The answer
(a) \(\dfrac{10 - n}{10}\)
(b) \(\dfrac{(10 - n)(9 - n)}{90}\)
(c)(i) shown
(ii) \(3\) yellow marbles
O-Level E-Math 2015 Paper 1 Question 24 · Verified worked solution by the Genius Plus Academy teaching team
What this question tests
This is Question 24 of the O-Level E-Math 2015 Paper 1. It tests probability without replacement, in the Probability (combined events) / Quadratic equations area. It is worth 7 marks: 1 + 1 + 2 + 3. It is a worded / diagram-based question, so open your Ten-Year Series (TYS) or the official paper at this question, then follow our full worked solution below.
There are 10 marbles in total, with \(n\) red and \((10 - n)\) yellow.
(a) \(P(\text{first is yellow}) = \dfrac{10 - n}{10}\).
(b) After one yellow is removed (not replaced), there are 9 marbles left with \((9 - n)\) yellow: \[P(\text{both yellow}) = \frac{10 - n}{10} \times \frac{9 - n}{9} = \frac{(10 - n)(9 - n)}{90}.\]
(c)(i) Set this equal to \(\dfrac{1}{15}\): \[\frac{(10 - n)(9 - n)}{90} = \frac{1}{15} \Rightarrow (10 - n)(9 - n) = \frac{90}{15} = 6.\] Expanding: \(90 - 19n + n^2 = 6\), so \(n^2 - 19n + 84 = 0\). (shown)
(ii) Factorise: \(n^2 - 19n + 84 = (n - 7)(n - 12) = 0\), so \(n = 7\) or \(n = 12\). Since there are only 10 marbles, \(n = 12\) is impossible, so \(n = 7\) red marbles. The number of yellow marbles is \(10 - 7 = 3\).
Answer: (a) \(\dfrac{10 - n}{10}\)
(b) \(\dfrac{(10 - n)(9 - n)}{90}\)
(c)(i) shown
(ii) \(3\) yellow marbles
Same structure, different numbers
Swap the constants, dress a quadratic as a length, hide a derivative inside an integral, and a student sees a brand new problem. The structure underneath is the same, and so is the method. Once a student can name the structure, a whole row of questions that look different start to open the same way.
That is where marks really leak: in choosing the method, not in the algebra that follows. We call it Lock and Key, name the lock, then the key follows.
Want more questions like this, with worked solutions?
Join our mailing list and we will send practice sets and worked solutions. One email, no spam.
Genius Plus Academy · O-Level & IP Mathematics
Our O-Level E-Math tuition trains the same recognise-the-structure method these worked solutions show, taught by a team that has marked these papers for years. It runs within our weekly Secondary Math programme, Sec 1 to 4 and IP.
It is a probability without replacement question from Probability (combined events) / Quadratic equations, worth 7 marks: 1 + 1 + 2 + 3.
Yes. IP (Integrated Programme) schools teach the same O-Level Mathematics content; they just sequence it differently and set their own internal exams, so these worked solutions apply to IP students too.
Yes. Every worked solution here is free to read, with no sign-up wall.
Browse E-Math and A-Math by year in our worked-solutions library at /resources/solutions/o-level/.
Book a free trial and diagnostic. We will look at a real paper and show you exactly where the marks are going.