Work out probabilities when the outcome of one event changes the chances of the next — from trees, Venn diagrams and tables. It is Higher only, aimed at grade 7. Free questions with answers are on this page. Worked examples show each line before you try the questions.
Conditional probability is P(A∣B)=P(B)P(A∩B): of the times B happens, how often is A also true? On a Venn diagram that is the overlap over the B circle. On a tree, it is a second-level branch, or a restricted set of end-routes over those routes that satisfy the condition.
Without replacement is conditional in action: after a red is drawn, P(red next) uses one fewer red and one fewer total. Independent events have P(A∣B)=P(A). Write the condition in a sentence ('given that it is a girl') so the denominator stays honest.
P(A∣B)=P(B)P(A∩B), or 'of the Bs, how many are A'. Without replacement, the second probability uses the reduced total. Independent: P(A∣B)=P(A).
Objectives
Calculate the probability of drawing a second counter given the first colour without replacement.
Calculate the probability that both counters drawn are of a given colour.
Calculate probabilities of drawing different colours without replacement.
Key points
Read 'given that' questions: the condition shrinks the total you divide by.
Draw a tree diagram for picking without replacement, adjusting the second-stage probabilities.
Multiply along branches for 'and'; add branches for 'or'.
Find conditional probabilities from a Venn diagram or two-way table by restricting to one set or row.
Worked examples
Example 1
Two counters are drawn without replacement. 5 red and 3 blue. What is the probability the second is red given that the first was red?
2 marks
Note that 1 red counter has been removed from the bag.
Slip: Use 6/9. Initially there are 6 red out of 10 counters.
Slip: Use 6/10 again. One red has been removed, leaving 5 red out of 9.
Slip: Leave answer as 30/90 without simplifying. 30/90 simplifies to 1/3.
Questions and answers
GCSE Maths conditional probability questions and answers. Try each one, then open the answer. Easy questions come first.
Question 1 Easy2 marks
Two counters are drawn without replacement. 6 red and 4 blue. What is the probability the second is red given that the first was red?
Show answer
Answer: 5/9
Red remaining: 6 − 1 = 5; total remaining: 10 − 1 = 9.
P = 5/9.
M1 — Red remaining: 6 − 1 = 5; total remaining: 10 − 1 = 9.
A1 — P = 5/9.
Question 2 Easy2 marks
Two counters are drawn without replacement. 4 red and 2 blue. What is the probability the second is red given that the first was red?
Show answer
Answer: 3/5
Red remaining: 4 − 1 = 3; total remaining: 6 − 1 = 5.
P = 3/5.
M1 — Red remaining: 4 − 1 = 3; total remaining: 6 − 1 = 5.
A1 — P = 3/5.
Question 3 Medium2 marks
Two counters are drawn without replacement. 3 red and 3 blue. What is the probability both are red?
Show answer
Answer: 1/5
First red: 3/6 = 1/2; second red: 2/5.
1/2 × 2/5 = 1/5.
M1 — First red: 3/6 = 1/2; second red: 2/5.
A1 — 1/2 × 2/5 = 1/5.
Question 4 Medium2 marks
Two counters are drawn without replacement. 4 red and 4 blue. What is the probability both are red?
Show answer
Answer: 3/14
First red: 4/8 = 1/2; second red: 3/7.
1/2 × 3/7 = 3/14.
M1 — First red: 4/8 = 1/2; second red: 3/7.
A1 — 1/2 × 3/7 = 3/14.
Question 5 Hard2 marks
Two counters are drawn without replacement. 5 red and 3 blue. What is the probability the second is blue given that the first was red?
Show answer
Answer: 3/7
Blue remaining: 3; total remaining: 7.
P = 3/7.
M1 — Blue remaining: 3; total remaining: 7.
A1 — P = 3/7.
Question 6 Hard2 marks
Two counters are drawn without replacement. 6 red and 4 blue. What is the probability the second is blue given that the first was red?
Show answer
Answer: 4/9
Blue remaining: 4; total remaining: 9.
P = 4/9.
M1 — Blue remaining: 4; total remaining: 9.
A1 — P = 4/9.
Question 7 Exam-style4 marks
A bag contains 7 red and 5 blue counters. Two counters are taken at random without replacement. Work out the probability that the two counters are different colours.
Show answer
Answer: 35/66
Red then blue: (7/12) × (5/11) = 35/132.
Blue then red: (5/12) × (7/11) = 35/132.
Add the two cases: 35/132 + 35/132 = 70/132.
Simplify: 70/132 = 35/66.
M1 — (7/12) × (5/11) or (5/12) × (7/11)
M1 — Both cases found
M1 — Adds the two cases
A1 — 35/66
Common mistakes
Most lost marks on this topic are one slip.
Using the original total on the second draw
Without replacement, the total number of items decreases by 1 on the second draw
Writing a rounded decimal instead of an exact fraction
Keep probabilities as exact simplified fractions
Adding branch probabilities instead of multiplying
To find the probability that both events occur, multiply the probabilities along the branch
Spot the slip
A bag contains 5 red and 3 blue counters. Two counters are drawn without replacement. The first counter is red. A student thinks the probability the second is red is still 5/8. What is the correct probability?
4/7
5/8
0.333
Show the slip
4/7 One red counter has been removed, leaving 4 red counters out of 7 total counters. The probability is 4/7.
Exam tips
Be ready to do this without a calculator. Paper 1 is the non-calculator paper.
If the question says “show that”, the result is given. The marks are for the steps that reach it.
The method is the same on AQA-style and Edexcel-style papers.
Write the probabilities on every branch of the tree, even the ones you do not use — this is usually a mark on its own.
Write 'given that' answers as a fraction with the restricted total as the denominator before simplifying, so the marker can see the reasoning.
Printable worksheet
The same questions are in a PDF you can print. Answers start on a later page.