Contingency tables and tree diagrams

The table as the engine

Every rule so far is a division of table entries. A contingency table (two-way table) lays two categorical variables against each other; the cells hold counts, the margins hold totals, and three kinds of probability read straight off it:

• Joint — a cell over the grand total: $P(A\cap B)$.
• Marginal — a row or column total over the grand total: $P(A)$.
• Conditional — a cell over its row or column total: $P(A\mid B)$.

Worked example — Cedar Valley Credit Union's 400 applications

Cedar Valley Credit Union took 400 loan applications last quarter: 200 in branch, 200 online. Of the branch applications 150 were approved and 50 declined; of the online ones 120 were approved and 80 declined.

              approved   declined   total
  branch         150         50       200
  online         120         80       200
  total          270        130       400

Fill the margins first — the column totals 270 and 130 are the step most learners skip and then need.

• $P(\text{approved})=270/400=0.675$ — marginal.
• $P(\text{online}\cap \text{declined})=80/400=0.20$ — joint.
• $P(\text{declined}\mid \text{online})=80/200=0.40$ — conditional; the online row.
• $P(\text{online}\mid \text{approved})=120/270=0.444$ — conditional; the approved column.

The last two use different denominators for the same 120-and-80 numbers. Say the condition aloud, find its total, divide.

The tree as the engine

When the information arrives as a sequence of conditionals rather than as counts, a tree diagram is the layout. Georgian Bay Mutual classifies claims: 60 % are auto, the rest property. Among auto claims 10 % are flagged fraudulent; among property claims, 4 %.

First split: auto 0.60, property 0.40. Second split, on each branch: fraud 0.10 / not 0.90 for auto; fraud 0.04 / not 0.96 for property. Each leaf is a joint probability, found by multiplying along the branch:

$$P(\text{auto}\cap \text{fraud})=0.60\times 0.10=0.060,P(\text{property}\cap \text{fraud})=0.40\times 0.04=0.016$$

The four leaves — 0.060, 0.540, 0.016, 0.384 — sum to 1. A marginal probability is found by adding across branches that end the same way:

$$P(\text{fraud})=0.060+0.016=0.076$$

This is the total probability of fraud: 7.6 % of all claims, a weighted average of the two rates with the claim-type shares as weights. A tree is a table drawn as a sequence, and either one can be converted into the other.

Which to draw

Counts in a cross-classified sample → table. A first-stage split followed by rates within each group → tree. If the question asks you to reverse a condition — a flagged claim, which type is it? — the tree's leaves are exactly what you need, and Lesson 5.6 finishes the job.