Conditional Probability – Grade 10 IGCSE Maths

Welcome to Conditional Probability!

Conditional probability asks: "Given that something has already happened, what is the probability of the next event?" It powers everything from medical testing to weather forecasting. The key idea — past events can change the probability of future ones.

P(A|B) = P(A∩B) / P(B) | Multiply along branches (AND) | Add branches (OR)

The Formula

P(A|B) definition and tree diagrams

Without Replacement

Adjusting denominators on second branch

Hard Tree Problems

Medical tests, three-event trees

False Positives

Sensitivity, specificity, prevalence

Total Probability

P(A) = P(A|B)P(B) + P(A|B')P(B')

Bayes' Theorem

Reverse conditional probabilities

1. Conditional Probability — The Formula

P(A|B) reads as "the probability of A given that B has occurred". It restricts the sample space to outcomes where B is true, then asks how likely A is within that restricted space.

P(A|B) = P(A∩B) / P(B) where P(B) > 0

Multiplication rule (rearrangement):
P(A∩B) = P(B) × P(A|B) i.e. "B first, then A given B"
Also: P(A∩B) = P(A) × P(B|A) i.e. "A first, then B given A"

Independence:
A and B are independent when P(A|B) = P(A). Knowing B happened gives no information about A.
For independent events: P(A∩B) = P(A) × P(B) — the simple multiplication rule.

2. Tree Diagrams — The Foundation

A tree diagram shows all possible outcomes of a multi-stage experiment. Write probabilities on branches, not at end-points.

Two rules for tree diagrams:
• AND (sequential outcomes): MULTIPLY along branches
• OR (alternative paths to same result): ADD the end-products
All end-products must sum to 1 — use this to check your tree.

Without Replacement — Critical Rule:
Bag: 5 Red, 3 Blue (8 total). Draw two without replacement.
First draw: P(R) = 5/8, P(B) = 3/8
Second draw (after drawing R): P(R|R already drawn) = 4/7, P(B|R) = 3/7
Second draw (after drawing B): P(R|B already drawn) = 5/7, P(B|B) = 2/7

End-products: RR = 5/8 × 4/7 = 20/56, RB = 5/8 × 3/7 = 15/56, BR = 3/8 × 5/7 = 15/56, BB = 3/8 × 2/7 = 6/56
Total = 56/56 = 1 ✓

When drawing WITHOUT replacement: the denominator on the second branch decreases by 1 (one item was taken). The numerator only decreases if you drew that colour — otherwise it stays the same.

3. Hard Tree Problems — Without Replacement

Bag: 4 green, 3 yellow. Draw two without replacement. Full tree:
P(GG) = 4/7 × 3/6 = 12/42
P(GY) = 4/7 × 3/6 = 12/42
P(YG) = 3/7 × 4/6 = 12/42
P(YY) = 3/7 × 2/6 = 6/42
Sum = 42/42 = 1 ✓

P(both same colour) = P(GG) + P(YY) = 12/42 + 6/42 = 18/42 = 3/7
P(at least one green) = 1 − P(YY) = 1 − 6/42 = 36/42 = 6/7

4. Medical Test Problems

These are conditional probability problems in disguise. Key vocabulary:

Sensitivity = P(positive test | disease) = true positive rate
False positive rate = P(positive test | no disease)
Prevalence = P(has disease) in the population

Example: Disease prevalence 1%, sensitivity 95%, false positive rate 2%.
P(disease) = 0.01, P(no disease) = 0.99
P(+ | disease) = 0.95, P(+ | no disease) = 0.02
P(+ test) = 0.01×0.95 + 0.99×0.02 = 0.0095 + 0.0198 = 0.0293
P(disease | positive) = 0.0095 / 0.0293 ≈ 0.324 (32.4%)
Surprisingly low! Low prevalence means most positives are false positives.

