| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2006 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | Name geometric distribution and parameter |
| Difficulty | Moderate -0.8 This is a straightforward question testing basic recognition and application of the geometric distribution. Part (i) requires naming the distribution (geometric with p=1/3) and stating independence as an assumption. Parts (ii)-(iii) involve direct formula application for P(X=4), P(X<4), and E(X)=3. Part (iv) extends slightly by requiring two failures followed by a success, but still uses standard geometric probability calculations. No novel problem-solving or complex multi-step reasoning required—purely routine application of a standard S1 topic. |
| Spec | 5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1) |
| Answer | Marks | Guidance |
|---|---|---|
| 8(i) | Geometric. Each attempt (or result or try) indep | B1: —; B1: 2 |
| 8(ii)(a) | \((^2/_{3})^x \times ^1/_{3}\) | M2: — |
| \(= ^9/_{81}\) or \(0.0988\) (3 sfs) | A1: 3 | |
| 8(ii)(b) | \(\frac{(^2/_{3})^3}{1 - (^2/_{3})^3}\) | M1: —; M1: — |
| \(= ^{19}_{27}\) or \(0.704\) (3sfs) | A1: 3 | |
| 8(iii) | \(3\) | B1f: 1 |
| 8(iv) | \(1 - \frac{^{19}_{27}}{(^1/_{27})^2 \times ^{19}_{27}}\) \((1 - 0.7037)\) or \(0.2963\) \(0.2963^2 \times 0.7037\) | M1: —; M1: — |
| \(= ^{1216}_{10683}\) \(= 0.0618\) (3 sfs) | A1: 3 | cao, allow art 0.0618 or 0.0617 |
8(i) | Geometric. Each attempt (or result or try) indep | B1: —; B1: 2 | In context. Not "events, trials, outcomes". Ignore extra |
8(ii)(a) | $(^2/_{3})^x \times ^1/_{3}$ | M2: — | $(^2/_{3})^x 1/_{3}$ or $(^2/_{3})^x 1/_{3}$; allow other numerical "p" $(0<p<1)$:M1 |
| $= ^9/_{81}$ or $0.0988$ (3 sfs) | A1: 3 | |
8(ii)(b) | $\frac{(^2/_{3})^3}{1 - (^2/_{3})^3}$ | M1: —; M1: — | not $(^2/_{3})^3 × \ldots$ or $^1/_{3} + ^2/_{3}x1/_{3} + (^2/_{3})^2 x 1/_{3}$ M2 $1 - (^2/_{3})^3$ or $1 - (^"q"^)^3$ M1 or 3 terms, with 2 correct M1 or 3 correct terms + 1 extra M1 or "p" + "qp" + "q²p" M1 or $1 - \sum$ of 3 correct terms M1 "p" means num value, not $^1/_{3}$ |
| $= ^{19}_{27}$ or $0.704$ (3sfs) | A1: 3 | |
8(iii) | $3$ | B1f: 1 | or $l'_rp''$ |
8(iv) | $1 - \frac{^{19}_{27}}{(^1/_{27})^2 \times ^{19}_{27}}$ $(1 - 0.7037)$ or $0.2963$ $0.2963^2 \times 0.7037$ | M1: —; M1: — | ft (b) for M1M1 must see method if ft Allow figs rounded to 2 sfs for M1M1 |
| $= ^{1216}_{10683}$ $= 0.0618$ (3 sfs) | A1: 3 | cao, allow art 0.0618 or 0.0617 |
**Total: 12**
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# **Total 72 marks**8 Henry makes repeated attempts to light his gas fire. He makes the modelling assumption that the probability that the fire will light on any attempt is $\frac { 1 } { 3 }$.
Let $X$ be the number of attempts at lighting the fire, up to and including the successful attempt.\\
(i) Name the distribution of $X$, stating a further modelling assumption needed.
In the rest of this question, you should use the distribution named in part (i).\\
(ii) Calculate
\begin{enumerate}[label=(\alph*)]
\item $\mathrm { P } ( X = 4 )$,
\item $\mathrm { P } ( X < 4 )$.\\
(iii) State the value of $\mathrm { E } ( X )$.\\
(iv) Henry has to light the fire once a day, starting on March 1st. Calculate the probability that the first day on which fewer than 4 attempts are needed to light the fire is March 3rd.
\end{enumerate}
\hfill \mbox{\textit{OCR S1 2006 Q8 [12]}}