| Exam Board | OCR |
|---|---|
| Module | Further Statistics (Further Statistics) |
| Session | Specimen |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | Conditional probability with geometric |
| Difficulty | Standard +0.8 This is a Further Maths statistics question requiring understanding of geometric distribution, conditional probability, and multi-step reasoning. Part (i) is routine, part (ii) requires using E(X)=1/p, but part (iii) is more challenging as it requires conditional probability calculation with a specific pattern constraint given X=6, which demands careful combinatorial thinking beyond standard textbook exercises. |
| Spec | 5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1)5.02h Geometric: mean 1/p and variance (1-p)/p^2 |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (i) | (a) |
| Answer | Marks |
|---|---|
| 4 | c |
| Answer | Marks |
|---|---|
| [1] | i |
| 2.5 | (cid:11)1(cid:12) |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (i) | (b) |
| Answer | Marks |
|---|---|
| 16384 | e |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 1.1 | (cid:11)3(cid:12)a (cid:16)(cid:11)3(cid:12)b |
| Answer | Marks |
|---|---|
| awrt 0.288 | Or (1(cid:16)q7)(cid:16)(1(cid:16)q3), |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (ii) | E(X)(cid:32)2(cid:159) p(cid:32) 1 |
| Answer | Marks |
|---|---|
| Hence w = 6 | M1 |
| Answer | Marks |
|---|---|
| [2] | 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (iii) | (cid:11)1(cid:12)4 |
| Answer | Marks |
|---|---|
| 2 16 | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.1a |
| 1.1 | (cid:11)3(cid:12)(cid:117)(cid:11)3(cid:12)4 (cid:121)(cid:11)3(cid:12)5 |
| Answer | Marks | Guidance |
|---|---|---|
| 6(ii) | 0 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| 6(iii) | 1 | 0 |
Question 6:
6 | (i) | (a) | (cid:11)1(cid:12)
X ~ Geo
4 | c
B1
[1] | i
2.5 | (cid:11)1(cid:12)
Accept Geo oe
4
6 | (i) | (b) | p
(cid:11)3(cid:12)3 (cid:11)3(cid:12)7
(cid:16)
4 4
S
(cid:32) 4725
16384 | e
M1
A1
[2] | 1.1
1.1 | (cid:11)3(cid:12)a (cid:16)(cid:11)3(cid:12)b
Expression of the form
4 4
with a < b
awrt 0.288 | Or (1(cid:16)q7)(cid:16)(1(cid:16)q3),
p(q3(cid:14)q4 (cid:14)q5(cid:14)q6)
6 | (ii) | E(X)(cid:32)2(cid:159) p(cid:32) 1
2
Hence w = 6 | M1
A1
[2] | 2.2a
2.2a
6 | (iii) | (cid:11)1(cid:12)4
(cid:32) 1
2 16 | M1
A1
[2] | 3.1a
1.1 | (cid:11)3(cid:12)(cid:117)(cid:11)3(cid:12)4 (cid:121)(cid:11)3(cid:12)5
Or, e.g.
4 8 4
--- 6(ii) ---
6(ii) | 0 | 2 | 0 | 02
--- 6(iii) ---
6(iii) | 1 | 0 | 1 | 0 | 26 A bag contains 3 green counters, 3 blue counters and $w$ white counters. Counters are selected at random, one at a time, with replacement, until a white counter is drawn.\\
The total number of counters selected, including the white counter, is denoted by $X$.\\
(i) In the case when $w = 2$,
\begin{enumerate}[label=(\alph*)]
\item write down the distribution of $X$,
\item find $P ( 3 < X \leq 7 )$.\\
(ii) In the case when $\mathrm { E } ( X ) = 2$, determine the value of $w$.\\
(iii) In the case when $w = 2$ and $X = 6$, find the probability that the first five counters drawn alternate in colour.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Statistics Q6 [7]}}