| Exam Board | OCR MEI |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2021 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | State assumptions for geometric model |
| Difficulty | Moderate -0.8 This question involves straightforward geometric series summation (part a), basic probability calculations with the geometric distribution (parts b, e), and conceptual explanations about model assumptions (parts c, d). All parts require only standard A-level techniques with no novel problem-solving or proof beyond showing a simple geometric series sums to 1. The modeling discussion is accessible and the calculations are routine. |
| Spec | 1.04j Sum to infinity: convergent geometric series |r|<12.04a Discrete probability distributions5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1) |
| Answer | Marks | Guidance |
|---|---|---|
| \(a = 0.01\) and \(r = 0.99\) | B1 | AO2.1 |
| \(a\) and \(r\) substituted in \(\frac{a}{1-r}\) seen | M1 | AO1.1 |
| \(\frac{0.01}{1-0.99}\) oe \(= 1\) | A1 | AO2.4 |
| Answer | Marks | Guidance |
|---|---|---|
| \((n =)\ 312\) | M1 | AO3.1a |
| \(\frac{0.01(1-0.99^n)}{1-0.99}\) or \(\sum_1^n 0.99^{n-1} \times 0.01\) evaluated | M1 | AO1.1 |
| awrt \(0.9565\) or \(0.9570 > 0.95\) so model predicts Layla will beat him within 6 years | A1 | AO3.4 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{0.01(1-0.99^n)}{1-0.99} = 0.95\) | M1 | |
| \(0.99^n = 0.05\) | M1 | |
| \(n = 298.1\) weeks \(< 312\) (or 313) so model predicts Layla will beat him within 6 years | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Layla thinks she will improve (by practising), so she should become increasingly likely to beat Kofi, or Layla thinks she is more likely to beat Kofi because he doesn't practise (but Layla does) | B1 | AO3.5a |
| Answer | Marks | Guidance |
|---|---|---|
| Probability of Layla winning increases as \(r\) increases | B1 | AO2.4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(k(1+4+9+16+25+36+49+64)=1\) | M1 | 3.3 |
| \(k=\frac{1}{204}\ (=0.00490...)\) | A1 | 1.1 |
| \(P(X \leq 6) = \frac{91}{204}\ (0.44607...) < 0.95\) so Layla's statement not consistent with her model oe | A1 | 3.5a |
# Question 15:
## Part (a):
$a = 0.01$ and $r = 0.99$ | **B1** | AO2.1 |
$a$ and $r$ substituted in $\frac{a}{1-r}$ seen | **M1** | AO1.1 | $a$ or $r$ must be correct
$\frac{0.01}{1-0.99}$ oe $= 1$ | **A1** | AO2.4 |
**[3 marks]**
## Part (b):
$(n =)\ 312$ | **M1** | AO3.1a | Allow 313
$\frac{0.01(1-0.99^n)}{1-0.99}$ or $\sum_1^n 0.99^{n-1} \times 0.01$ evaluated | **M1** | AO1.1 | $n = 312$ or $313$; condone $n=6$
awrt $0.9565$ or $0.9570 > 0.95$ so model predicts Layla will beat him within 6 years | **A1** | AO3.4 |
**[3 marks]**
*Alternative:*
$\frac{0.01(1-0.99^n)}{1-0.99} = 0.95$ | **M1** | | Allow $>$ or $\geq$ instead of $=$
$0.99^n = 0.05$ | **M1** | | Attempt to simplify as far as "$0.99^n =$"
$n = 298.1$ weeks $< 312$ (or 313) so model predicts Layla will beat him within 6 years | **A1** |
**[3 marks]**
## Part (c):
Layla thinks she will improve (by practising), so she should become increasingly likely to beat Kofi, or Layla thinks she is more likely to beat Kofi because he doesn't practise (but Layla does) | **B1** | AO3.5a |
**[1 mark]**
## Part (d):
Probability of Layla winning increases as $r$ increases | **B1** | AO2.4 | **B0** for e.g. probability of Layla winning increases exponentially
**[1 mark]**
## Question 15(e):
| Answer | Mark | Guidance |
|--------|------|----------|
| $k(1+4+9+16+25+36+49+64)=1$ | M1 | 3.3 |
| $k=\frac{1}{204}\ (=0.00490...)$ | A1 | 1.1 |
| $P(X \leq 6) = \frac{91}{204}\ (0.44607...) < 0.95$ so Layla's statement not consistent with her model **oe** | A1 | 3.5a |
**Total: [3]**
---15
\begin{enumerate}[label=(\alph*)]
\item Show that $\sum _ { r = 1 } ^ { \infty } 0.99 ^ { r - 1 } \times 0.01 = 1$.
Kofi is a very good table tennis player. Layla is determined to beat him.\\
Every week they play one match of table tennis against each other. They will stop playing when Layla wins the match for the first time.\\
$X$ is the discrete random variable "the number of matches they play in total".
Kofi models the situation using the probability function\\
$\mathrm { P } ( \mathrm { X } = \mathrm { r } ) = 0.99 ^ { \mathrm { r } - 1 } \times 0.01 \quad r = 1,2,3,4 , \ldots$
Kofi states that he is $95 \%$ certain that Layla will not beat him within 6 years.
\item Determine whether Kofi's statement is consistent with his model.
In between matches, Layla practises, but Kofi does not.
\item Explain why Layla might disagree with Kofi's model.
Layla models the situation using the probability function\\
$\mathrm { P } ( \mathrm { X } = \mathrm { r } ) = \mathrm { kr } ^ { 2 } \quad r = 1,2,3,4,5,6,7,8$.
\item Explain how Layla's model takes into account the fact that she practises between matches, but Kofi's does not.
Layla states that she is $95 \%$ certain that she will beat Kofi within the first 6 matches.
\item Determine whether Layla's statement is consistent with her model.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 2 2021 Q15 [11]}}