| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2011 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Definitions |
| Type | Sequential events and tree diagrams |
| Difficulty | Standard +0.3 This is a standard geometric probability question using tree diagrams with repeated trials. While it requires careful tracking of multiple players and understanding of conditional probability, the calculations are straightforward applications of multiplying probabilities (5/6)^n × (1/6). Part (iii) requires slight insight about complementary events, but overall this is a routine S1 question testing basic sequential probability concepts without requiring novel problem-solving approaches. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space |
| Answer | Marks |
|---|---|
| Answer: Incorrect p (eg "cubical die means 18 sides hence \(p = \frac{1}{18}\)"): can gain all B & M marks. | — |
| Answer | Marks |
|---|---|
| Answer: \(\frac{28}{216}\) oe or \(0.116\) (3 sfs) | B1, 1, M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: \((\frac{1}{6})^3 \times \frac{1}{6}\) alone | M2 | M1 for \((\frac{1}{6})^8 \times \frac{1}{6}\) alone; \(= 0.0465\) (3 sfs) or \(\frac{78125}{1679616}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: \((\frac{1}{6})^6\) oe alone | M1 | \(1 - P(X \leq 8)\), with exactly 8 correct terms; \(= 0.233\) (3 sfs) or \(\frac{390625}{1679616}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: NB If more than 5 products are added (eg \(P(1 \leq X \leq 12)\)): no marks | — | \((\frac{1}{6})^3(\frac{5}{6})^7 + (\frac{1}{6})^{10}(\frac{5}{6})^2\) or \((1 - (\frac{1}{6})^3[1 - (\frac{5}{6})^9])\) or \(\pm[(\frac{1}{6})^9 - (\frac{1}{6})^{13}]\) or \(\pm[1-(\frac{1}{6})^{13} - [1 - (\frac{1}{6})^9]]\) |
**Answer:** Incorrect p (eg "cubical die means 18 sides hence $p = \frac{1}{18}$"): can gain all B & M marks. | — |
## Question 8 (i)
**Answer:** $\frac{28}{216}$ oe or $0.116$ (3 sfs) | **B1, 1, M1** |
## Question 8ii
**Answer:** $(\frac{1}{6})^3 \times \frac{1}{6}$ alone | **M2** | M1 for $(\frac{1}{6})^8 \times \frac{1}{6}$ alone; $= 0.0465$ (3 sfs) or $\frac{78125}{1679616}$ | **A1, 3** |
## Question 8iii
**Answer:** $(\frac{1}{6})^6$ oe alone | **M1** | $1 - P(X \leq 8)$, with exactly 8 correct terms; $= 0.233$ (3 sfs) or $\frac{390625}{1679616}$ | **A1, 2** | NOT $1 - (\frac{5}{6})^8$, NOT $(\frac{5}{6})^8 \times \ldots$
## Question 8iv
**Answer:** NB If more than 5 products are added (eg $P(1 \leq X \leq 12)$): no marks | — | $(\frac{1}{6})^3(\frac{5}{6})^7 + (\frac{1}{6})^{10}(\frac{5}{6})^2$ or $(1 - (\frac{1}{6})^3[1 - (\frac{5}{6})^9])$ or $\pm[(\frac{1}{6})^9 - (\frac{1}{6})^{13}]$ or $\pm[1-(\frac{1}{6})^{13} - [1 - (\frac{1}{6})^9]]$ | **M3, M2, M1** | $(\frac{1}{6})^9 - (\frac{1}{6})^{13}$ or $1 - (\frac{1}{6})^{13} - [1 - (\frac{1}{6})^9]$ | M2; $\pm[(\frac{1}{6})^9 - (\frac{1}{6})^{13}]$ or $\pm[1-(\frac{1}{6})^{13} - [1 - (\frac{1}{6})^9]]$ | M1; Allow 0.1 with wking | **A1, 4** |
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**Total: 72 marks**8 Ann, Bill, Chris and Dipak play a game with a fair cubical die. Starting with Ann they take turns, in alphabetical order, to throw the die. This process is repeated as many times as necessary until a player throws a 6 . When this happens, the game stops and this player is the winner.
Find the probability that\\
(i) Chris wins on his first throw,\\
(ii) Dipak wins on his second throw,\\
(iii) Ann gets a third throw,\\
(iv) Bill throws the die exactly three times.
\hfill \mbox{\textit{OCR S1 2011 Q8 [10]}}