Turn a without-replacement probability into an equation in n, solve the quadratic, and reject impossible roots. It is Higher only, aimed at grade 8. Free questions with answers are on this page. Worked examples show each line before you try the questions.
Unknown probabilities on a tree or Venn become algebra. Branches from a point still add to 1, so if one branch is x the other is 1−x. Along a route you multiply, so an 'and' path might be x(1−x). Set that equal to a given probability and solve the quadratic or linear equation. Reject roots that are not between 0 and 1, or that make another branch negative.
A two-way setup with P(A∩B)=0.12 and independence P(A)P(B)=0.12 is an equation in one unknown if one of them is x. Show the rejection of the illegal root — that is an accuracy mark. Write a probability sentence at the end, not just x=0.3.
Without replacement, P(both)=nk×n−1k−1. Set that equal to the given fraction and clear denominators to get a quadratic. Reject n≤0 or n too small to hold the coloured counters. Show-that questions stop at the unsimplified product.
Objectives
Write the two-draw probability as a product.
Clear the equation to a quadratic.
Reject the root that cannot be a number of items.
Key points
Write the probability of two picks without replacement as a product of two fractions whose denominators drop by 1.
Set the product equal to the given probability and cross-multiply to clear the fractions.
Expand, collect terms and rearrange into an2+bn+c=0.
Solve by factorising, and reject any root that is negative, not a whole number, or too small for the context.
Worked examples
Example 1
A bag has n counters, 6 of them red. Two are taken without replacement. P(both red) = 1/3. Find n.
2 marks
(6/n) * (5/(n-1)) = 1/3
30 * 3 = n(n-1), so n(n-1) = 90
n = 10
Answer: 10
Common slips
Slip: (6/n) * (6/n) = 1/3 The second denominator is n - 1.
Slip: n = 6 6 is the number of red counters, not the total. Solve n(n-1) = 90.
Slip: n = -9 Only 10 is a possible number of counters.
Example 2
A tin has n sweets, 3 of them mint. P(two mints without replacement) = 1/2. Find n.
2 marks
(3/n)*(2/(n-1)) = 1/2
12 = n(n-1)
n = 4
Answer: 4
Common slips
Slip: (3/n)*(2/n) = 1/2 Without replacement.
Slip: 6 = n Cross-multiplying gives n(n-1) = 12, not n = 6.
Slip: n = -3 A number of sweets cannot be negative, so only n = 4 works.
Example 3
There are n tickets, 5 of them winners. P(two winners without replacement) = 2/9. Find n.
3 marks
(5/n)*(4/(n-1)) = 2/9
20 * 9 = 2 n(n-1), so n(n-1) = 90
n = 10
Answer: 10
Common slips
Slip: 5/n = 2/9 Both draws appear in the product.
Slip: n(n-1) = 20 Divide by the probability carefully.
Slip: n = -9 Reject the negative root.
Questions and answers
GCSE Maths probability equation questions questions and answers. Try each one, then open the answer. Easy questions come first.
Question 1 Easy2 marks
n counters, 6 red. P(both red, no replacement) = 1/3. How many counters are in the bag?
Show answer
Answer: 10
Product (6/n)(5/(n-1)).
Reject the negative root.
M1 — Product (6/n)(5/(n-1)).
A1 — Answer 10
Question 2 Easy2 marks
n sweets, 3 mint. P(both mint, no replacement) = 1/5. Find n.
Show answer
Answer: 6
(3/n)(2/(n-1)) = 1/5.
n(n-1) = 30.
M1 — (3/n)(2/(n-1)) = 1/5.
A1 — Answer 6
Question 3 Medium2 marks
n cards, 4 red. P(both red, no replacement) = 3/14. Find n.
Show answer
Answer: 8
(4/n)(3/(n-1)) = 3/14.
n(n-1) = 56.
M1 — (4/n)(3/(n-1)) = 3/14.
A1 — Answer 8
Question 4 Medium2 marks
A bag has 6 red and 10 counters in total. The first draw is red and is not replaced. What is the probability the second is also red?
Show answer
Answer: 5/9
Five red counters remain.
Nine counters remain, so the probability is 5/9.
M1 — Five red counters remain.
A1 — Answer 5/9
Question 5 Hard2 marks
A box has n pens, 5 of them blue. Two pens are taken without replacement and P(both blue) = 2/9. Find n, and say why the other root fails.
Show answer
Answer: 10
Set up (5/n)(4/(n-1)) = 2/9.
The other root is negative, so it is not a number of pens.
M1 — Set up (5/n)(4/(n-1)) = 2/9.
A1 — Answer 10
Question 6 Hard2 marks
n stickers, 6 red. P(both red without replacement) = 1/7. Find n.
Show answer
Answer: 15
(6/n)(5/(n-1)) = 1/7.
n(n-1) = 210.
M1 — (6/n)(5/(n-1)) = 1/7.
A1 — Answer 15
Question 7 Exam-style5 marks
There are n tickets and 7 are winning tickets. Two tickets are taken without replacement. P(both winning) = 1/5. Find n.
Show answer
Answer: 15
(7/n)(6/(n-1)) = 1/5.
42 * 5 = n(n-1).
n^2 - n - 210 = 0.
n = 15 or n = -14.
n = 15.
M1 — Two-draw product
M1 — Clear the fraction
M1 — Quadratic
A1 — Positive root
A1 — Negative root rejected
Common mistakes
Most lost marks on this topic are one slip.
Using the same denominator for both draws
Without replacement the second denominator is one less
Keeping the negative root of the quadratic
A number of counters must be positive
Stopping at n(n-1) = 90 without solving
Rearrange to n^2 - n - 90 = 0 and solve it
Spot the slip
A bag has n counters and the quadratic gives n = 10 or n = -9. Which root do you keep?
10
-9
Both
Show the slip
10 A number of counters cannot be negative.
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.
Draw a quick tree diagram with the algebraic fractions on the branches — it earns method marks and stops you losing the '−1'.
In a 'show that' part, aim for the exact equation printed in the question; if yours differs by a factor, divide or multiply through and say so.
Printable worksheet
The same questions are in a PDF you can print. Answers start on a later page.