Reverse the conditioning in a dice table

Use 36 equally likely ordered pairs. The first throw labels columns and the second labels rows; each table entry is their sum. Total 6 occurs at (1,5),(2,4),(3,3),(4,2),(5,1), so $P(\text{total six})=5/36$. The four displayed panels retain every sum-table entry and its relevant shading.

Columns 2,4,6 each contain 6 outcomes, giving 18 in total. Hence $P(\text{first even})=18/36=1/2$.

The two outcomes satisfying both conditions are (2,4) and (4,2), so the numerator from the outcome table is 2/36. The denominator is the result 1/2 from part (ii). Thus $P(\text{total six}\mid\text{first even})=(2/36)/(1/2)=1/9$. The source recommends this probability formula.

First identify the 18 even-first outcomes as the reduced sample space. Then mark the 2 total-six outcomes within it. Therefore the count ratio is 2/18=1/9. Follow the Teacher guidance: identify the reference set first, then its favourable subset. In the diagram grey/yellow marks the reference set and yellow marks favourable cells. Counting works here because those 18 outcomes are equally likely; it is not a universal shortcut for arbitrary outcome spaces.

The overlap is still 2/36, but the condition is now total six, with probability 5/36 from part (i). Hence $P(\text{first even}\mid\text{total six})=(2/36)/(5/36)=2/5$, incorporating the Teacher completion.

First identify the 5 total-six outcomes as the reference set, then the 2 even-first outcomes among them. The reduced-space count is 2/5. The same overlap produces a different conditional probability because the reference set changed.