P9 - Conditional probability

P9 studies how a condition changes the group of possible outcomes. You will calculate conditional probabilities using expected frequencies, two-way tables, Venn diagrams and trees, and distinguish a condition from its reverse. Each part develops Higher content.

Restricting the sample space (Higher)

If $30$ pupils play music and $12$ of them play sport, what is the chance that a music player plays sport? The key is to choose from the $30$ music players, rather than from the whole school.

You already know

Probability is successful outcomes divided by all eligible outcomes when selection is uniform. An intersection $A\cap B$ means both events; a two-way table or Venn diagram keeps overlapping counts organised.

A conditional probability uses new information to restrict the possible group. Write $P(A\mid B)$ for “probability of $A$ , given $B$ ”. The event after the bar tells you who remains eligible. The numerator counts eligible outcomes that also meet $A$ .

Conditional probability

For $P(B)>0$ ,

P(A\mid B)=\frac{P(A\cap B)}{P(B)}

Not on the formula sheet: derive it from the conditional multiplication rule $P(A\cap B)=P(A\mid B)P(B)$ .

That multiplication rule is on the Higher formula sheet (2026 version; the 2027 sheet will have the same content). Dividing both sides by $P(B)$ gives the displayed quotient. An impossible condition has no eligible outcomes, explaining why its denominator cannot be zero.

Worked example 1: choosing from music players

Question. Thirty pupils play music, including twelve who play sport. A music player is chosen uniformly. Find the probability that they play sport.

Plan. Restrict the denominator to music players, then count sport within that group.

\begin{aligned} \text{eligible count}&=30 && \text{given music}\\ \text{successful count}&=12 && \text{sport within music}\\ P(\text{sport}\mid\text{music})&=\frac{12}{30}=\frac25 && \text{conditional share} \end{aligned}

Why this step? We know the pupil plays music, so pupils outside that group cannot be selected.

Answer. $2/5$ , or $0.4$ .

Check. The other $18$ music players give $18/30=3/5$ , and the shares total one.

Keep the same eligible group and change only the requested event to not playing sport.

Worked example 2: complement within the condition

Question. Find the probability that the chosen music player does not play sport.

Plan. Subtract the sport count within the group of thirty.

\begin{aligned} \text{music, not sport}&=30-12=18 && \text{complement within music}\\ P(\text{not sport}\mid\text{music})&=\frac{18}{30}=\frac35 && \text{same eligible group} \end{aligned}

Why this step? The condition remains music; only the successful subset changes.

Answer. $3/5$ .

Check. $1-2/5=3/5$ .

The music-player data alone cannot give a joint probability for the whole school: its total is missing. The reverse condition can also be very different. Among $90$ club members, $36$ play badminton, $25$ squash and $15$ both. Then $P(\text{squash}\mid\text{badminton})=15/36=5/12$ , while $P(\text{badminton}\mid\text{squash})=15/25=3/5$ . The joint probability uses everyone: $15/90=1/6$ .

Independence has a precise interpretation here: $P(A\mid B)=P(A)$ when $A,B$ are independent and $P(B)>0$ . If the conditional chance differs, they are dependent.

Watch out

Wrong: $P(\text{squash}\mid\text{badminton})=15/90$ . That is the joint chance of both among everyone. Given badminton, use its $36$ members: $15/36$ . Reversing the condition uses $25$ , not $36$ .

Keep the sport count and condition from example 2, changing only the music group size from thirty to forty.

Conditional probabilities in two-way tables (Higher)

A condition selects a whole row or column of a two-way table, and the crossing cell supplies its successful count.

Here the complete population contains $100$ bookings. Given telephone, use the telephone row total; given late, use the late column total.

Booking	On time	Late	Total
Online	$54$	$6$	$60$
Telephone	$28$	$12$	$40$
Total	$82$	$18$	$100$

The same cell can answer two reversed questions, but with different denominators.

Worked example 3: selecting a row

Question. A telephone booking is selected uniformly. Find the probability it arrived late.

Plan. Restrict to the telephone row and take the late share.

\begin{aligned} \text{telephone total}&=28+12=40 && \text{the condition}\\ P(\text{late}\mid\text{telephone})&=\frac{12}{40}=\frac3{10} && \text{late cell within the row} \end{aligned}

Why this step? On-time telephone bookings remain in the denominator even though they are unsuccessful.

Answer. $3/10$ .

Check. The on-time share is $28/40=7/10$ , its complement.

Change only the direction of the condition: select from late arrivals instead.

Worked example 4: selecting a column

Question. A late booking is selected uniformly. Find the probability it was made by telephone.

Plan. Use the late column total instead of the telephone row total.

\begin{aligned} \text{late total}&=6+12=18 && \text{new condition}\\ P(\text{telephone}\mid\text{late})&=\frac{12}{18}=\frac23 && \text{telephone share of late arrivals} \end{aligned}

Why this step? The successful cell stays twelve, but the eligible group has changed from forty to eighteen.

Answer. $2/3$ .

Check. Online given late is $6/18=1/3$ , so both booking methods total one.

When given probabilities rather than counts, build expected frequencies in a convenient reference population. Multiplying every count by the same scale leaves the conditional proportion unchanged; these model frequencies need not be actual future counts.

Worked example 5: expected frequencies for damaged parcels

Question. A model sends $45\%$ of parcels by route $X$ , the rest by $Y$ . Damage probabilities are $0.08$ for $X$ and $0.04$ for $Y$ . Find the probability a damaged parcel used $X$ .

Plan. Use $1000$ reference parcels, build the damaged column, then restrict to it.

\begin{aligned} X\text{ total},Y\text{ total}&=450,550 && \text{split the reference population}\\ X\text{ damaged}&=450(0.08)=36 && \text{route X damage rate}\\ Y\text{ damaged}&=550(0.04)=22 && \text{route Y damage rate}\\ \text{all damaged}&=36+22=58 && \text{complete condition group}\\ P(X\mid\text{damaged})&=\frac{36}{58}=\frac{18}{29} \end{aligned}

Why this step? A damaged parcel could have used either route, so both routes contribute to the denominator.

Answer. $18/29$ .

Check. The remaining expected counts are $414$ undamaged on $X$ and $528$ on $Y$ ; all four cells total $1000$ .

Change only which route is requested, keeping the damage condition.

Worked example 6: the other route within damaged parcels

Question. For the same model, find $P(Y\mid\text{damaged})$ .

Plan. Keep the damaged denominator and use its route-Y count.

\begin{aligned} P(Y\mid\text{damaged})&=\frac{22}{58} && \text{Y within damaged parcels}\\ &=\frac{11}{29} && \text{simplify} \end{aligned}

Why this step? The two routes exhaust the damaged group, so only the numerator changes.

Answer. $11/29$ .

Check. $18/29+11/29=1$ .

The condition is what matters, whether counts were observed or generated from a model.

Watch out

Wrong: “ $P(X\mid\text{damaged})=0.08$ .” That is damage given $X$ , the reverse direction. Within all damaged parcels the X share is $36/(36+22)=18/29$ .

Keep the damage rates fixed and change only route X's share from $45\%$ to $40\%$ .

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account

Last Accessed

Papers 1, 2 and 3

P9 - Conditional probability

Restricting the sample space (Higher)

Conditional probabilities in two-way tables (Higher)

Continue this lesson.

Learn faster with Nova