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Conditional probability

Free Printable Conditional probability PDF Worksheets, Revision Guide & GCSE Past Paper Questions (Grade 7)

Download free conditional probability worksheets (PDF), GCSE maths past paper questions and worked solutions for Conditional probability (7S). This is a Grade 7 probability topic from the GCSE questions by topic collection for Higher tier GCSE Maths (Edexcel, AQA, OCR and IGCSE).

The free PDF worksheets table below brings together printable question booklets (Qs) and worked answers or mark schemes (Ans) from Maths Genie, Maths4Everyone, Corbett Maths, Maths Made Easy, Dr Frost, 1st Class Maths, Dr Austin, E Classroom and PMT, along with past paper question booklets.

Below the worksheet links you can read our conditional probability revision notes and worked examples with diagrams, try real GCSE past paper conditional probability questions with step-by-step solutions, or download a PDF of all conditional probability worksheet links.

Free Conditional probability PDF Worksheets & Answers

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GCSE Maths Grade 7 Revision Notes & Method

Guide to Conditional probability: Explanation, Diagrams & Worked Examples

Download Printable One Page Guide (PDF)

Explanation:

Conditional probability occurs when one event changes the probabilities for the next event, most commonly when items are chosen without replacement. On a tree diagram, always reduce both the total denominator by 1 on the second set of branches and the numerator by 1 for whichever item was picked first.

Multiply along branches for AND, and add the results of separate paths for OR. For "at least one" or "two of the same colour", you can also use 1 − P(opposite event) as a fast check. For Venn diagrams or two-way tables, restrict the denominator strictly to the "given" group.

Example 1: A bag has 5 red and 3 blue counters. Two are taken at random without replacement. Find P(one of each colour).

5/8 3/8 R B 4/7 3/7 5/7 2/7 R B → 15/56 R → 15/56 B

Solution:

P(R, B) = 58 × 37 = 1556;  P(B, R) = 38 × 57 = 1556
Add both paths: 1556 + 1556 = 3056
Answer: 3056 (or 1528)

Example 2: In a group of 50 students, 28 study French, 20 study Spanish, and 9 study both. Given a student studies French, find P(studies Spanish).

ξ = 50 F (28) S (20) 19 9 11 11

Solution:

Restrict sample space to the 28 French students:
Number in French circle also in Spanish = 9
P(Spanish | French) = n(F ∩ S)n(F) = 928
Answer: 928
Scanned GCSE Exam Questions & Worked Solutions

GCSE Past Paper Conditional probability Questions & Answers

Past Paper Example 1 GCSE Nov 2023 2H Q21 4 marks
Vicky has a collection of medals.
The Venn diagram gives information about the number of medals in her collection where
[INDENT]\mathscr{E} = {all medals}
[INDENT]A = {English medals}
[INDENT]B = {gold medals}
ℰAB7x – 11x5x – 334
Vicky is going to take at random a medal from her collection.
Given that the medal is gold, the probability that the medal is English is 211
Work out the number of medals in Vicky’s collection.
Worked Solution & Mark Scheme (4 marks)
Total gold medals n(B) = x + (5x – 3) = 6x – 3. P(A | B) = x / (6x – 3) = 2/11 M2
⇒ 11x = 12x – 6 M1
⇒ x = 6. Total medals = (7(6) – 11) + 6 + (5(6) – 3) + 34 = 31 + 6 + 27 + 34 = 98. A1
Answer: 98
Past Paper Example 2 AQA Nov 2020 2H Q27 4 marks
These 20 discs are in a bag.
Two of the discs are taken at random from the bag.
1111111122222222222233333333333333444444
Work out the probability that the first disc has a smaller number than the second disc.
Worked Solution & Mark Scheme (4 marks)
Disc counts: four 11, six 22, seven 33, three 44 M1
Total = 20
Pairs where 1st < 2nd:
P(11, 22) = 4 × 620 × 19 = 24380 M1
P(11, 33) = 4 × 720 × 19 = 28380
P(11, 44) = 4 × 320 × 19 = 12380
P(22, 33) = 6 × 720 × 19 = 42380 M1
P(22, 44) = 6 × 320 × 19 = 18380
P(33, 44) = 7 × 320 × 19 = 21380
Sum = 24 + 28 + 12 + 42 + 18 + 21380 = 145380 = 2976. A1
Answer: Answer 2976
Past Paper Example 3 OCR Jun 2025 4H Q22 5 marks
A school has 90 students in Year 11.
• 50 of the students study French.
• 59 of the students study Spanish.
• 4 of the students do not study either of these languages.
One of the students is chosen at random.
Find the probability that the student studies French given that they study Spanish.
You may use the grid **or** the Venn diagram to help present your working if you wish.
ℰ
Worked Solution & Mark Scheme (5 marks)
Number studying French or Spanish = 90 − 4 = 86 M2
Number studying both French and Spanish = 50 + 59 − 86 = 23 M2
P(French | Spanish) = 23 / 59. A1
ℰSS'F232750F'36440593190FS2723364
Answer: 2359

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