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The conditional probability rule

Covers AS91585

What conditioning does

A conditional probability is the probability of one event given that another has already happened, or is known to be true.
Written $P(A \mid B)$ and read "the probability of $A$ given $B$ ".
Conditioning shrinks the sample space. Instead of considering everyone, you consider only the members of $B$ — and ask what fraction of those are also in $A$ .

The formula

Given in the Formulae and Tables Booklet:

$P(A \mid B) = \frac{P(A \text{ and } B)}{P(B)}$

The denominator is the conditioning event. Whatever comes after the vertical bar goes on the bottom.
Rearranged, this is the multiplication rule used along a tree path:

$P(A \text{ and } B) = P(B) \times P(A \mid B)$

Both forms are worth knowing: the first extracts a conditional from joint information, the second builds joint information from conditionals.

Reading the question

The bar is in the opposite order to the way English states it, which is a constant source of error:

In words	Symbols
"Given that it rained, the ferry was cancelled"	$P(\text{cancelled} \mid \text{rain})$
"Of the students who cycle, how many are in Year 13?"	$P(\text{Year 13} \mid \text{cycle})$
"Among smokers, the rate of the disease is…"	$P(\text{disease} \mid \text{smoker})$
"What proportion of those with the disease smoke?"	$P(\text{smoker} \mid \text{disease})$

The conditioning event is the group being described, not the outcome being counted. Underline it before you write anything.

Getting a conditional from each representation

Two-way table: the cell for " $A$ and $B$ " divided by the total for $B$ (its row or column total).
Venn diagram: the intersection region divided by everything inside the $B$ circle.
Tree: it is already there — the second-stage branches are conditional probabilities, provided the conditioning event is the first stage.
From probabilities alone: apply the formula directly.

Conditioning and independence

If $A$ and $B$ are independent, then conditioning changes nothing:

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

So a quick way to test independence is to compare a conditional with its marginal. If they are equal, the events are independent; if $P(A \mid B) > P(A)$ , knowing $B$ makes $A$ more likely.

The chain rule for three stages

The multiplication rule extends:

$P(A \text{ and } B \text{ and } C) = P(A) \times P(B \mid A) \times P(C \mid A \text{ and } B)$

This is exactly what multiplying along a three-stage tree does, which is why trees handle sequential problems so cleanly.

The law of total probability

To find $P(A)$ when you only know it conditionally, add over the ways $A$ can happen:

$P(A) = P(A \mid B) \times P(B) + P(A \mid B') \times P(B')$

On a tree this is simply adding the path probabilities that end in $A$ . You do not need to memorise the formula if you draw the tree — but recognise it when a question presents the information without a diagram.

Tips and Tricks

Whatever follows the bar is the denominator. Write $P(\_ \mid \_)$ with the conditioning event filled in first, before you look for numbers.

Check the answer's size against its marginal. If you calculate $P(\text{disease} \mid \text{positive test})$ and get a number smaller than $P(\text{disease})$ , you have inverted something.

A conditional probability is not symmetric. $P(A \mid B)$ and $P(B \mid A)$ are different numbers with different meanings, and confusing them is the single most examined error in this standard — see the next page.

State the answer with its group named: "of the residents with children at home, 66% support the extension" rather than a bare 0.66. Relating findings to the context is explicitly part of the Merit criterion.

Worked ExampleConditional probabilities from a description

A courier company finds that 12% of its parcels are sent by the overnight service, the rest by standard. Of the overnight parcels, 3% arrive late. Of the standard parcels, 9% arrive late.

(a) Find the probability a randomly chosen parcel is sent overnight and arrives late. (b) Find the probability a randomly chosen parcel arrives late. (c) Find the probability a parcel arrives on time, given it was sent by the standard service. (d) The company advertises "97% of overnight parcels arrive on time". Comment on whether this is consistent with the figures, and on what it omits.

(a) Overnight and late — the multiplication rule

Step 1 — Identify what is given. The 3% is a conditional probability: it applies to overnight parcels only.

$P(\text{overnight}) = 0.12 \qquad P(\text{late} \mid \text{overnight}) = 0.03$

Step 2 — Apply the multiplication rule.

$P(\text{overnight and late}) = P(\text{overnight}) \times P(\text{late} \mid \text{overnight})$

$= 0.12 \times 0.03 = 0.0036$

Note how much smaller this is than the 3% figure. The 3% applies within the overnight group; only 12% of parcels are in that group.

(b) Late overall — the law of total probability

Step 1 — Identify the two ways a parcel can be late.

$P(\text{standard}) = 1 - 0.12 = 0.88 \qquad P(\text{late} \mid \text{standard}) = 0.09$

Step 2 — Compute both path probabilities.

$P(\text{overnight and late}) = 0.12 \times 0.03 = 0.0036$

$P(\text{standard and late}) = 0.88 \times 0.09 = 0.0792$

Step 3 — Add.

$P(\text{late}) = 0.0036 + 0.0792 = 0.0828$

Observe that this is much closer to the standard service's 9% than to the overnight service's 3%, because standard parcels make up 88% of the total. The overall rate is a weighted average, dominated by the larger group.

(c) On time given standard

Step 1 — Use the complement within the conditioning group. Given a parcel is standard, it is either late or on time:

$P(\text{on time} \mid \text{standard}) = 1 - P(\text{late} \mid \text{standard}) = 1 - 0.09 = 0.91$

Note that the complement is taken within the conditioning group. The two branches leaving the "standard" node sum to 1; this is not the same as taking the complement of a joint probability.

(d) Evaluating the advertisement

Step 1 — Check the claim is consistent.

$P(\text{on time} \mid \text{overnight}) = 1 - 0.03 = 0.97$

Step 2 — Identify what it omits — the conditioning group.

Step 3 — Explain why this matters in context.

Step 4 — A fair version.