P9 - Conditional probability (Higher)

Conditions in two-way tables (Higher)

A detector sounds for $90\%$ of faulty components. Does an alarm mean that a component is probably faulty? Not necessarily: we also need to know how common faults are and how often good components trigger alarms. We will settle this with counts at the end.

Suppose $27$ of $36$ cyclists wear helmets. If you know a randomly selected pupil is a cyclist, only those $36$ cyclists remain possible. The probability of a helmet is $\frac{27}{36}$, even if the school has many more pupils.

Read the condition first. In $P(A\mid B)$, the group $B$ becomes the denominator. The notation means nothing in this calculation if $B$ has probability zero: there is no possible selected member of that group.

The next example changes only the direction of the condition. The numerator stays the same, but the denominator changes.

Conditional probabilities in opposite directions are not generally equal. The count-based rule used here is not printed separately on the formulae sheet; learn it. The Higher sheet gives an equivalent multiplication identity, used with trees later in this lesson.

Read the condition carefully in this new table.

Conditions in Venn diagrams (Higher)

A condition on a Venn diagram selects a region that becomes your new whole.

In a group of $30$ pupils, $12$ attend art only, $8$ attend both art and badminton, $7$ attend badminton only, and $3$ attend neither. Let $A$ mean art and $B$ mean badminton.

[DIAGRAM: asset_name: Conditional club membership Venn diagram; asset_slug: aqa8300_p04_conditional_venn; caption: Given badminton, use the 15 pupils in circle B as the whole; description: The universal-set rectangle contains 30 pupils. Art-only has 12, the intersection of art and badminton has 8, badminton-only has 7 and neither has 3. Circle A has total 20 and circle B total 15.]

Given $B$, ignore everything outside circle $B$. Given $B'$, do the opposite: use the art-only and neither regions. A complement is always taken within the universal set before applying the condition. The notation and counting rule are not printed on the formulae sheet: learn them.

The next example keeps the condition $B$ and changes only the requested event to not art.

A different condition creates a different denominator. For example, $P(A\mid B')=\frac{12}{12+3}=\frac45$. The art-only region is the favourable count, and the entire outside-badminton region is the possible count.

Try a new pair of clubs, with the condition on the first set instead.

Conditions in probability trees (Higher)

Second-stage tree branches already describe probabilities given the first-stage result.

A pupil travels by bus with probability $0.6$. Given bus travel, the probability of being late is $0.2$. Given no bus travel, it is $0.05$. These are different second-stage probabilities, so bus travel and lateness are dependent in this model.

[DIAGRAM: asset_name: Bus travel and lateness probability tree; asset_slug: aqa8300_p04_bus_tree; caption: The late branches give probabilities within different first-stage groups; description: The first split is bus 0.6 and no bus 0.4. After bus, late is 0.2 and not late 0.8. After no bus, late is 0.05 and not late 0.95. End path probabilities are bus and late 0.12, bus and not late 0.48, no bus and late 0.02, no bus and not late 0.38.]

A branch is a fraction of its parent group. A complete path is a fraction of the starting whole. Multiplying converts the conditional branch chance into a joint chance.