P9: Conditional Probability
Calculate and interpret conditional probabilities; use two-way tables, tree diagrams and Venn diagrams
📋 Key Concepts
Definition: Conditional probability is the probability of an event occurring GIVEN that another event has already occurred. Written as P(A|B) - "probability of A given B".
Key Formula
Conditional Probability Formula:
P(A|B) = P(A ∩ B) ÷ P(B)
In words: P(A given B) = P(A and B) ÷ P(B)
P(A|B) = P(A ∩ B) ÷ P(B)
In words: P(A given B) = P(A and B) ÷ P(B)
Key Terms
| Notation | Meaning |
|---|---|
| P(A|B) | Probability of A given that B has occurred |
| P(A ∩ B) | Probability of A and B both occurring |
| P(A ∪ B) | Probability of A or B or both occurring |
📝 Understanding Conditional Probability
Key Idea: When we know B has happened, we restrict our sample space to only outcomes where B occurred.
Example 1
In a class of 30 students: 18 study Maths, 12 study English, 6 study both. Find P(studies Maths | studies English).
Solution:
P(Maths | English) = P(Maths ∩ English) ÷ P(English)
= (6/30) ÷ (12/30)
= 6/12 = 1/2
Alternative: Among the 12 who study English, 6 study Maths. So 6/12 = 1/2
📝 Two-Way Tables
Method: Use the row or column totals to find conditional probabilities. Restrict to one row or column.
Example 2
A survey of 100 people:
| Left-handed | Right-handed | Total | |
|---|---|---|---|
| Male | 8 | 42 | 50 |
| Female | 4 | 46 | 50 |
| Total | 12 | 88 | 100 |
Find:
a) P(Male | Left-handed)
b) P(Left-handed | Male)
Solutions:
a) P(Male | Left-handed) = 8/12 = 2/3
Among left-handed people (12), 8 are male
b) P(Left-handed | Male) = 8/50 = 4/25
Among males (50), 8 are left-handed
📝 Tree Diagrams for Conditional Probability
Method: The second set of branches shows conditional probabilities. Probabilities change based on what happened first.
Example 3
A bag has 5 red and 3 blue balls. Two balls are drawn without replacement. Find P(second is red | first was blue).
Solution:
After drawing a blue ball, 5 red and 2 blue remain.
P(second is red | first was blue) = 5/7
This is read directly from the tree diagram branch.
Example 4
Using the same bag, find P(first was blue | second is red).
Solution:
P(B|R) = P(B ∩ R) / P(R)
P(B ∩ R) = P(B first) × P(R second | B first) = 3/8 × 5/7 = 15/56
P(R) = P(RR) + P(BR) = (5/8 × 4/7) + (3/8 × 5/7) = 20/56 + 15/56 = 35/56
P(B|R) = (15/56) ÷ (35/56) = 15/35 = 3/7
📝 Venn Diagrams and Conditional Probability
Method: P(A|B) is the proportion of B that also contains A. Find values from the Venn diagram.
Example 5
60 students: 35 study Maths, 28 study Physics, 15 study both. Find P(studies Physics | studies Maths).
Solution:
Maths only = 35 - 15 = 20
Physics only = 28 - 15 = 13
Neither = 60 - 20 - 15 - 13 = 12
P(Physics | Maths) = P(Physics ∩ Maths) / P(Maths)
= 15/35 = 3/7
Among 35 Maths students, 15 also study Physics
📝 Independent Events and Conditional Probability
Test for Independence: If P(A|B) = P(A), then A and B are independent. The probability of A doesn't change when we know B.
Independent Events: P(A|B) = P(A)
Also: P(A ∩ B) = P(A) × P(B)
Also: P(A ∩ B) = P(A) × P(B)
Example 6
From the earlier two-way table, are "being male" and "being left-handed" independent?
Solution:
P(Left-handed) = 12/100 = 0.12
P(Left-handed | Male) = 8/50 = 0.16
Since 0.16 ≠ 0.12, the events are NOT independent
📝 Multiplication Rule
Formula: P(A ∩ B) = P(A) × P(B|A) = P(B) × P(A|B)
Example 7
P(A) = 0.6 and P(B|A) = 0.3. Find P(A ∩ B).
Solution:
P(A ∩ B) = P(A) × P(B|A) = 0.6 × 0.3 = 0.18
❓ Practice Questions
Q1: In a group: P(A) = 0.4, P(B) = 0.5, P(A ∩ B) = 0.2. Find P(A|B).
Q2: A box has 4 red and 6 blue counters. Two are drawn without replacement. Find P(second red | first blue).
Q3: In a survey: P(likes tea) = 0.6, P(likes coffee) = 0.5, P(likes both) = 0.3. Find P(likes coffee | likes tea).
