replacement is equivalent to revealing a uniformly random permutation of seven labelled balls. Each draw position is equally likely to contain any label, so its colour has the original proportions. ThusFor example the law of total probability gives $\mathbb{P}(B_2)=(4/7)(3/6)+(3/7)(4/6)=4/7$. Equal marginal probabilities express exchangeability, rather than independence between successive draws.

The first black draw has probability $4/7$. Given that event, three of the remaining six balls are black. The multiplication rule therefore givesUsing $(4/7{)}^2$ would incorrectly treat the draws as independent.

Any two distinct draw positions have the same pair law by exchangeability. The marginal and pair probabilities in sampling without replacement therefore giveDivide this joint probability by the relevant marginal probability to get the two conditional probabilities:Conditioning on the third draw can change the probability of the second draw; conditional probability does not require the conditioning event to occur earlier in time.

Use the inclusion-exclusion principle for two events:As a check, its complement is two white draws, with probability $(3/7)(2/6)=1/7$.

Let $I_j$ be the indicator random variable of a black ball at draw $j$. The black count after three draws is $N(B_3)=I_1+I_2+I_3$. By linearity of expectation,Independence is not required for this step. This is also the mean of the corresponding hypergeometric distribution.

Distinguish the $N$ balls by labels. For a fixed position $n$, each label appears there in exactly $(N-1)!$ of the $N!$ equally likely permutations. Since $N_B$ labels are black,This is the same marginal law at every position, even the last one. The argument averages over earlier draw outcomes; it does not assert that a conditional probability after a specified history remains $N_B/N$. The original PDF uses $B_n$, correcting the transcription's uppercase subscript $B_N$.