Higher June 2023 Paper 3 Q24
24 There is a total of \(y\) counters in a box.
There are \(x\) pink counters and 5 blue counters in the box.
The rest of the counters are green.
\(x : y = 1 : 3\)
Freda takes at random two counters from the box.
Find, in terms of \(x\), an expression for the probability that Freda takes two counters of the same colour.
Give your answer as a fraction in the form \(\dfrac{ax^2 + bx + c}{dx^2 + ex}\) where \(a\), \(b\), \(c\), \(d\) and \(e\) are integers. (5)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(\dfrac{5x^2 - 23x + 50}{9x^2 - 3x}\) | P1 | for using \(y = 3x\) or \(x = \dfrac{1}{3}y\) to obtain an expression for a probability in one variable |
| P1 | for a correct second probability eg \(\dfrac{x - 1}{y - 1}\) or \(\dfrac{x - 1}{3x - 1}\) or \(\dfrac{4}{y - 1}\) or \(\dfrac{4}{3x - 1}\) or \(\dfrac{y - x - 6}{y - 1}\) or \(\dfrac{2x - 6}{3x - 1}\) | |
| P1 | for forming a correct product eg \(\dfrac{x}{y} \times \dfrac{x - 1}{y - 1}\) or \(\dfrac{x}{3x} \times \dfrac{x - 1}{3x - 1}\) or \(\dfrac{5}{y} \times \dfrac{4}{y - 1}\) or \(\dfrac{5}{3x} \times \dfrac{4}{3x - 1}\) or \(\dfrac{y - x - 5}{y} \times \dfrac{y - x - 6}{y - 1}\) or \(\dfrac{2x - 5}{3x} \times \dfrac{2x - 6}{3x - 1}\) | |
| P1 | for adding the 3 correct probabilities eg \(\dfrac{x}{y} \times \dfrac{x - 1}{y - 1} + \dfrac{5}{y} \times \dfrac{4}{y - 1} + \dfrac{y - x - 5}{y} \times \dfrac{y - x - 6}{y - 1}\) or \(\dfrac{x}{3x} \times \dfrac{x - 1}{3x - 1} + \dfrac{5}{3x} \times \dfrac{4}{3x - 1} + \dfrac{2x - 5}{3x} \times \dfrac{2x - 6}{3x - 1}\) | |
| A1 | \(\dfrac{5x^2 - 23x + 50}{9x^2 - 3x}\) |
Additional guidance
This may be awarded at any time
Can be seen after processing of algebra