Q.Using the bivariate frequency table of Worked Example 1, find the conditional distribution of $y$ given $x = 2$, and express it in relative (proportional) form.

A conditional distribution of $y$ given a fixed $x$ is just that one row of the table, read on its own.

Row $x=2$: $y=1 \to 2$, $y=2 \to 2$, $y=3 \to 1$, total $5$.

Relative form: $\tfrac{2}{5}, \tfrac{2}{5}, \tfrac{1}{5}$ i.e. $0.4, 0.4, 0.2$.

Step 1 — Isolate the row for the fixed value $x = 2$. From the table of Worked Example 1, the row $x = 2$ reads:

These frequencies alone are the conditional distribution of $y$ given $x = 2$ — they describe how $y$ is spread out among only the 5 units that have $x = 2$.

Step 2 — Convert to relative (proportional) form. Divide each frequency by the row total, 5:

Independent check. The proportions add to $0.4 + 0.4 + 0.2 = 1.0$, exactly as any complete relative distribution must — confirming the division by the correct total (5, the row total, not the grand total 12).

The conditional distribution answers a different question from the marginal distribution of $y$ (column totals $4, 5, 3$): the marginal describes $y$ over all 12 units, while this conditional describes $y$ over only the 5 units with $x = 2$. Comparing a conditional distribution against the marginal is, informally, how one begins to see whether $x$ and $y$ are related.

Dividing the row frequencies by the grand total (12) instead of the row total (5) is the classic error — a conditional distribution is always relative to the total of its own row or column, so its proportions must sum to 1, not to the row total over $N$. Reading a column instead of a row (which would give the conditional distribution of $x$ given a fixed $y$) also answers the wrong question.