| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | First success before/after trial n |
| Difficulty | Standard +0.3 This is a straightforward application of geometric distribution with clearly defined success probabilities. Part (i) requires basic geometric probability calculations P(X=6) and P(X<6). Part (ii) involves recognizing a pattern and applying the geometric series formula, but the setup is given and the algebra is routine. All steps follow standard S1 procedures with no novel insight required. |
| Spec | 4.06b Method of differences: telescoping series5.02f 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 |
|---|---|---|
| Geo stated or implied; \(0.9^5 \times 0.1\) alone \(= 0.059\ldots\) (2 sfs) | M1, M1, A1 | eg by \(0.9^n \times 0.1\) or \(0.1^n \times 0.9\) alone, \(p>1\); all correct |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.9^5\) or \(0.59\ldots\) (NB cf ans to (i)(a)!!); \(1 - 0.9^5\) | M1, M1 | \(0.1 + 0.9 \times 0.1 + \ldots 0.9^4 \times 0.1\) : M2; 1 term wrong or omit or extra; or \(1 -\) (all terms correct): M1; or \(1 - 0.9^6\): M1; M0M0A0 for \(0.9^n \times 0.1\) |
| \(= 0.4095\) or \(0.410\) (3 sfs) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.05 + 0.95^2 \times 0.05 = \frac{761}{8000}\) or \(0.0951\) (3 sfs) | M1, A1 | All correct; NB!! \(2 \times 0.95 \times 0.05 = 0.095\) M0A0 |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.05, 0.95^2 \times 0.05, \ldots\) or \(\frac{1}{20}, \frac{361}{8000}, \ldots\) oe | M1 | \(\geq 2\) terms. Not nec'y added; May be implied by next line; or \(r = 0.95^2\) stated or implied |
| \(\dfrac{0.05}{1-0.95^2}\) or \(\dfrac{0.05}{1-0.9025}\) oe \(= \dfrac{20}{39}\) or \(0.513\) (3 sfs) | M1, A1 | or \(\frac{0.05}{(1-(1-0.5))^2}\) or \(\frac{0.05}{2 \times 0.05 - 0.05^2}\) or \(\frac{1}{1.95}\) oe; NB \(\frac{0.05}{1-0.5 \times 0.05} = 0.0513\) M0A0 |
# Question 9(i)(a):
Geo stated or implied; $0.9^5 \times 0.1$ alone $= 0.059\ldots$ (2 sfs) | M1, M1, A1 | eg by $0.9^n \times 0.1$ or $0.1^n \times 0.9$ alone, $p>1$; all correct
**Total: [3]**
---
# Question 9(i)(b):
$0.9^5$ or $0.59\ldots$ (NB cf ans to (i)(a)!!); $1 - 0.9^5$ | M1, M1 | $0.1 + 0.9 \times 0.1 + \ldots 0.9^4 \times 0.1$ : M2; 1 term wrong or omit or extra; or $1 -$ (all terms correct): M1; or $1 - 0.9^6$: M1; M0M0A0 for $0.9^n \times 0.1$
$= 0.4095$ or $0.410$ (3 sfs) | A1 |
**Total: [3]**
---
# Question 9(ii)(a):
$0.05 + 0.95^2 \times 0.05 = \frac{761}{8000}$ or $0.0951$ (3 sfs) | M1, A1 | All correct; NB!! $2 \times 0.95 \times 0.05 = 0.095$ M0A0
**Total: [2]**
---
# Question 9(ii)(b):
$0.05, 0.95^2 \times 0.05, \ldots$ or $\frac{1}{20}, \frac{361}{8000}, \ldots$ oe | M1 | $\geq 2$ terms. Not nec'y added; May be implied by next line; or $r = 0.95^2$ stated or implied
$\dfrac{0.05}{1-0.95^2}$ or $\dfrac{0.05}{1-0.9025}$ oe $= \dfrac{20}{39}$ or $0.513$ (3 sfs) | M1, A1 | or $\frac{0.05}{(1-(1-0.5))^2}$ or $\frac{0.05}{2 \times 0.05 - 0.05^2}$ or $\frac{1}{1.95}$ oe; NB $\frac{0.05}{1-0.5 \times 0.05} = 0.0513$ M0A0
**Total: [3]**9 (i) A clock is designed to chime once each hour, on the hour. The clock has a fault so that each time it is supposed to chime there is a constant probability of $\frac { 1 } { 10 }$ that it will not chime. It may be assumed that the clock never stops and that faults occur independently. The clock is started at 5 minutes past midnight on a certain day. Find the probability that the first time it does not chime is
\begin{enumerate}[label=(\alph*)]
\item at 0600 on that day,
\item before 0600 on that day.\\
(ii) Another clock is designed to chime twice each hour: on the hour and at 30 minutes past the hour. This clock has a fault so that each time it is supposed to chime there is a constant probability of $\frac { 1 } { 20 }$ that it will not chime. It may be assumed that the clock never stops and that faults occur independently. The clock is started at 5 minutes past midnight on a certain day.\\
(a) Find the probability that the first time it does not chime is at either 0030 or 0130 on that day.\\
(b) Use the formula for the sum to infinity of a geometric progression to find the probability that the first time it does not chime is at 30 minutes past some hour.
\end{enumerate}
\hfill \mbox{\textit{OCR S1 2012 Q9 [11]}}