A2 June 2025 Q1
1. Irina is practising her serves in badminton and counts the number of her serves that are faults. She finds that 15% of her serves are faults.
Assuming that each serve is independent,
With practice, Irina reduces her proportion of faults, \(p\), so that the mean number of serves until her 4th fault is at least 32
| Scheme | Marks | AO |
|---|---|---|
| (i) \(X \sim \mathrm{NB}(4, 0.15)\) | M1 | 3.3 |
| \(\displaystyle\binom{19}{3}(0.15)^4(1 - 0.15)^{16} = 0.03642\) awrt 0.0364 | A1 | 1.1b |
| (2) | ||
| (ii) \(X \sim \mathrm{B}(18, 0.15)\) | M1 | 3.3 |
| \(\displaystyle\binom{18}{4}(0.15)^4(1 - 0.15)^{14} = 0.15920\ldots\) awrt 0.159 | A1 | 1.1b |
| (2) |
Notes
(i) M1: selecting NB(4, 0.15) or use of a correct method, may be implied by a correct answer
A1: awrt 0.0364
(ii) M1: selecting B(18, 0.15) or use of a correct method, may be implied by a correct answer
A1: awrt 0.159
| Scheme | Marks | AO |
|---|---|---|
| Uses \(\dfrac{4}{p}\) as an expression for the mean | B1 | 3.4 |
| \(\dfrac{4}{p} \geqslant 32\) | M1 | 1.1b |
| \(p = \dfrac{1}{8}\) or 0.125 or 12.5% | A1 | 1.1b |
| (3) | ||
| (7 marks) |
Notes
B1: [\(\mathrm{E}(X)\)] \(= \dfrac{4}{p}\)
M1: for an equation or inequality with their expression for mean and 32
allow mean = 32 or mean \(\gt\) 32
mean \(\leqslant\) 32 or mean \(\lt\) 32 is M0
A1: 0.125 or exact equivalent
Must not come from clearly incorrect working
NB: \(p \leqslant \dfrac{1}{8}\) is B1M1A0
M0 implies A0 (method mark cannot be implied here)