| Exam Board | OCR |
|---|---|
| Module | Further Statistics (Further Statistics) |
| Year | 2018 |
| Session | September |
| Marks | 10 |
| Topic | Combinations & Selection |
| Type | Repeated trials with selection |
| Difficulty | Standard +0.8 This is a multi-part Further Maths statistics question requiring hypergeometric probability (part i), binomial distribution parameters (part ii), and sophisticated counting with inclusion-exclusion for arrangements with restrictions (part iii). Part (iii) particularly requires careful combinatorial reasoning about mutually exclusive events and is non-routine for A-level, placing this solidly above average difficulty. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems5.02d Binomial: mean np and variance np(1-p) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{^7C_x \times ^7C_2}{^{12}C_4}\) | M1 | 1.1a |
| M1 | 1.1a | Divide by \(^{12}C_4\) |
| \(= \frac{21 \times 10}{495} = \frac{14}{33}\) AG | A1 [3] | 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(np = 99 \times \frac{14}{33} = 42\) | B1 [1] | 1.1 |
| (b) \(np(1-p) = 99 \times \frac{14}{33} \times \frac{19}{33} = \frac{266}{11}\) or 24.18... | B1 [1] | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(x\) R \(x\) R \(x\) R \(x\) R \(x\) R \(x\) R \(x\) with Bs in 5 x's | M1 | 3.1a |
| \(^5C_5\) \([= 56]\) | M1 | 1.1a |
| \(+8\) | M1 | 1.1a |
| \((56 + 8) \times ^{12}C_5\) | M1 | 3.3 |
| \(= \frac{8}{99}\) or 0.080808... | A1 [5] | 3.4 |
## (i)
$\frac{^7C_x \times ^7C_2}{^{12}C_4}$ | M1 | 1.1a | Two "$^7C$" multiplied in numerator
| M1 | 1.1a | Divide by $^{12}C_4$
$= \frac{21 \times 10}{495} = \frac{14}{33}$ AG | A1 [3] | 2.2a | Convincingly obtain AG | E.g. 21, 10 and 495 seen
## (ii)
(a) $np = 99 \times \frac{14}{33} = 42$ | B1 [1] | 1.1 | 42 or 42.0 only
(b) $np(1-p) = 99 \times \frac{14}{33} \times \frac{19}{33} = \frac{266}{11}$ or 24.18... | B1 [1] | 1.1 | Exact or awrt 24.2
## (iii)
$x$ R $x$ R $x$ R $x$ R $x$ R $x$ R $x$ with Bs in 5 x's | M1 | 3.1a | Method for Bs separate | Or $(^7C_5 + 8) \times ^{12}C_5$
$^5C_5$ $[= 56]$ | M1 | 1.1a | Allow $7!x^5P_s$ or $7!x^5P_s$, $n \in \{6, 7, 8\}$ | Or; $7! \times P_s [= 5040 \times 6720]$
$+8$ | M1 | 1.1a | Allow 7! for 8! or 3! for 5! | $8! \times 5! [= 40320 \times 120]$
$(56 + 8) \times ^{12}C_5$ | M1 | 3.3 | Add two terms, divide by $^{12}C_5$ | $(3386800 + 4838400) \div 12!$
$= \frac{8}{99}$ or 0.080808... | A1 [5] | 3.4 | Any equivalent exact single fraction, or awrt 0.0808
---6 A bag contains 7 red counters and 5 blue counters.\\
(i) Fred chooses 4 counters at random, without replacement. Show that the probability that Fred chooses exactly 2 red counters is $\frac { 14 } { 33 }$.\\
(ii) Lina chooses 4 counters at random from the bag, records whether or not exactly 2 red counters are chosen, and returns the counters to the bag. She carries out this experiment 99 times.
\begin{enumerate}[label=(\alph*)]
\item Find the mean of the number of experiments that result in choosing exactly 2 red counters.
\item Find the variance of the number of experiments that result in choosing exactly 2 red counters.\\
(iii) Alex arranges all 12 counters in a random order in a straight line.
A is the event: no two blue counters are next to one another.
B is the event: all the blue counters are next to one another.
Find $\mathrm { P } ( A \cup B )$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Statistics 2018 Q6 [10]}}