Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / Paper 2 / 12

Write $N=g+r$. Successive conditional probabilities account for the changing bag under sampling without replacement:On this event no equality of the two colour counts has occurred, so the game's stopping rule does not interrupt these three draws.

After one draw the two counts are either $(1,0)$ or $(0,1)$ and cannot agree. ThereforeThis is the elementary parity restriction behind the first return time interpretation of this game.

Success after two draws occurs on the two mutually exclusive events $GR$ and $RG$. With $N=g+r$, sampling without replacement gives

The equality must be a first return time: an order starting $GR$ or $RG$ would already have stopped at draw two. The only successful four-draw orders are therefore $GGRR$ and $RRGG$. Their probabilities under sampling without replacement add toHere $N=g+r$. Counting all balanced four-draw orders would include games that have already stopped.

The two possible colour orders are $GGRR$ and $RRGG$. Thus the answer isThey correspond to the positive and negative four-step excursions with first return time four.

Suppose the first draw is green. The excess of green over red must remain positive after draws one through five and return to zero at draw six. After deleting the first green and final red, the remaining four steps must form a Dyck path: they have equal colour counts and never go below zero relative to the new starting level. There are $C_2=2$ such orders, $GGRR$ and $GRGR$, giving $GGGRRR$ and $GGRGRR$ in the original game.

Interchanging green and red gives $RRRGGG$ and $RRGRGG$. Thus there areorders. More generally, the Catalan number $C_{m-1}$ counts each sign of excursion with first return time $2m$.

Let $X_1$ and $X_2$ denote the first and second labels. Sampling with replacement restores the bag before the second draw, so conditioning on $X_1=x$ only specifies the threshold. For $1\le x\le g$, there are $x$ qualifying green balls and $x$ qualifying red balls. For $g<x\le r$, every green ball and the first $x$ red balls qualify. Hence the requested conditional probability isThe formulas agree at $x=g$ and give one at $x=r$. Labels outside the possible first-draw range do not define a conditioning event of positive probability.

Only a green ball carries label four. After it is removed, nine of the remaining nineteen balls are green. The conditional probability is

The specified colour tells us that one green ball has been removed. Thus sampling without replacement again leaves nine green balls among nineteen:The numerical label supplies no further change once this first colour is known.

There are two equally likely balls bearing label eight, one of each colour. The conditional probability that the removed ball was green is therefore $1/2$. By the law of total probability,Both possible first colours must be included because the observed label alone does not determine the colour.