What is conditional probability?
Conditional probability is the probability of an event happening given that another event has already happened. It's written P(A | B), read as "the probability of A given B".
- P(A | B) = P(A ∩ B) ÷ P(B) — the formal formula
- Conditional probabilities can also be read directly from a two-way table (by restricting to one row or column), a Venn diagram (by restricting to one circle), or a tree diagram (the second set of branches)
- P(A | B) is not generally the same as P(B | A) — the order matters
Conditional probability from a two-way table
Worked Example 1
The table shows car and bike ownership for 100 people. Find P(owns a bike | owns a car).
| Owns a bike | No bike | Total | |
|---|---|---|---|
| Owns a car | 15 | 45 | 60 |
| No car | 25 | 15 | 40 |
| Total | 40 | 60 | 100 |
1
"Given owns a car" means restrict to the "Owns a car" row only — total 60 people
2
Of those 60, 15 also own a bike
3
P(bike | car) = 15 ÷ 60 = 1/4
Answer1/4
Worked Example 2
Using the same table, find P(owns a car | does not own a bike).
1
"Given no bike" means restrict to the "No bike" column — total 60 people
2
Of those 60, 45 own a car
3
P(car | no bike) = 45 ÷ 60 = 3/4
Answer3/4
Conditional probability from a tree diagram
Worked Example 3
A bag contains 3 red and 2 blue counters. Two are picked without replacement. Find P(second is red | first is red).
1
If the first counter is red, 2 red and 2 blue counters remain — 4 counters in total
2
P(second red | first red) = 2 ÷ 4 = 1/2
3
This is exactly the second-branch probability on the tree diagram — conditional probability is built into every tree diagram with dependent events
Answer1/2
The second set of branches shows conditional probabilities — P(red | red) = 2/4, but P(red | blue) = 3/4, because removing a counter changes what's left in the bag.
Using the conditional probability formula
Worked Example 4 — Formula directly
P(A ∩ B) = 0.12 and P(B) = 0.3. Find P(A | B).
1
Use P(A | B) = P(A ∩ B) ÷ P(B)
2
P(A | B) = 0.12 ÷ 0.3
3
= 0.4
Answer0.4
Practice questions
Work through each question before checking the answers.
Foundation (Grade 3–5)
Q1
The table shows sport preference by gender for 50 students. Find P(prefers football | is a boy).
Foundation| Football | Tennis | Total | |
|---|---|---|---|
| Boys | 18 | 7 | 25 |
| Girls | 9 | 16 | 25 |
Q2Using the data from Q1, find P(is a girl | prefers tennis).Foundation
Q3P(A ∩ B) = 0.2 and P(B) = 0.5. Find P(A | B).Foundation
Q4True or false: P(A | B) means the probability of B given A. Explain your answer.Foundation
Q5In a two-way table, 40 out of 60 people in a category satisfy a condition. Find the conditional probability.Foundation
Higher (Grade 5–7)
Q6
The table shows exam results for 90 students. Find P(passed | revised).
Higher| Passed | Failed | Total | |
|---|---|---|---|
| Revised | 42 | 8 | 50 |
| Didn't revise | 15 | 25 | 40 |
Q7Using the data from Q6, find P(failed | didn't revise).Higher
Q8Using the data from Q6, find P(revised | passed).Higher
Q9A bag has 3 red and 2 blue counters. Two are drawn without replacement. Find P(second is red | first is red).Higher
Q10Using the bag from Q9, find P(second is blue | first is red).Higher
Higher — Hard (Grade 8–9)
Q11P(A ∩ B) = 0.18 and P(A | B) = 0.45. Find P(B).Grade 8–9
Q12P(A) = 0.6, P(B) = 0.5, P(A ∩ B) = 0.3. Determine whether A and B are independent, using P(A | B).Grade 8–9
Q13P(studies French) = 0.4, P(studies French and Spanish) = 0.15. Find P(studies Spanish | studies French).Grade 8–9
Q14A bag has n red and 4 blue counters. Two are drawn without replacement. P(second is red | first is red) = 3/7. Find n.Grade 8–9
Q15Explain, using an example, why P(A | B) is not generally equal to P(B | A).Grade 8–9
Answers
Foundation (Q1–Q5)
Q118/25(restrict to the "Boys" row, total 25)
Q216/23(restrict to the "Tennis" column, total 23)
Q30.4(0.2 ÷ 0.5)
Q4False(P(A | B) means the probability of A given B — B has already happened)
Q52/3(40 ÷ 60)
Higher (Q6–Q10)
Q621/25(42 ÷ 50)
Q75/8(25 ÷ 40)
Q814/19(42 ÷ 57 — restrict to the "Passed" column)
Q91/2(2 ÷ 4 — 2 red left out of 4 remaining counters)
Q101/2(2 ÷ 4 — 2 blue left out of 4 remaining counters)
Higher — Hard (Q11–Q15)
Q110.4(P(B) = P(A∩B) ÷ P(A|B) = 0.18 ÷ 0.45)
Q12Independent(P(A|B) = 0.3 ÷ 0.5 = 0.6, which equals P(A) — B occurring doesn't change the probability of A)
Q130.375(0.15 ÷ 0.4)
Q14n = 4((n−1)/(n+3) = 3/7 → 7n − 7 = 3n + 9 → 4n = 16)
Q15Example: let A = "it is raining" and B = "the ground is wet". P(ground wet | raining) is very high — almost 1. But P(raining | ground wet) is much lower, since the ground could be wet for other reasons (a sprinkler, a spilled drink). The two conditional probabilities describe different situations and are generally not equal.
Common mistakes
Common Mistake 1
Dividing by the whole total instead of the "given" total
For P(A | B), you must divide by the total number in B (the given condition), not the overall total. Restrict your table, Venn diagram, or sample space to B first, then work out the fraction within that restricted group.
Common Mistake 2
Reversing P(A | B) and P(B | A)
These are generally different values. Always double-check which event is the "given" one (it comes after the vertical bar) — reading the notation the wrong way round is one of the most common errors on this topic.
Common Mistake 3
Forgetting to update probabilities in a "without replacement" tree
Once the first item is removed, both the total and the count of the relevant outcome change. The second-branch probabilities on the tree are conditional probabilities — they must be recalculated, not copied from the first branch.
Common Mistake 4
Misapplying the formula P(A | B) = P(A ∩ B) ÷ P(B)
Make sure P(A ∩ B) is the probability of both events happening, not just P(A). Confusing P(A ∩ B) with P(A) in this formula is a common algebraic slip, especially under exam pressure.
Exam tips
💡 Exam Tip 1
Identify the method before calculating
Decide first whether the question is best solved using a two-way table, a Venn diagram, or a tree diagram — the right representation makes the conditional probability far easier to see and calculate correctly.
💡 Exam Tip 2
Underline or circle the "given" condition in the question
Before writing any working, mark which event comes after the word "given" or "if" — this single habit prevents the most common error on this topic (reversing which event is conditional on which).
💡 Exam Tip 3
State the formula, even when using a table or diagram
Writing "P(A | B) = P(A ∩ B) ÷ P(B)" alongside your table or tree working shows the examiner you understand the underlying method, and helps you check your answer makes sense.
💡 Exam Tip 4
Use independence as a final check where relevant
If a question asks about independence, remember: A and B are independent exactly when P(A | B) = P(A) — calculating both sides and comparing is a clean, reliable way to answer this type of question.
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