5. Three-Event Trees

Three-stage experiment — 8 branches:
Each branch is found by multiplying three branch probabilities.
All 8 end-products must sum to 1.

Example: Fair coin three times. P(exactly two heads):
HHT: ½×½×½ = 1/8 HTH: 1/8 THH: 1/8
P(exactly two heads) = 3/8

For three-event trees: draw all 8 paths, calculate each product, then identify and add the ones matching your target outcome. Always verify all 8 sum to 1.

6. Law of Total Probability

When B and B' partition the sample space (B or not-B covers all possibilities):

P(A) = P(A|B) × P(B) + P(A|B') × P(B')

Example: Factory: Machine X makes 60% of items, Machine Y makes 40%. X has 3% defect rate, Y has 5% defect rate.
P(defective) = P(def|X)×P(X) + P(def|Y)×P(Y) = 0.03×0.6 + 0.05×0.4 = 0.018 + 0.02 = 0.038

Extended total probability (3 partitions B₁, B₂, B₃):
P(A) = P(A|B₁)P(B₁) + P(A|B₂)P(B₂) + P(A|B₃)P(B₃)
Use when the sample space is split into 3 groups.

7. Bayes' Theorem

Bayes' theorem "reverses" a conditional probability — finding P(B|A) when you know P(A|B).

P(B|A) = P(A|B) × P(B) / P(A)

Using the factory example above:
Given an item is defective, what is P(it came from Machine X)?
P(X|def) = P(def|X)×P(X) / P(def) = 0.018 / 0.038 ≈ 0.474

So 47.4% of defective items came from Machine X — even though X makes 60% of all items, it has a lower defect rate.

At IGCSE level, Bayes is usually done step-by-step using the total probability law in the denominator — you may not need to state "Bayes' theorem" explicitly. Just use the conditional probability formula with P(A) computed first.

8. Venn Diagram Connection

From a Venn diagram:
P(A|B) = n(A∩B) / n(B) (count of elements in both A and B, divided by count in B)

Example: 30 students. 18 study Maths, 12 study Science, 5 study both.
P(studies Maths | studies Science) = 5/12
P(studies Science | studies Maths) = 5/18

Example 1 — Without Replacement (Fractions)

Bag: 6 red, 4 blue. Two drawn without replacement. Find P(both same colour).

P(RR) = 6/10 × 5/9 = 30/90 = 1/3

P(BB) = 4/10 × 3/9 = 12/90 = 2/15

P(same) = 1/3 + 2/15 = 5/15 + 2/15 = 7/15

Example 2 — Medical Test

Disease P=0.02. P(+|disease)=0.90, P(+|no disease)=0.05. Find P(disease|positive test).

P(positive) = 0.02×0.90 + 0.98×0.05 = 0.018 + 0.049 = 0.067

P(disease|+) = 0.018/0.067 ≈ 0.269 (26.9%)

Example 3 — Three-Event Tree

P(rain on any day) = 0.3. What is P(exactly one rainy day in 3 days)?

RNN: 0.3×0.7×0.7 = 0.147 NRN: 0.7×0.3×0.7 = 0.147 NNR: 0.7×0.7×0.3 = 0.147

P(exactly 1 rainy) = 3 × 0.147 = 0.441

Example 4 — P(A|B) from Venn Diagram

40 students: 25 play football, 18 play tennis, 8 play both.

P(football | tennis) = 8/18 = 4/9

P(tennis | football) = 8/25

Example 5 — Bayes-Style Calculation

Box A: 3 red, 2 blue. Box B: 1 red, 4 blue. Choose a box at random (50/50), then draw one ball. Given the ball is red, find P(it came from Box A).

P(red) = 0.5×3/5 + 0.5×1/5 = 0.3 + 0.1 = 0.4

P(Box A | red) = (0.5 × 3/5)/0.4 = 0.3/0.4 = 3/4

Example 6 — "At Least One" Using Complement

Bag: 3 white, 7 black. Draw two with replacement. P(at least one white).

