SPEC P8, P9 ★ HIGHER TIER ONLY
Dependent Events & Conditional Probability
"Without replacement" trees, conditional probability from trees, Venn diagrams, and two-way tables — the grade 7-9 probability toolkit
Dependent Events
Dependent Events (Higher)
Two events are dependent when the outcome of the first event changes the probability of the second. The typical signal in a question is the phrase "without replacement".
Independent (Lesson 5.03)
With replacement: the ball is put back. Probabilities on the second branch are the same as the first.
P(B|A) = P(B)
Dependent (this lesson)
Without replacement: the first ball is NOT put back. Probabilities on the second branch change based on the first outcome.
P(B|A) ≠ P(B)
Spec signal phrases
"Without replacement", "does not replace", "takes two counters together", or "given that the first was red" all signal dependent events. Treat them identically — you must update the probabilities on the second stage.
Update BOTH Numerator and Denominator
The most common exam error is only updating one of the two. After the first draw:
The two updates
Denominator (total): goes down by 1 — one fewer item in the bag.
Numerator (that colour): goes down by 1 only if the first was that colour — otherwise stays the same.
Ex
Bag: 5 red, 3 blue — first draw is red. What are the second-draw probabilities?
Before first draw: 5 red, 3 blue, total 8. So P(R) = 5/8, P(B) = 3/8.
After drawing a red (without replacement): 4 red left, 3 blue left, total 7.
P(R on 2nd | R on 1st) = 47, P(B on 2nd | R on 1st) = 37
Red numerator went 5 → 4 (one red taken). Total went 8 → 7. Blue stayed at 3.
Simple Worked Example — P(both red)
Ex
A bag has 5 red and 3 blue balls. Two are drawn without replacement. Find P(both red).
Method: Multiply along the path: first red, then second red given the first was red.
First draw: P(red) = 5/8.
Second draw given first was red: P(red | red) = 4/7 (updated).
P(R R) = P(R) × P(R | R) = 58 × 47 = 2056 = 514
Contrast with "with replacement"
If the first ball were replaced, P(R R) would be 5/8 × 5/8 = 25/64 — a different answer. The "without replacement" assumption matters.
Tree Diagram for Dependent Events
A tree for dependent events looks the same as an independent one — same branches, same nodes. The key difference: the probabilities on the second stage change depending on which first-stage branch you came from.
Bag: 4 red, 6 blue — spot the updates
Orange fractions (3/9 top, 5/9 bottom) are where a ball of that colour was removed — numerator drops by 1. Purple fractions (6/9, 4/9) are where the other colour was removed — numerator stays, denominator still drops. Denominator is always 9 on the second stage (one fewer total).
MCE Worked Example — Part A
Ex
A bag has 4 red and 6 blue marbles. Two are drawn without replacement. Find P(both red).
M — Method: Dependent events. Use the tree from the previous slide and multiply along the R,R path.
C — Calculate:
First draw: P(R) = 4/10
Second draw given first red: P(R | R) = 3/9 (one red removed, one fewer total)
P(R R) = 410 × 39 = 1290 = 215
E — Evaluate: P(both red) = 2/15.
Compare: with replacement gives 4/10 × 4/10 = 16/100 = 4/25
The without-replacement answer 2/15 ≈ 0.133 is smaller than the with-replacement answer 4/25 = 0.16 — because removing a red makes the second red less likely.
MCE Worked Example — Part B
Ex
Same bag (4R, 6B). Find P(exactly one red).
M — Method: Two paths: R,B or B,R. Multiply along each, then add.
C — Calculate:
P(R, B) = 4/10 × 6/9 = 24/90
P(B, R) = 6/10 × 4/9 = 24/90
P(exactly one R) = 2490 + 2490 = 4890 = 815
E — Evaluate: P(exactly one red) = 8/15.
Notice something interesting
Both paths give the same probability (24/90) because the arithmetic works out: 4 × 6 = 6 × 4. The numerators on the second branches are "what's left of the colour we want" — and those multiply identically regardless of order.
3+ stage dependent trees
The same method scales. If three marbles are drawn without replacement from the 4R 6B bag, the third-stage probabilities update again. For example, P(R, R, R) = 4/10 × 3/9 × 2/8 = 24/720 = 1/30. Each draw decreases both the relevant colour's count and the total.
What is Conditional Probability?
Conditional Probability (P9 Higher only)
The probability that event B occurs given that event A has already occurred. Written P(B | A) and read "the probability of B given A".