Q4: P(A|B) = 0.4 and P(B) = 0.3. Find P(A ∩ B).
Q5: Two dice are rolled. Given that the sum is 8, find P(at least one die shows 4).
✅ Answers
- P(A|B) = P(A ∩ B) / P(B) = 0.2 / 0.5 = 0.4
- After blue drawn: 4 red, 5 blue left. P(second red) = 4/9
- P(coffee | tea) = P(both) / P(tea) = 0.3 / 0.6 = 0.5
- P(A ∩ B) = P(A|B) × P(B) = 0.4 × 0.3 = 0.12
- Sum = 8: (2,6), (3,5), (4,4), (5,3), (6,2) - 5 outcomes. At least one 4: (4,4) only. P = 1/5
🎯 Exam Tips
- Read "given that" or "if" as the conditional event (the one after |)
- P(A|B) ≠ P(B|A) - the order matters!
- In two-way tables, conditional probability restricts to one row or column
- If P(A|B) = P(A), events are independent
- Use P(A ∩ B) = P(A) × P(B|A) when you need the probability of both events
🧠 Problem-Solving Strategies
Problem-Solving
For conditional probability: (1) Identify what is "given" — this restricts your sample space, (2) P(A|B) ≠ P(B|A) — the order matters!, (3) In two-way tables, restrict to one row or column when finding conditional probability, (4) For Venn diagrams, P(A|B) = P(A ∩ B)/P(B) — divide the overlap by the "given" set, (5) If P(A|B) = P(A), the events are independent.
Multi-Step Problem
A survey of 80 people shows: 50 own a phone, 35 own a tablet, 20 own both. Find P(owns tablet | owns phone) and P(owns phone | owns tablet). Show they are different.
Solution: P(tablet|phone) = 20/50 = 2/5 = 0.4. P(phone|tablet) = 20/35 = 4/7 ≈ 0.571. These are different — among phone owners, 40% also own a tablet; among tablet owners, 57% also own a phone. This shows P(A|B) ≠ P(B|A).
⚠️ Common Errors
Watch Out!
1. Wrong: Assuming P(A|B) = P(B|A) — swapping the condition and the event Correct: P(A|B) = P(A∩B)/P(B) while P(B|A) = P(A∩B)/P(A) — these are usually different unless P(A) = P(B)
2. Wrong: Dividing by the total population instead of dividing by P(B) when finding P(A|B) Correct: P(A|B) = P(A∩B)/P(B) — you must restrict the denominator to those where B occurred
3. Wrong: Confusing independent events with mutually exclusive events Correct: Independent means P(A|B) = P(A); mutually exclusive means P(A∩B) = 0. Mutually exclusive events are NEVER independent (unless one has probability 0).
✍️ 6-Mark Exam Question
Extended Answer
6 marks: A school has 120 students in Year 12. 75 study Maths, 50 study Physics, 30 study both. A student is chosen at random. (a) Find P(studies Physics | studies Maths). (b) Find P(studies Maths | studies Physics). (c) Are "studying Maths" and "studying Physics" independent? Justify your answer using conditional probability.
(a) P(Physics|Maths) = P(Physics ∩ Maths)/P(Maths) = 30/75 = 2/5 = 0.4.
(b) P(Maths|Physics) = P(Maths ∩ Physics)/P(Physics) = 30/50 = 3/5 = 0.6.
(c) For independence, P(Physics|Maths) must equal P(Physics). P(Physics) = 50/120 = 5/12 ≈ 0.417. Since P(Physics|Maths) = 0.4 ≠ 0.417, the events are NOT independent. Studying Maths changes the probability of studying Physics.
Mark scheme: M1 for correct formula, A1 for 2/5, M1 for correct formula, A1 for 3/5, M1 for comparing P(Physics|Maths) with P(Physics), A1 for "not independent" with numerical justification
📊 AO3: Reason & Interpret
Reasoning and Interpretation
A company has 200 employees. The table shows attendance at a training course by department:
| Attended | Not Attended | Total | |
|---|---|---|---|
| Sales | 40 | 60 | 100 |
| Tech | 70 | 30 | 100 |
| Total | 110 | 90 | 200 |
(a) Find P(Attended | Tech department).
(b) Find P(Tech department | Attended).
(c) The manager says "Tech employees are more likely to attend than Sales employees." Is this correct? Does this mean being in Tech causes higher attendance? Explain.
Answers: (a) 70/100 = 0.7 (b) 70/110 = 7/11 ≈ 0.636 (c) Yes — P(Attended|Tech) = 0.7 vs P(Attended|Sales) = 0.4, so Tech employees are more likely to attend. However, correlation ≠ causation — there may be other factors (e.g. Tech roles require training, or Tech managers prioritise it).
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