P(no white) = P(BB) = 7/10 × 7/10 = 49/100

P(at least one white) = 1 − 49/100 = 51/100 = 0.51

Common Mistakes to Avoid

Mistake 1 — Confusing P(A|B) with P(B|A)

Mistake 2 — Unchanged Second-Branch Fractions (Without Replacement)

When drawing without replacement, the denominator MUST decrease by 1 on every second branch. If you drew a red ball, the red numerator also decreases by 1. If you drew blue, only the denominator decreases. A very common error is leaving all second-branch fractions identical to first-branch fractions.

Mistake 3 — Wrong Denominator in P(A|B)

P(A|B) = P(A∩B)/P(B). Students often divide by the wrong thing — e.g. dividing by P(A) instead of P(B), or using the total 1 as the denominator. The denominator must be P(B), the probability of the "given" event.

Mistake 4 — Missing a Branch in Total Probability

P(A) = P(A|B)P(B) + P(A|B')P(B'). Students sometimes include only one term. Remember B' is "not B" — these two cases together must cover all possibilities. For 3 partitions, there are 3 terms.

Mistake 5 — Assuming Independence When Events Are Not Independent

"Without replacement" creates dependence. If the problem says "without replacement", the second draw always depends on the first. You cannot just use P(A) × P(B) = P(A∩B) for dependent events — that formula only works for independent events.

Key Formulas — Conditional Probability

Conditional probability: P(A|B) = P(A∩B) / P(B)

Multiplication rule: P(A∩B) = P(B) × P(A|B) = P(A) × P(B|A)

Independence: P(A|B) = P(A) ⟺ P(A∩B) = P(A)×P(B)

Law of Total Probability: P(A) = P(A|B)P(B) + P(A|B')P(B')

Bayes' Theorem: P(B|A) = P(A|B)×P(B) / P(A)

Tree diagram rules:
• Write probabilities ON branches (not at end-nodes)
• Multiply along branches for AND (joint probability)
• Add branch-products for OR (union via different paths)
• All end-products must sum to 1

Without replacement checklist:
1. Note total n and category counts
2. First branch: category/n
3. Second branch: if same colour drawn → (count−1)/(n−1); if different → count/(n−1)
4. Denominator always (n−1) on second branches

Complement shortcut:
P(at least one A) = 1 − P(no A) = 1 − P(A')ⁿ for n independent trials with replacement.

Two-Stage Tree Diagram Builder

Adjust sliders to set P(A), P(B|A) and P(B|A'). See the full tree drawn with all probabilities and computed results.

P(A) = 0.40

P(B|A) = 0.70

P(B|A') = 0.30

Exercise 1 — Without Replacement

Give all answers as decimals to 3 decimal places (or exact fractions converted to 3 d.p.).

1. Bag: 4 red, 3 blue (7 total). Draw 2 without replacement. P(both red) = ?

2. Same bag. P(both blue) = ?

3. Same bag. P(first red, second blue) = ?

4. Same bag. P(one of each colour, in any order) = ?

5. Bag: 5 green, 5 yellow (10 total). Draw 2 without replacement. P(both green) = ?

6. Same 5G/5Y bag. P(one green, one yellow) = ?

7. Bag: 3 red, 5 blue (8 total). P(second draw is red | first draw was red) = ?

8. Bag: 6 white, 4 black (10 total). Draw 2 without replacement. P(at least one white) = ?

Exercise 2 — Using the Conditional Formula

Use P(A|B) = P(A∩B)/P(B). Give answers to 3 decimal places.

1. P(A∩B)=0.2, P(B)=0.5. Find P(A|B).

2. P(A∩B)=0.12, P(B)=0.4. Find P(A|B).

3. P(A∩B)=0.15, P(A)=0.3. Find P(B|A).

4. P(B)=0.6, P(A|B)=0.35. Find P(A∩B).

5. 50 students: 30 passed Maths, 20 passed Science, 12 passed both. Find P(passed Maths | passed Science).

