Conditional probability and the multiplication rule

Shrinking the sample space

Maple Ledger's partner already knows the drawn invoice was late. What is the chance it also has an error? Knowing "late" shrinks the sample space from 50 invoices to the 12 late ones; three of those have an error:

$$P(\text{error}\mid \text{late})=\frac{3}{12}=0.25$$

Read the bar as "given". The event after the bar is the condition; its row (or column) becomes the new denominator. In probabilities rather than counts,

$$P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{0.06}{0.24}=0.25$$

The order matters

Reverse the condition and the denominator changes:

$$P(\text{late}\mid \text{error})=\frac{3}{8}=0.375$$

Same overlap, different conditioning row, different answer. Among late invoices a quarter have errors; among erroneous invoices three-eighths are late. The paper will hand you one and ask for the other (L5.6); the first defence is to write the condition down before dividing.

The error the marker sees most

$P(\text{error}\mid \text{late})=3/50=0.06$ divides by the grand total. That is the joint probability $P(\text{late}\cap \text{error})$, not the conditional. The denominator of a conditional is never the whole table — it is the row or column of the condition.

The multiplication rule

Rearrange the definition and the joint probability is a product:

$$P(A\cap B)=P(B)P(A\mid B)$$

This is the engine for sequential draws. Two invoices are drawn without replacement. The chance both are late:

$$P(\text{both late})=P(\text{1st late})\times P(\text{2nd late}\mid \text{1st late})=\frac{12}{50}\times \frac{11}{49}=\frac{132}{2,450}=0.0539$$

After one late invoice leaves the file, 11 late remain among 49. With replacement the file is restored between draws, the second probability is again $12/50$, and

$$P(\text{both late})=0.24\times 0.24=0.0576$$

The two differ by a little here and by a lot when the population is small. The complement from L5.1 now has its partner: $P(\text{at least one late})=1-P(\text{neither late})=1-\frac{38}{50}\times \frac{37}{49}=1-0.5739=0.4261$ without replacement.

Writing it on the paper

A conditional answer earns its marks in three lines: the condition named, the denominator as the count (or probability) of that condition, the division. "$3/12=0.25$" with the words "given late" is full marks; "$0.25$" alone is not.