"Given that" — the key phrase
"Given that" restricts the sample space to only those outcomes where A has happened. It's the language AQA uses to flag a conditional probability question.
Examples of P(B | A)
"Given that the first was red, P(second is blue)"
"If a student plays football, P(they also play tennis)"
"Given that a person bought trainers, P(they bought a hoodie)"
Not conditional
"P(both red)" — asking for the joint event, not conditional.
"P(at least one blue)" — an unrestricted probability.
P9 says "calculate AND interpret"
Interpreting conditional probability means explaining what it tells you in context. For example, if P(pass | studied) = 0.9 and P(pass) = 0.5 overall, then studying is strongly associated with passing. If P(B | A) = P(B) (no difference), then A and B are not related — they're independent.
The Conditional Probability Formula
The formula defines conditional probability precisely:
P(B | A) = P(A B)P(A)
How to read it
"Of the outcomes where A happened, what fraction also had B?" The denominator restricts to A; the numerator counts the overlap with B.
Rearranged form (general multiplication rule)
Multiply both sides by P(A):
P(A B) = P(A) × P(B | A)
The formula also works the other way: P(A B) = P(B) × P(A | B). This is exactly what you do along a tree path — multiply first branch by conditional second branch.
Independence — test & link to 5.03
If A and B are independent, P(B | A) = P(B), so:
P(A B) = P(A) × P(B)
The converse is a test: if P(B | A) equals P(B), then A and B are independent. If they differ, the events are dependent.
Conditional Probability from a Tree
Tree diagrams are the natural home for conditional probability — every probability on the second stage is already conditional.
Ex
Using the 4R 6B tree: given that the first marble was red, find P(second is blue).
Method: Read directly off the tree. "Given first red" restricts to the top half of the tree (the "first R" branch). The second-stage blue branch from there is labelled 6/9.
P(B | R) = 69 = 23
Why tree probabilities ARE conditional
Every second-stage probability on a tree diagram is conditional on the first-stage outcome you followed to reach it. P(B|R) = 6/9 is already written on the tree — you just read it off.
Conditional Probability from a Venn Diagram
For conditional from a Venn, the rule is: count only within the region given. The numbers inside the Venn are expected frequencies — the P9 spec term for counts that represent a probability distribution.
Ex
In a class of 50, 20 study Maths (M), 18 study English (E), and 8 study both.
First fill the Venn: Only M = 20 − 8 = 12. Only E = 18 − 8 = 10. Both = 8. Neither = 50 − 30 = 20.
Question: Given that a student studies English, find P(they also study Maths).
Method: Restrict to just the E circle (10 + 8 = 18 students). Count how many of those also study M — that's the overlap: 8.
P(M | E) = n(M E)n(E) = 818 = 49
Conditional Probability from a Two-Way Table
Two-way tables make conditional probability look almost trivial — just divide the inner cell by the relevant row or column total. The inner counts are again expected frequencies (P9 spec term).
100 households: do they own a Cat (C) and/or Dog (D)?
Owns DogNo DogTotal
Owns Cat203050
No Cat153550
Total3565100
Ex
Two conditional probabilities from the table
(a) Given a household owns a dog, P(they also own a cat):
Restrict to the dog column (35 households). Of these, 20 also own a cat.
P(C | D) = n(C D)n(D) = 2035 = 47
(b) Given a household owns a cat, P(they also own a dog):
Restrict to the cat row (50 households). Of these, 20 also own a dog.
P(D | C) = 2050 = 25
P(A|B) ≠ P(B|A)
Notice: P(C | D) = 4/7 but P(D | C) = 2/5. These are different numbers. Pay close attention to which event is "given" and which is the one you're finding.
Addition Rule (Non-Mutually-Exclusive Events)
At Higher tier, you're expected to compute P(A B) for events that can both happen. The formula is the probability version of the inclusion-exclusion rule from Lesson 5.02.
P(A B) = P(A) + P(B) − P(A B)
Why we subtract the overlap
When you add P(A) + P(B), anyone in the overlap is counted twice. Subtract P(A B) once to get the correct total.
Ex
Using the Maths/English Venn data
P(M) = 20/50 = 2/5. P(E) = 18/50 = 9/25. P(M E) = 8/50 = 4/25.
P(M E) = 25 + 925 − 425 = 1025 + 925 − 425 = 1525 = 35
Verify via count: n(M E) = 12 + 8 + 10 = 30. 30/50 = 3/5. ✓
Link to 5.01: mutually exclusive special case
If A and B are mutually exclusive, then P(A B) = 0, so the formula reduces to P(A B) = P(A) + P(B) — the simple addition rule for disjoint events.