6. Same group. Find P(passed Science | passed Maths).

7. P(A)=0.45, P(B)=0.3, A and B are independent. Find P(A|B).

8. P(A∩B)=0.08, P(A)=0.2. Find P(B|A).

Exercise 3 — Law of Total Probability

Use P(A)=P(A|B)P(B)+P(A|B')P(B'). Give answers to 3 d.p.

1. P(B)=0.6, P(A|B)=0.4, P(A|B')=0.1. Find P(A).

2. P(B)=0.3, P(A|B)=0.8, P(A|B')=0.2. Find P(A).

3. P(B)=0.5, P(A|B)=0.6, P(A|B')=0.3. Find P(A).

4. Factory: Machine X makes 70%, Machine Y 30%. X defect rate 2%, Y defect rate 6%. Find P(defective).

5. P(B)=0.4, P(A|B)=0.5, P(A|B')=0.2. Find P(A).

6. Weather: P(rain)=0.35. P(late|rain)=0.5, P(late|no rain)=0.1. Find P(late).

7. P(B)=0.25, P(A|B)=0.9, P(A|B')=0.3. Find P(A).

8. P(B)=0.55, P(A|B)=0.7, P(A|B')=0.4. Find P(A).

Exercise 4 — Three-Event Trees

Each question involves 3 stages. Give answers to 3 decimal places.

1. P(success) = 0.6 each attempt, independent. Three attempts. P(all three succeed).

2. Same setup. P(exactly two successes).

3. Same setup. P(at least one success).

4. P(success) = 0.7 each trial, independent. P(exactly one success in three trials).

5. Fair coin three times. P(exactly two heads) = ?

6. Fair coin three times. P(at least two heads) = ?

7. P(A)=0.4 each trial. Three independent trials. P(exactly zero A events).

8. P(rain) = 0.3 each day. Three days. P(no rain on any day).

Exercise 5 — Bayes-Style Calculations

Use the total probability law then the conditional formula. Give answers to 3 d.p.

1. P(B)=0.4, P(A|B)=0.5, P(A|B')=0.2. Find P(B|A).

2. P(B)=0.6, P(A|B)=0.3, P(A|B')=0.7. Find P(B|A).

3. P(disease)=0.05, P(+|disease)=0.92, P(+|no disease)=0.04. Find P(disease|+).

4. P(B)=0.3, P(A|B)=0.8, P(A|B')=0.2. Find P(B|A).

5. Two boxes. Box A: 2R 3B. Box B: 4R 1B. Choose box at random. Draw one ball. It is red. Find P(it came from Box A).

6. P(B)=0.25, P(A|B)=0.9, P(A|B')=0.3. Find P(B|A).

7. P(disease)=0.02, P(+|disease)=0.95, P(+|healthy)=0.06. Find P(disease|positive).

8. P(B)=0.5, P(A|B)=0.6, P(A|B')=0.3. Find P(B|A).

Practice — 25 Questions

Mixed conditional probability. Give all answers to 3 decimal places.

