June 2025 Paper 3 Q6
6. The discrete random variable \(R\) takes even integer values from 2 to \(2n\) inclusive.
The probability distribution of \(R\) is given by
\[\mathrm{P}(R = r) = \frac{r}{k} \qquad r = 2, 4, 6, \ldots, 2n\]where \(k\) is a constant.
When \(n = 20\)
When \(n = 20\), a random value \(g\) of \(R\) is taken and the quadratic equation in \(x\)
\[x^2 + gx + 3g = 5\]is formed.
| Scheme | Marks | AO |
|---|---|---|
| Use sum of probs = 1: \(1 = \dfrac{1}{k}(2 + 4 + \ldots + 2n)\) Allow \(1 = \dfrac{\mathrm{S}_r}{k}\) where \(\mathrm{S}_r = \displaystyle\sum_{r=1}^{n} 2r\) | M1 | 3.1a |
| Sum e.g. \(\dfrac{n}{2}[4 + 2(n - 1)]\) or \(\dfrac{n}{2}[2 + 2n]\) or use \(\displaystyle\sum_{1}^{n} r = \dfrac{n(n+1)}{2}\) | M1 | 2.1 |
| Use sum formula: e.g. \(1 = \dfrac{1}{k} \times \dfrac{n}{2}[4 + 2(n - 1)]\) or \(1 = \dfrac{n}{2}\left[\dfrac{2}{k} + \dfrac{2n}{k}\right]\) | A1 | 1.1b |
| \(k = n(n + 1)\) * | A1*cso | 1.1b |
| (4) |
Notes
M1: for clear attempt/intention to use sum of probs = 1 for \(n\) terms.
Must see \(k\) and “= 1” Condone missing … \(\displaystyle\sum_{2}^{2n} r\) is M0 since not \(n\) terms
M1: for using an arithmetic series or \(\Sigma r\) to find sum to \(n\) terms. No need for \(k\)
A1: (dep on M1M1) for a correct working leading to a correct equation in \(k\) and \(n\)
A1*: cso for correct solution with both Ms clearly scored and no incorrect working seen
SC: Start with \(1 = \dfrac{n}{2}\left[\dfrac{2}{k} + \dfrac{2n}{k}\right]\) or \(1 = \dfrac{1}{k} \times \dfrac{n}{2}[4 + 2(n - 1)]\) and no mention of \(a\) and \(l\) or \(a\) and \(d\)
Score SC B2 (M1M0A1A0) 1st B1 for the start and 2nd B1cso for completing to printed answer.
| Scheme | Marks | AO |
|---|---|---|
| Cases: \(R = 16\), 18, 20, 22 and 24 | M1 | 3.4 |
| Probability \(= \left[\dfrac{16 + 18 + 20 + 22 + 24}{20 \times 21}\right] = \dfrac{5}{21}\) | A1 | 1.1b |
| (2) |
Notes
M1: for identifying the correct values of \(R\)
A1: for a correct probability (any exact equivalent) Allow \(0.\dot{2}3809\dot{5}\)
Correct answer with no working scores 2 marks
Acc: If scored A0 for an answer of awrt 0.238 in (b), allow A1 in (c) for awrt 0.0429
| Scheme | Marks | AO |
|---|---|---|
| [No real roots \(\Rightarrow \Delta < 0\)] so \(g^2 - 4(3g - 5)[< 0]\) | M1 | 3.1a |
| \(\left[g^2 - 12g + 20 < 0 \Rightarrow\right]\ (g - 10)(g - 2)[< 0]\) | M1 | 2.1 |
| So require \(2 < g < 10\) or \(4 \leqslant g \leqslant 8\) | M1 | 1.1b |
| So require \(g = 4\), 6, 8 only | dM1 | 1.1b |
| Probability \(= \left[\dfrac{4 + 6 + 8}{20 \times 21}\right] = \dfrac{18}{420} = \dfrac{3}{70}\) | A1 | 3.2a |
| (5) | ||
| (11 marks) |
Notes
M1: for attempt to use discriminant [\(< 0\) not needed here]. Condone \(g^2 - 12g\)
M1: for attempting to factorise or find critical values for their 3TQ (ignore their \(<\) & allow \(= 0\))
M1: for choosing the correct (inside) region, ft their critical values from a 3TQ with \(< 0\)
dM1: (dep on 3rd M1) for choosing/identifying the appropriate values of \(g\)
(ft their critical values from 3TQ with \(< 0\) used)
A1: for an exact probability in any form. Allow \(0.0\dot{4}2857\dot{1}\)
A correct answer with no incorrect working will score full marks
Acc: If scored A0 for an answer of awrt 0.238 in (b), allow A1 in (c) for awrt 0.0429