| Exam Board | OCR |
|---|---|
| Module | Further Statistics AS (Further Statistics AS) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Direct binomial probability calculation |
| Difficulty | Standard +0.3 Part (i) is a standard hypergeometric probability calculation requiring summation of four terms, which is routine but slightly more involved than basic binomial questions. Part (ii) requires setting up an inequality involving adjacent arrangements (2(n-1)!/n! < 0.1), which demands some combinatorial reasoning beyond pure recall, but the algebra is straightforward once the setup is correct. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \dfrac{335}{8398}\) or \(\mathbf{0.039890\ldots}\) | M1 M1 B1 A1 [4] | M1: Any one correct pair of \(^nC_r\) multiplied. M1: Three pairs added. B1: \(^{20}C_9\) or 167960 seen anywhere; all correct A1. A1: Exact or awrt 0.0399 |
| Answer | Marks | Guidance |
|---|---|---|
| \(n_{\min} = 21\) | M1 M1 M1 A1 [4] | SC: any other method: \(\dfrac{2}{n}\) B2. Compare their \(\dfrac{2}{n}\) with 0.1, needs division. Not \(\geq 21\). SC: \(\dfrac{2}{n+1}\), 22: B1+B1. SC: \(\dfrac{1}{n}\), 11: B1+B1. T&I: any one of below B2: \(\frac{19}{0.10}\ \frac{20}{0.1}\ \frac{21}{0.09}\ \frac{22}{0.09}\ \frac{}{5}\ \frac{}{\ }\ \frac{}{5}\ \frac{}{1}\). Final answer 21: B2. Ignore inequalities until final answer only |
## Question 5:
### Part (i)
$(^8C_6 \times {}^{12}C_3) + (^8C_7 \times {}^{12}C_2) + (1 \times {}^{12}C_1)$
$[= 28\times220 + 8\times66 + 12 = 6160 + 528 + 12]$
$\div {}^{20}C_9 \quad [= 167960]$
$= \dfrac{335}{8398}$ or $\mathbf{0.039890\ldots}$ | **M1** M1 B1 A1 [4] | M1: Any one correct pair of $^nC_r$ multiplied. M1: Three pairs added. B1: $^{20}C_9$ or 167960 seen anywhere; all correct A1. A1: Exact or awrt 0.0399
### Part (ii)
$2\times(n-1)!$
$\div n!$
$\dfrac{2}{n} < 0.1$
$n_{\min} = 21$ | **M1** M1 M1 A1 [4] | SC: any other method: $\dfrac{2}{n}$ B2. Compare their $\dfrac{2}{n}$ with 0.1, needs division. Not $\geq 21$. SC: $\dfrac{2}{n+1}$, 22: B1+B1. SC: $\dfrac{1}{n}$, 11: B1+B1. T&I: any one of below B2: $\frac{19}{0.10}\ \frac{20}{0.1}\ \frac{21}{0.09}\ \frac{22}{0.09}\ \frac{}{5}\ \frac{}{\ }\ \frac{}{5}\ \frac{}{1}$. Final answer 21: B2. Ignore inequalities until final answer only
---5 (i) A team of 9 is chosen at random from a class consisting of 8 boys and 12 girls.\\
Find the probability that the team contains no more than 3 girls.\\
(ii) A group of $n$ people, including Mr and Mrs Laplace, are arranged at random in a line. The probability that Mr and Mrs Laplace are placed next to each other is less than 0.1 . Find the smallest possible value of $n$.
\hfill \mbox{\textit{OCR Further Statistics AS 2018 Q5 [8]}}