11.02c, 11.02f - Conditional probability | OCR GCSE Mathematics

A student is chosen at random, and then you learn that they play football. Should your probability calculation still include everybody? Under 11.02f you will restrict the sample space, use the conditional multiplication rule and test independence. Under 11.02c you will choose and construct tables, Venn diagrams or trees when the question supplies no structure.

Restricting the possible outcomes

“Given that” is new information. It removes outcomes that no longer fit. If you know the chosen student plays football, the possible student is now somewhere inside the football group, rather than anywhere in the whole school.

Conditional probability

$p(A\text{ given }B)$ means the probability of $A$ when event $B$ is known to have happened. Restrict to $B$ , then ask what fraction of that group also satisfies $A$ . This requires $p(B)>0$ .

Suppose 40 students have the following two properties. The table separates all four possible combinations, so the totals remain visible.

	Music	Not music	Total
Football	10	14	24
Not football	6	10	16
Total	16	24	40

The table concerns one randomly chosen student. All students are equally likely to be chosen.

Worked example 1: conditioning on football

Question. Using the table, find the probability that a chosen student makes music, given that they play football.

Plan. Keep only the football row, then identify its music members.

\begin{aligned} \text{possible football students} &=24 \\ \text{also making music} &=10 \\ p(\text{music given football}) &=\frac{10}{24}=\frac5{12} \end{aligned}

Answer. $\frac5{12}$ .

Check. Within the restricted group, music and not music have probabilities $\frac{10}{24}$ and $\frac{14}{24}$ , which sum to one.

Reverse only the direction of the information. The overlap stays ten, but the group that is now known changes.

Worked example 2: reversing the condition

Question. Find the probability that the student plays football, given that they make music.

Plan. Keep the music column instead of the football row.

\begin{aligned} \text{possible music students} &=16 \\ \text{also playing football} &=10 \\ p(\text{football given music}) &=\frac{10}{16}=\frac58 \end{aligned}

Answer. $\frac58$ .

Check. The six music students not playing football account for the remaining $\frac6{16}=\frac38$ , complementing $\frac58$ .

This answers the opening question: the denominator must reflect the information given. Reversing a condition generally changes the probability; it does not merely change the wording.

Watch out

$\frac{10}{40}$ is the probability of football and music among all students. It is neither of these conditional probabilities. If the conditioning group has no possible members, division by its zero size is undefined, not a probability of zero.

Restrict a new population.

The conditional multiplication rule

A conditional probability describes a fraction within another event. To recover a fraction of the whole population, multiply by the fraction that reached that group. This is the same fraction-of-a-fraction reasoning used along a tree path.

Imagine 100 trainees, of whom 60 attend a workshop. If 80% of those 60 pass, the workshop-and-pass group contains $0.8\times60=48$ trainees. It is 48% of all 100, rather than 80% of all 100.

Conditional multiplication

For $p(B)>0$ ,

p(A\text{ given }B)=\frac{p(A\text{ and }B)}{p(B)},

p(A\text{ and }B)=p(A\text{ given }B)\,p(B).

The multiplication formula is supplied on the Higher 2027 formula sheet.

The quotient comes from using the size of $B$ as the denominator: divide both favourable and restricted counts by the whole population size. Multiplying by $p(B)$ reverses that restriction. This rule does not require independence.

Worked example 3: a fraction of the workshop group

Question. A trainee attends a workshop with probability $0.60$ . Given that they attend, their probability of passing is $0.80$ . Find the probability that they attend and pass.

Plan. Multiply the workshop probability by the passing probability within that group.

\begin{aligned} p(\text{attend and pass}) &=p(\text{pass given attend})p(\text{attend}) \\ &=0.80\times0.60=0.48 \end{aligned}

Answer. $0.48$ .

Check. A probability-model population of 100 has 60 attendees and $0.8\times60=48$ attendees who pass. The combined event is also no more likely than attending alone.

Change only the workshop attendance probability. The success rate within that group stays 80%.

Worked example 4: changing the first group

Question. Attendance probability is now $0.35$ , with passing probability given attendance still $0.80$ . Find the probability of attendance and passing.

Plan. Take the same fraction of the smaller attendance group.

\begin{aligned} p(\text{attend and pass}) &=0.80\times0.35 \\ &=0.28 \end{aligned}

Answer. $0.28$ .

Check. Among 100 model trainees, 35 attend and 28 of them pass; $\frac{28}{35}=0.80$ , recovering the stated conditional probability.

You can reverse the order consistently: $p(A\text{ and }B)=p(B\text{ given }A)p(A)$ when $p(A)>0$ . You cannot replace a conditional factor by an unconditional one without checking independence. The examples do not tell us the pass rate of non-attendees, so they do not determine the overall pass probability.

Watch out

$p(\text{pass given attend})$ and $p(\text{pass})$ answer different questions. A conditional rate of 80% does not assert that 80% of all trainees pass.

Multiply a new conditional path.

Testing independence

To test independence, compare like with like: the chance of event $A$ in the whole population and the chance of $A$ within event $B$ . Equal group counts are not the test; the corresponding proportions must agree.

The independence criterion

Provided $p(B)>0$ , events $A$ and $B$ are independent if and only if

p(A\text{ given }B)=p(A).

Equivalently, $p(A\text{ and }B)=p(A)p(B)$ . The criterion is not separately printed on the Higher 2027 sheet: learn it.

Why is the product test equivalent? Substitute the independence equality into the conditional multiplication rule. Conversely, divide the product equality by the positive number $p(B)$ to recover the conditional criterion. This proves both directions.

Worked example 5: unchanged proportions

Question. Of $20$ equally likely students, $12$ play sport, $10$ make music and $6$ do both. Are sport and music independent for the chosen student?

Plan. Compare the music proportion within sport with its proportion in the whole group.

\begin{aligned} p(\text{music given sport}) &=\frac6{12}=\frac12 \\ p(\text{music}) &=\frac{10}{20}=\frac12 \end{aligned}

Answer. Yes. The conditional and unconditional probabilities agree, so the events are independent.

Check. $p(\text{sport})p(\text{music})=\frac{12}{20}\times\frac{10}{20}=\frac3{10}=\frac6{20}$ , the actual probability of both.

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11.02c, 11.02f - Conditional probability

Restricting the possible outcomes

The conditional multiplication rule

Testing independence

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