MCE Practice Question
A bag contains 4 yellow and 3 purple marbles. Two marbles are drawn at random without replacement.
(a) Draw and complete a tree diagram for the two draws. [2 marks]
(b) Find P(both marbles are yellow). [2 marks]
(c) Find P(exactly one marble is yellow). [3 marks]
(d) Given that the first marble drawn was yellow, find the probability that the second marble is purple. [2 marks]
Total: [9 marks]
M
Dependent events (without replacement). Build a tree: first draw has P(Y)=4/7, P(P)=3/7. Second draw probabilities depend on the first. For (d), read directly off the top "given first Y" branch.
C
(a) Tree: First: Y (4/7), P (3/7). If first Y: Y (3/6), P (3/6). If first P: Y (4/6), P (2/6).
(b) P(Y Y) = 4/7 × 3/6 = 12/42 = 2/7.
(c) P(Y,P) + P(P,Y) = (4/7 × 3/6) + (3/7 × 4/6) = 12/42 + 12/42 = 24/42 = 4/7.
(d) P(second P | first Y) = 3/6 = 1/2.
E
(b) 2/7, (c) 4/7, (d) 1/2. Sense-check all paths: 2/7 + 2/7 + 2/7 + 1/7 = 7/7 = 1. ✓
Examiner Alert — Denominator & Numerator Updates
The most costly exam error at Higher tier
Students treat "without replacement" questions like "with replacement" and forget to update the probabilities on the second branch. One mis-labelled branch → every downstream answer is wrong.
After first red drawn (from 4R 6B)WrongCorrect
P(red on 2nd)4/10 (same as first)3/9 (numerator & denominator both drop)
P(blue on 2nd)6/10 (same as first)6/9 (numerator same, denominator drops)
Two-step check
(1) Did the total go down by 1? (2) Did the colour I'm computing go down by 1 only if it was the one drawn? If yes to both, your branch is correct. If the probabilities on one node don't sum to 1, one branch is wrong.
Examiner Alert — "Given that" Misread
P(A | B) ≠ P(A B)
Students confuse conditional P(A | B) with joint P(A B). These are different:
P(A B): "The probability that both A and B occur" — numerator over the full total.
P(A | B): "Given B happened, the probability that A also occurred" — numerator over just n(B).
P(A | B) and P(B | A) are usually different
From the cat/dog table: P(Cat | Dog) = 20/35 = 4/7. P(Dog | Cat) = 20/50 = 2/5. Read the question carefully to know which is which.
Decoding AQA wording
Look for these conditional signals: "given that...", "if... then P(...)", "a person who... is chosen". All three restrict the sample space.
Common Errors Summary
Students often writeCorrect versionWhy it matters
P(R,R) without replacement = 4/10 × 4/10= 4/10 × 3/9Both numerator and denominator must update
Update denominator but not numeratorUpdate both, if the colour was takenHalf-updating gives wrong answer
P(A | B) = P(A B)P(A | B) = P(A B) / P(B)Must divide by P(condition)
P(A | B) = P(B | A)Usually different — read carefullyWhich event is "given" matters
P(A B) = P(A) + P(B) alwaysSubtract P(A B) unless mutually exclusiveAvoids double-counting the overlap
Treat dependent events as independentCheck for "without replacement"Totally different methods
✓ Checkpoint — Module 5 Complete
Core Skills from This Lesson
Identify dependent events — spot "without replacement" and related phrases.
Update both numerator and denominator on the second tree branch.
Build a dependent-event tree diagram.
Apply P(B | A) = P(A B) / P(A) to calculate conditional probabilities.
Read conditional probabilities off trees, Venn diagrams, and two-way tables (P9).
Use the Higher addition rule: P(A B) = P(A) + P(B) − P(A B).
Distinguish P(A | B) from P(B | A) and from P(A B).
Module 5 Probability — Full Toolkit Recap
5.01: Probability scale, theoretical vs experimental, expected frequency, mutually exclusive.
5.02: Sample spaces, grids, two-way tables, Venn diagrams, set notation.
5.03: Tree diagrams for independent events; multiply along path, add between paths.
5.04: Dependent events, conditional probability, addition rule. (Higher only)
Next module
Module 6 covers Statistics — collecting and representing data, averages, cumulative frequency, histograms, and scatter graphs. The probability toolkit you've just built will help with data-based questions involving frequency tables.
Next: Dependent Events