Higher June 2025 Paper 1 Q25
25 There are \(n\) counters in a box.
6 of the counters are black.
Two counters are chosen at random without replacement.The probability that both counters are black is \(\dfrac{1}{8}\)
Use an algebraic method to work out the value of \(n\). [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\dfrac{6}{n}\) or \(\dfrac{5}{n - 1}\) | M1 | may be seen on a tree diagram |
| \(\dfrac{6}{n} \times \dfrac{5}{n - 1} = \dfrac{1}{8}\) | M1dep | oe eg \(n(n - 1) = 240\) |
| Correctly rearranges their equation, which must be correct or of the form \(\dfrac{6}{n} \times \dfrac{a}{n + b} = \dfrac{1}{8}\), where \(a\) and \(b\) are integers, into a quadratic equation with brackets expanded and no unknowns in denominators | M1dep | dep on 1st M1 \(n^2 - n = 240\) or \(n^2 - n - 240 = 0\) implies M1M1M1 |
| For their three-term quadratic, correctly factorises or correctly substitutes into the quadratic formula or correctly completes the square to the form \(n = \ldots\) for their quadratic or \(-15\) and 16 | M1 | eg \((n + 15)(n - 16)\ (= 0)\) eg \(\dfrac{--1 \pm \sqrt{(-1)^2 - 4 \times 1 \times -240}}{2 \times 1}\) eg \(\dfrac{1}{2} \pm \sqrt{240.25}\) |
| 16 with at least the first two marks awarded | A1 | SC1 16 with no other marks awarded and not from incorrect working |
Additional guidance
| Accept the use of any letter throughout | |
| 16 from trial and improvement, or without working | SC1 |
| \(\dfrac{6}{n} \times \dfrac{5}{n} = \dfrac{1}{8}\), \(n^2 = 240\), \(n = 4\sqrt{15}\) | M1M0M1 M0A0 |
| The second mark may be awarded for work done in stages eg \(\dfrac{6}{n} \times \dfrac{5}{n - 1} = \dfrac{36}{n(n - 1)}\) then \(\dfrac{36}{n(n - 1)} = \dfrac{1}{8}\) implies \(\dfrac{6}{n} \times \dfrac{5}{n - 1} = \dfrac{1}{8}\) (36 should be 30, which does not affect these marks but means they cannot get the third mark) | M1M1 |