What denominator do you use for a conditional probability?

The restricted set named by the condition, not the original total. When a question says given something, you only count the outcomes that satisfy that condition, so the denominator shrinks to the conditioning group. For the JUN24 Paper 1 Question 5b Venn of 78 students, P(wears glasses given left-handed) uses only the left-handed students: 7 wear glasses out of 7 + 15 = 22 left-handers, so the answer is 7/22, not 7/78.

Why does without replacement change the denominator?

Because the first item is gone, so the set you draw from is smaller. Drawing two notes from £5, £10, £20, £50 without replacement gives six different pairs, not the sixteen ordered outcomes you would get if repeats were allowed. Four of the six pairs total at least £30, so the probability is 4/6 = 2/3. Treating it as with replacement counts impossible repeats and gives the wrong 5/8.

If the first disc is given to be red, how many discs are in the denominator next?

24, not 25. A given outcome is a fixed fact, so you remove it before working out the rest. With 25 discs (11 red, 9 blue, 5 yellow) and the first given as red, 24 remain (10 red, 9 blue, 5 yellow). For all three different colours, the other two must be one blue and one yellow in either order: (9/24)(5/23) times 2 = 15/92. Keeping 25 and multiplying by 11/25 double prices the given red and gives the wrong 33/460.

GCSE Maths Higher · AQA · Combined & conditional probability

Conditional denominator is the restricted set

The moment a question says given something, or without replacing, the world shrinks. Your denominator becomes what is left or allowed, never the original total. For the JUN24 Venn of 78 students, $P(\text{glasses} \mid \text{left-handed})$ counts only the left-handers: $\frac{7}{7 + 15} = \frac{7}{22}$ , not $\frac{7}{78}$ .

Keeping the original total is the single biggest grade 7 to 9 mark-loser in conditional and without-replacement questions. The condition has already removed everything outside the restricted set, so dividing by everyone answers a different question entirely.

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How to spot it in your own work

You divided by the whole total when the question said given a condition, instead of by the conditioning group.
You treated a without replacement draw as if the item could be drawn again, opening up impossible repeats.
You multiplied in a probability for an outcome that was already given as a fixed fact.
Your conditional answer came out smaller than the matching unconditional one, because the denominator was too big.

An exam question that triggers it

Here is the structure of JUN24 Paper 1 Question 5b:

A Venn diagram records 78 students.

21 wear glasses only, 7 wear glasses and are left-handed, 15 are left-handed only, and 35 are neither.

Find the probability a student wears glasses, given they are left-handed.

The misconception keeps the whole year, writing $\frac{7}{78}$ . The fix: given restricts the world to the left-handed group, $7 + 15 = 22$ , so $P(\text{glasses} \mid \text{left-handed}) = \frac{7}{22}$ .

Why students fall for this

The original total is the number students wrote down first, so it stays anchored in mind. When the condition arrives, the instinct is to keep that familiar denominator and only adjust the numerator, when in fact the condition has already changed the whole sample space.

The word given is quiet. It does not look like an instruction to throw away most of the data, but that is exactly what it does: everything outside the conditioning group stops counting.

Without replacement hides the same shrink. Students picture the full bag at every draw, so they keep the starting total instead of reducing it by one after each pick, and a given first outcome gets priced as a chance rather than fixed as a fact.

The fix: Name the restricted set first, then count inside it

Step 1: read the restriction. The words given, of those, or without replacing all shrink the world. Decide which group you are now choosing from.

Step 2: count the restricted total. For given left-handed, the total is the left-handed group: $7 + 15 = 22$ . For a without-replacement draw, drop the total by one for each item already taken.

Step 3: count the favourable outcomes inside that set. Here 7 of the 22 left-handers wear glasses.

Step 4: divide. $P(\text{glasses} \mid \text{left-handed}) = \frac{7}{22}$ . Sense-check: a conditional probability should never share the original total unless the condition changed nothing.

Worked example

JUN24·1·5b Venn: 78 students, 7 wear glasses and are left-handed, 15 are left-handed only. Find $P(\text{glasses} \mid \text{left-handed})$ .

Restrict the total. Given left-handed, the denominator is $7 + 15 = 22$ .
Divide.
$P(\text{glasses} \mid \text{left-handed}) = \frac{7}{22}$
Trap: $\frac{7}{78}$ divides by everyone and ignores the word given.

NOV24·3·13 without replacement: a bag holds £5, £10, £20, £50. Two notes are drawn without replacement. Find P(total is at least £30).

List the genuine pairs. Without replacement there are six: $15, 25, 55, 30, 60, 70$ .
Count and divide. Four reach £30 (30, 55, 60, 70), so $\frac{4}{6} = \frac{2}{3}$ . Trap: with replacement gives 16 ordered outcomes and $\frac{10}{16} = \frac{5}{8}$ , but repeats are impossible here.

NOV24·3·21 conditional draw: 25 discs, 11 red, 9 blue, 5 yellow. Three are drawn without replacement and the first is red. Find P(all three different colours).

Fix and remove the given red. 24 discs remain: 10 red, 9 blue, 5 yellow.
Price the other two. All different needs one blue and one yellow in either order: $\frac{9}{24} \times \frac{5}{23} \times 2 = \frac{90}{552} = \frac{15}{92}$ . Trap: keeping 25 and multiplying by $\frac{11}{25}$ gives $\frac{33}{460}$ , double pricing the given red.

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Common questions

What denominator do you use for a conditional probability?: The restricted set named by the condition. For $P(\text{glasses} \mid \text{left-handed})$ on the 78-student Venn, the denominator is the 22 left-handers, so the answer is $\frac{7}{22}$ , not $\frac{7}{78}$ .
Why does without replacement change the denominator?: Because the first item is gone, so the set shrinks. Drawing two of £5, £10, £20, £50 without replacement gives six pairs, four reaching £30, so $\frac{4}{6} = \frac{2}{3}$ , not the with-replacement $\frac{5}{8}$ .
If the first disc is given to be red, how many discs are in the denominator next?: 24, not 25. A given outcome is fixed and removed, leaving 24 discs, so $P(\text{all different}) = \frac{9}{24} \times \frac{5}{23} \times 2 = \frac{15}{92}$ , not the keep-25 answer $\frac{33}{460}$ .

Related misconceptions

AND events multiply along the branchAlong a branch, and means multiply, never add. The same branch multiplication prices the without-replacement draws once the denominator has shrunk.
At least one is one minus noneFor at least one problems, take 1 minus the chance it never happens, rather than adding the winning branches, which double counts the overlap.

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