1. P(A∩B)=0.3, P(B)=0.6. Find P(A|B).

2. P(A∩B)=0.18, P(A)=0.45. Find P(B|A).

3. Bag: 3R 2B (5). Draw 2 without replacement. P(both red).

4. Same bag. P(first red, second blue).

5. P(B)=0.5, P(A|B)=0.4, P(A|B')=0.2. Find P(A).

6. Same as Q5. Find P(B|A).

7. P(success)=0.5 per trial. Three independent trials. P(all three succeed).

8. Same. P(at least one success).

9. P(B)=0.7, P(A|B)=0.5, P(A|B')=0.1. Find P(A).

10. Bag: 5W 5B (10). Draw 2 without replacement. P(both white).

11. Same. P(one of each).

12. P(disease)=0.01, P(+|d)=0.99, P(+|h)=0.02. Find P(disease|+).

13. P(A)=0.4, P(B)=0.5, independent. Find P(A|B).

14. P(B)=0.35, P(A|B)=0.6, P(A|B')=0.3. Find P(A).

15. P(rain each day)=0.25. Three independent days. P(no rain at all).

16. P(B)=0.45, P(A|B)=0.7, P(A|B')=0.3. Find P(B|A).

17. Bag: 7R 3B (10). Draw 2 without replacement. P(both same colour).

18. P(A∩B)=0.06, P(B)=0.15. Find P(A|B).

19. P(success)=0.4 per trial. P(exactly two successes in three trials).

20. P(B)=0.6, P(A|B)=0.4, P(A|B')=0.1. Find P(B'|A).

21. 80 people: 50 like tea, 40 like coffee, 20 like both. Find P(likes tea | likes coffee).

22. P(B)=0.3, P(A|B)=0.5, P(A|B')=0.4. Find P(A).

23. P(B)=0.55, P(A|B)=0.8, P(A|B')=0.2. Find P(B|A).

24. Bag: 2R 4B (6). Draw 2 without replacement. P(at least one red).

25. P(B)=0.4, P(A|B)=0.6, P(A|B')=0.3. Find P(A).

Challenge — 12 Questions

Harder multi-step problems. Give answers to 3 d.p.

1. Bag: 3R 2G 1B (6). Draw 2 without replacement. P(both red).

2. Same bag. P(first red, second green).

3. P(B)=0.4, P(A|B)=0.5, P(A|B')=0.2. Find P(B'|A').

4. Disease P=0.003, sensitivity=0.98, false positive=0.01. Find P(disease|positive).

5. Three boxes (equal prob). Box1: 3R1B, Box2: 2R2B, Box3: 1R3B. Draw 1 ball, it's red. P(it came from Box1).

6. P(success)=0.6. Four independent trials. P(exactly two successes).

7. P(A)=0.5, P(B|A)=0.6, P(B|A')=0.3. Find P(A|B').

8. Bag: 4R 3B (7). Draw three without replacement. P(all three red).

9. P(B)=0.4, P(A|B)=0.7, P(A|B')=0.2. Find P(B|A).

10. 100 students: 60 study French, 50 study Spanish, 25 study both. Find P(studies Spanish | does NOT study French).

11. P(success)=0.7 per trial. P(first success on exactly the third trial) = P(FF S) = ?

12. P(B)=0.3, P(A|B)=0.9, P(A|B')=0.2. Find P(B'|A').

Exam Style Questions — 5 Questions

Cambridge IGCSE style. Show all working on paper. Enter your final answer to 3 d.p.

Q1. A bag contains 5 red and 3 blue counters. Two counters are drawn without replacement.
(a) Draw a complete probability tree diagram (on paper).
(b) Find the probability that both counters are the same colour. (3 marks)
Enter P(both same colour):

Q2. P(it rains on Monday) = 0.6. If it rains on Monday, P(it rains on Tuesday) = 0.7. If it does not rain on Monday, P(it rains on Tuesday) = 0.3.
Find P(it rains on Tuesday). (3 marks)
Enter P(rain Tuesday):

Q3. Using Q2 data: given that it rains on Tuesday, find the probability that it rained on Monday. (3 marks)
Enter P(Monday rain | Tuesday rain):

Q4. 200 people tested for a condition. 40 have the condition. Test gives positive for 36 of those 40. Test also gives positive for 12 of the 160 healthy people.
A person tests positive. Find P(they have the condition). (3 marks)
Enter the probability:

Q5. Bag A: 4 red, 1 blue. Bag B: 2 red, 3 blue. A fair coin is tossed: heads → draw from A; tails → draw from B.
Find P(red ball drawn). (2 marks)
Enter P(red):