Pearson Edexcel · GCSE · Mathematics
Probability
Conditional Probability asks 'what is the chance of this happening, given that something else already has?' It's a high-value topic that separates top-tier candidates, requiring you to master tree diagrams, Venn diagrams, and the crucial formula P(A|B) = P(A∩B)/P(B).
- 5 min read
- 3 worked examples
- 5 practice questions
- 6 key terms
Study Notes
Overview
Conditional probability is a fascinating and crucial area of Mathematics that deals with how the likelihood of an event changes when we have extra information. Think about it: the probability of someone wearing sunglasses is quite low generally, but given that it is a sunny day, that probability increases significantly. This is conditional probability in action.
In your GCSE exams, conditional probability is a staple of the Higher tier papers. It tests not just your ability to crunch numbers, but your logical reasoning and your capacity to interpret changing scenarios. Examiners love to test this through 'without replacement' problems (like taking sweets from a bag without putting them back), Venn diagrams, and two-way tables.
Mastering this topic is essential because it connects deeply with other areas of statistics and provides a foundation for A-Level Mathematics. Typical exam questions will ask you to complete a tree diagram and calculate combined probabilities, or to interpret a Venn diagram to find the probability of one event given another.
Listen to our comprehensive audio guide for a detailed walkthrough of the key concepts:
Key Concepts
Concept 1: The Concept of 'Given'
The fundamental idea behind conditional probability is that knowing an event has occurred changes our sample space (the total number of possible outcomes). If we are looking for the probability of Event A given Event B, we restrict our entire world just to the times when Event B happens.
Why this works: Probability is always (Number of successful outcomes) / (Total possible outcomes). When we add a condition, we are shrinking the denominator (the total possible outcomes) to only include the cases where the condition is met.
Example: In a class of 30 students, 12 play football, 10 play tennis, and 5 play both. If a student is chosen at random, the probability they play football is 12/30. But if we are told the student plays tennis (this is the condition), we only look at the 10 tennis players. Out of those 10, 5 play football. So the probability they play football given they play tennis is 5/10.
Concept 2: Probability Tree Diagrams (Without Replacement)
Tree diagrams are brilliant for visualising sequential events. When events are dependent (conditional), the probabilities on the second set of branches change based on what happened on the first branch.
Why this works: If you have 5 red sweets and 3 blue sweets, and you eat a red one, the bag now contains 4 red sweets and 3 blue sweets. The total number of sweets has decreased from 8 to 7, and the number of red sweets has decreased from 5 to 4. The second branch must reflect this new reality.
Example: A bag has 6 green and 4 yellow counters. You take two without replacing the first.
- P(Green on 1st pick) = 6/10
- If you picked Green first, the bag now has 5 green and 4 yellow (9 total). So P(Green on 2nd pick | Green on 1st pick) = 5/9.
- To find P(Green AND Green), you multiply along the branches: (6/10) × (5/9) = 30/90 = 1/3.
Concept 3: Venn Diagrams and Conditional Probability
Venn diagrams are incredibly powerful for solving conditional probability problems when events happen simultaneously rather than sequentially.
Why this works: The condition tells you which circle to focus on. If the question asks for the probability of A given B, you completely ignore everything outside circle B. Your new denominator is the total of circle B. Your numerator is the part of A that is inside B (the intersection).
Example: If circle A has 15, circle B has 20, the intersection has 8, and the outside has 5.
- Total in B = 20.
- Number in A that are also in B = 8.
- Therefore, P(A|B) = 8/20 = 2/5.
Mathematical Relationships
The most important formula for this topic is the conditional probability formula:
**P(A|B) = P(A ∩ B) / P(B)**Where:
- P(A|B) means 'the probability of A occurring, given that B has already occurred'.
- P(A ∩ B) means 'the probability of both A and B occurring together' (the intersection).
- P(B) means 'the probability of B occurring'.
Note: This formula is often NOT given on the formula sheet. You must memorise it!
You can also rearrange this formula to find the probability of both events occurring:
**P(A ∩ B) = P(A|B) × P(B)**This is exactly what you are doing when you multiply along the branches of a tree diagram!
Practical Applications
Conditional probability is used extensively in the real world:
- Medical Testing: If a patient tests positive for a disease, what is the probability they actually have the disease? (This depends on the false positive rate of the test and the rarity of the disease in the population).
- Spam Filters: Given that an email contains the word 'lottery' and 'winner', what is the probability that it is spam?
- Insurance: Given that a driver is under 25 and drives a sports car, what is the probability they will make a claim?
Visual Resources
2 diagrams and illustrations
Interactive Diagrams
2 interactive diagrams to visualise key concepts
Conceptual Flow Outline
Decision flowchart for tackling probability questions.
Conceptual Flow Outline
The logical flow of conditional probability.
Worked Examples
3 worked examples — open one to explore the question and available guidance.
Practice Questions
Test your understanding — click to reveal model answers
A bag contains 5 red balls and 4 green balls. Two balls are drawn at random without replacement. Calculate the probability that the second ball is green, given that the first ball was red.
Hint: Think about what is left in the bag AFTER the red ball is removed.
In a sixth form of 150 students, 85 study Maths, 60 study Physics, and 40 study both. A student is chosen at random. Given that the student studies Physics, find the probability that they do not study Maths.
Hint: Draw a quick Venn diagram or use the formula. What is the denominator?
A box contains 10 chocolates. 6 are milk chocolate and 4 are dark chocolate. Three chocolates are chosen at random and eaten. Calculate the probability that exactly two of the chocolates eaten are dark chocolate.
Hint: List all the possible combinations that give exactly two dark chocolates (e.g., DDM, DMD, MDD). Calculate each pathway.
Events A and B are such that P(A) = 0.5, P(B) = 0.6 and P(A ∪ B) = 0.8. Find P(A|B).
Hint: Use the addition rule P(A ∪ B) = P(A) + P(B) - P(A ∩ B) to find the intersection first.
A diagnostic test for a virus is 95% accurate (if you have the virus, it tests positive 95% of the time; if you don't, it tests negative 95% of the time). 2% of the population actually has the virus. A person is chosen at random and tests positive. Calculate the probability they actually have the virus.
Hint: Draw a tree diagram. First branches: Has Virus / Doesn't Have Virus. Second branches: Tests Positive / Tests Negative.