| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | Game theory with alternating players |
| Difficulty | Standard +0.3 This is a standard S1 geometric distribution question with straightforward applications of probability rules. Parts (i)(a-b) are routine geometric probability calculations, (i)(c) requires combining geometric scenarios, and part (ii) involves independent events but follows predictable patterns. All techniques are standard textbook exercises requiring careful calculation rather than novel insight. |
| 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 |
|---|---|
| \(0.9^4 × 0.1\) = \(\frac{6561}{100000}\) or 0.0656 (3sf) | M1, A1, [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.9^3\); = \(\frac{59049}{100000}\) or 0.59 (2 sf) | M1, A1, [2] | Allow 0.9³ or 1−0.9³ :M1; Allow without "1 −" OR omit last term; NB 0.9³×0.1 = 0.0590 M0A0 |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.1×0.1 + [0.1×0.1×0.9 +0.1×0.9×0.1]\) oe; or \(0.1×0.1 + [0.1×0.1×0.9 +0.1×0.9×0.1]\) oe; or \(+ 0.9×0.1×0.1\) oe | M1, M1, M1, A1, [4] | M1M1 two correct terms, no incorrect multiples; M1 all correct; Ans 0.027 probably M0M1M1A0 but check working; SC if no M-mks scored: SSF, SSS, FSS, SFS or SS, FSS, SFS seen or implied: B1; 1 − one or both terms with no further wking: M1(dep M1); eg 1 − 0.9³ alone M1M0M1; This method only scores using "1 − "; 0.9³; 3×0.1²×0.1 no incorrect multiples; M1; M1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.9× 0.8 × 0.1\) = \(\frac{9}{125}\) or 0.072 | M1, A1, [2] | NOT 0.9×0.8×0.1×0.2 = 0.0144: M0A0; NOT 0.9×0.8×0.2 = 0.144: M0A0 |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.9^{gr10} × 0.8^{gr10}×0.1\) (or ×0.2, not ×0.1×0.2) = 5.2×10⁻³ or 0.0052 (2 sf) | M1, M1, A1, [3] | allow \(0.9^{gr10} × 0.8^{gr10}×0.1 ±^{18,19,20}C_1\); fully correct |
## (i)(a)
$0.9^4 × 0.1$ = $\frac{6561}{100000}$ or 0.0656 (3sf) | M1, A1, [2]
## (i)(b)
$0.9^3$; = $\frac{59049}{100000}$ or 0.59 (2 sf) | M1, A1, [2] | Allow 0.9³ or 1−0.9³ :M1; Allow without "1 −" OR omit last term; NB 0.9³×0.1 = 0.0590 M0A0
## (i)(c)
$0.1×0.1 + [0.1×0.1×0.9 +0.1×0.9×0.1]$ oe; or $0.1×0.1 + [0.1×0.1×0.9 +0.1×0.9×0.1]$ oe; or $+ 0.9×0.1×0.1$ oe | M1, M1, M1, A1, [4] | M1M1 two correct terms, no incorrect multiples; M1 all correct; Ans 0.027 probably M0M1M1A0 but check working; SC if no M-mks scored: SSF, SSS, FSS, SFS or SS, FSS, SFS seen or implied: B1; 1 − one or both terms with no further wking: M1(dep M1); eg 1 − 0.9³ alone M1M0M1; This method only scores using "1 − "; 0.9³; 3×0.1²×0.1 no incorrect multiples; M1; M1
---
## (ii)(a)
$0.9× 0.8 × 0.1$ = $\frac{9}{125}$ or 0.072 | M1, A1, [2] | NOT 0.9×0.8×0.1×0.2 = 0.0144: M0A0; NOT 0.9×0.8×0.2 = 0.144: M0A0
## (ii)(b)
$0.9^{gr10} × 0.8^{gr10}×0.1$ (or ×0.2, not ×0.1×0.2) = 5.2×10⁻³ or 0.0052 (2 sf) | M1, M1, A1, [3] | allow $0.9^{gr10} × 0.8^{gr10}×0.1 ±^{18,19,20}C_1$; fully correct | SC Consistent use of 0.8 for both girls: (ii)(a) 0.128 (ii)(b) 0.00360 or 0.9 for both girls: (ii)(a) 0.081 (ii)(b) 0.0150 If both these ans seen, allow (a) 0 (b) B1 | If ans = 0.00360 or 0.0150 see SC below
---Sandra makes repeated, independent attempts to hit a target. On each attempt, the probability that she succeeds is 0.1.
\begin{enumerate}[label=(\roman*)]
\item Find the probability that
\begin{enumerate}[label=(\alph*)]
\item the first time she succeeds is on her 5th attempt, [2]
\item the first time she succeeds is after her 5th attempt, [2]
\item the second time she succeeds is before her 4th attempt. [4]
\end{enumerate}
Jill also makes repeated attempts to hit the target. Each attempt of either Jill or Sandra is independent. Each time that Jill attempts to hit the target, the probability that she succeeds is 0.2. Sandra and Jill take turns attempting to hit the target, with Sandra going first.
\item Find the probability that the first person to hit the target is Sandra, on her
\begin{enumerate}[label=(\alph*)]
\item 2nd attempt, [2]
\item 10th attempt. [3]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{OCR S1 2013 Q8 [13]}}