Mu6, Mu7 - Conditional probability and two-way data | OCR A-Level Mathematics B (MEI)

Calculate conditional probabilities from the tables, trees, Venn regions and sample-space information in Mu6. Then use Mu7 to decide whether a probability stays unchanged after conditioning, giving a numerical argument for independence.

A condition selects the relevant group

Can six people represent both $3/8$ and $1/3$ of a group? In a library survey, six people have online membership and borrow fiction. There are sixteen fiction borrowers and eighteen online members. The six are the same people, but the group used as the denominator changes.

You already know

$A\cap B$ means both events and $A'$ means not $A$ . With equally likely outcomes, a probability is a favourable count divided by the count of the relevant sample space. A probability need not remain the same when that space is restricted.

“Given fiction” removes the non-fiction borrowers from consideration before any division takes place.

Conditional probability

$\mathrm P(A\mid B)$ means the probability of $A$ given $B$ . It measures $A$ within the outcomes satisfying $B$ . The event after the bar is the condition, so its probability supplies the denominator.

A person is selected uniformly from this complete survey of 40. Let $A$ mean online membership and $B$ mean borrowing fiction.

	$B$	$B'$	Total
$A$	6	12	18
$A'$	10	12	22
Total	16	24	40

The joint event has probability $\mathrm P(A\cap B)=6/40$ . A conditional probability is a different fraction.

Conditional probability from a joint event

For $\mathrm P(B)>0$ ,

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

A zero denominator does not define a conditional probability by this rule; an empty conditioned group supplies no usable proportion.

In the formula sheet.

The table lets you choose the required group directly, without treating a conditional as a joint probability.

Worked example 1: membership within fiction borrowers

Question. Use the complete library table to find $\mathrm P(A\mid B)$ and $\mathrm P(A'\mid B)$ .

Plan. Keep only the fiction column, then divide each relevant count by its total.

\begin{aligned} \mathrm P(A\mid B)&=\frac6{16}=\frac38 &&\text{members among fiction borrowers}\\ \mathrm P(A'\mid B)&=\frac{10}{16}=\frac58 &&\text{non-members in the same group} \end{aligned}

Why this step? The information given has already restricted the selection to sixteen people.

Answer. The probabilities are $3/8$ and $5/8$ .

Check. They sum to 1. The formula gives the same result: $(6/40)/(16/40)=3/8$ .

Change only the condition to not borrowing fiction. Keep the membership event and the complete table unchanged.

Worked example 2: membership within non-fiction borrowers

Question. From the same table find $\mathrm P(A\mid B')$ and $\mathrm P(A'\mid B')$ .

Plan. Use the non-fiction column and its total of 24.

\begin{aligned} \mathrm P(A\mid B')&=\frac{12}{24}=\frac12 &&\text{members within }B'\\ \mathrm P(A'\mid B')&=\frac{12}{24}=\frac12 &&\text{non-members within }B' \end{aligned}

Why this step? The condition selects the sample space before the event count is divided by its size.

Answer. Both probabilities are $1/2$ .

Check. The two counts fill all 24 non-fiction borrowers. A denominator of 40 would instead give the corresponding joint probabilities.

The complete table also contains six fiction borrowers among eighteen online members, giving $6/18=1/3$ . That is the opening puzzle: changing the conditioned group changes the denominator. It does not justify exchanging the order of event symbols in an answer.

Watch out

Wrong: $\mathrm P(A\mid B)=6/40$ . Right: $6/16$ . The numerator concerns both events, while the denominator concerns the given event alone.

Use the same restricted-group reasoning on a fresh survey.

Later branches contain conditional probabilities

A later branch describes the situation after the earlier branch has occurred; it is not the probability of the whole path.

Write $G_1,Y_1$ for first green or yellow, and $G_2,Y_2$ for the corresponding second-draw events. After $Y_1$ , the branch to green is $\mathrm P(G_2\mid Y_1)$ .

A complete conditional path

For $\mathrm P(A)>0$ ,

\mathrm P(A\cap B)=\mathrm P(A)\mathrm P(B\mid A).

This works for dependent events too. The conditional second factor becomes $\mathrm P(B)$ only when independence applies.

In the formula sheet.

Outgoing branches from each node are disjoint and exhaustive, so their probabilities sum to 1. Distinct final paths are disjoint and can be added.

Green-yellow branches and complete paths for two draws without replacement

Worked example 3: a branch and its joint path

Question. A bag contains four green and five yellow counters. Two are drawn without replacement. Find $\mathrm P(G_2\mid Y_1)$ , $\mathrm P(Y_1\cap G_2)$ , and the probability of different colours.

Plan. Use the remaining bag after yellow, then add the two mixed-colour paths.

\begin{aligned} \mathrm P(G_2\mid Y_1)&=\frac48=\frac12 &&\text{four green remain}\\ \mathrm P(Y_1\cap G_2)&=\frac59\times\frac48=\frac5{18} &&\text{complete the path}\\ \mathrm P(G_1\cap Y_2)&=\frac49\times\frac58=\frac5{18} &&\text{the other order}\\ \mathrm P(\text{different})&=\frac59 &&\text{add disjoint paths} \end{aligned}

Why this step? The branch already assumes first yellow; the joint event must include the chance of reaching that branch.

Answer. The three probabilities are $1/2$ , $5/18$ and $5/9$ .

Check. GG has probability $1/6$ , YY has probability $5/18$ , and all four path probabilities sum to 1.

Change only replacement. Restore and thoroughly mix the counter between independent draws; keep the bag and requested events fixed.

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Mu6, Mu7 - Conditional probability and two-way data

A condition selects the relevant group

Later branches contain conditional probabilities

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