Conditional Probability, Total Probability and Bayes' Rule

A two-way grid of counts, one row per value of the first variable and one column per value of the second. Interior cell $(i,j)$ carries the number of cases with both values; the right margin carries the row totals and the bottom margin the column totals, with the grand total in the corner. Dividing the interior by the grand total gives the joint distribution, the margins give the two marginal distributions, and dividing a cell by its own row or column total gives a conditional.

This reading answers questions about which set is being divided by. Conditioning becomes a physical operation, cover the rows that did not occur, then renormalise what is left, rather than a formula to recall. Because $P(A\mid B)$ divides by a row total and $P(B\mid A)$ by a column total, the asymmetry of conditioning is visible as two fractions sharing a numerator: the same 45 students over 60 in one direction and over 55 in the other.

It is also the form that makes base rates impossible to overlook. A rare condition is a short row, so the false positives drawn from the long row can outnumber the true positives even when each is individually unlikely, and the comparison is a glance at two cells rather than an inference from percentages. Counting a concrete population, 10,000 people rather than probabilities, turns Bayes' rule into one count divided by a total of counts.

Independence is equally direct: the variables are independent exactly when every interior cell equals its row total times its column total divided by the grand total, so a single cell disagreeing with that product settles the question.

What the table does not expose is anything beyond two discrete variables with few values. Continuous quantities have no cells, a third variable has no axis, and the layout silently suggests that every probability is a tally waiting to be counted, which is false for a genuinely unrepeatable event, where the counts that would fill the grid do not exist.

The student table is one shape of problem. Two more, to separate the method from the example.

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1. A partition with three cases. A factory takes components from three suppliers. Supplier A provides 50% of them and 2% of those are defective; B provides 30% at 4% defective; C provides 20% at 5% defective.

The overall defect rate, by the law of total probability over the partition $\{A,B,C\}$:

The three cases are mutually exclusive and cover every component, which is what licenses the sum.

Reversing the conditioning. A defective component is found. Which supplier is it most likely from?

summing to 1. B is the most likely source despite not having the worst defect rate, because it supplies half again as many components as C. And A, with the best rate, is exactly as likely a source as C with the worst, because it ships two and a half times the volume. Neither the rate nor the volume decides this alone; the product does.

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2. A table with a different shape. 200 job applicants, classified by whether they were referred by an employee and whether they were hired.

Referral doubles the hire rate. But reversing the conditioning asks a different question:

so most hires were not referred, even though referral doubled an individual's chances. Both statements are true of the same table. The first divides by a row, the second by a column, and they answer different questions: one is about an applicant's prospects, the other about the composition of the hired group.

Dependence check. $P(\text{referred})P(\text{hired})=0.30\times 0.26=0.078$ against the actual joint $\frac{24}{200}=0.12$. Not equal, so the two are dependent.

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What carries across both. The arithmetic never changed: identify the conditioning event, restrict to it, renormalise. What changed was which quantity the question wanted. In the supplier case the tempting error is to answer with the defect rates; in the hiring case it is to read a doubled individual rate as a statement about who the hires are. Both errors are the same error, answering the conditional that was easiest to see rather than the one that was asked.

$P(A\mid B)$ and $P(B\mid A)$ describe the same relationship, so a test that detects a condition 99% of the time makes a positive result 99% likely to be correct.