| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2009 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Reverse conditional probability |
| Difficulty | Moderate -0.3 This is a straightforward tree diagram problem with clear branches and standard conditional probability calculations. Part (i) requires adding probabilities from two paths, while parts (ii) and (iii) are direct applications of conditional probability using Bayes' theorem structure. The arithmetic is simple (halves and sixths) and the problem-solving is routine for S1 level, making it slightly easier than average. |
| Spec | 2.03a Mutually exclusive and independent events2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{1}{6} + 3 \times (\frac{1}{6})^2 = \frac{1}{4}\) | M2, A1 3 | or \(3 \times (\frac{1}{6})^2\) or \(\frac{1}{6} + (\frac{1}{6})^2\) or \(\frac{1}{6} + 2(\frac{1}{6})^2\) or \(\frac{1}{6} + 4(\frac{1}{6})^2\): M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{1}{3}\) | B1 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| 3 routes clearly implied | M1 | |
| out of 18 possible (equiprobable) routes | M1 | or \(\frac{1}{3} \times \frac{1}{6} \times 3\): M2; or \(\frac{1}{3}x\frac{1}{6}\) or \(\frac{1}{6}x\frac{1}{6}x3\) or \(\frac{1}{3}x\frac{1}{3}x3\) or \(\frac{1}{4} - \frac{1}{6}\): M1; but \(\frac{1}{6} \times \frac{1}{6} \times 2\): M0 |
| \(\frac{(\frac{1}{6})^2 \times 3}{\frac{1}{2}}\) or \(\frac{\frac{1-1}{4-6}}{\frac{1}{2}}\) or \(\frac{\frac{1\times1}{2\cdot6}}{\frac{1}{2}}\) oe | M2 | or \(\frac{P(4\&\text{twice})}{P(\text{twice})}\) stated or \(\frac{\text{prob}}{\frac{1}{2}}\): M1; Whatever \(1^{st}\), only one possibility on \(2^{nd}\): M2 |
| \(\frac{1}{6}\) | A1 3 | \(\frac{1}{6}\), no working: M1M1A1; \(\frac{1}{12}\), no working: M0 |
## Question 8:
**Part (i):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{6} + 3 \times (\frac{1}{6})^2 = \frac{1}{4}$ | M2, A1 3 | or $3 \times (\frac{1}{6})^2$ or $\frac{1}{6} + (\frac{1}{6})^2$ or $\frac{1}{6} + 2(\frac{1}{6})^2$ or $\frac{1}{6} + 4(\frac{1}{6})^2$: M1 |
**Part (ii):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{3}$ | B1 1 | |
**Part (iii):**
| Answer | Mark | Guidance |
|--------|------|----------|
| 3 routes clearly implied | M1 | |
| out of 18 possible (equiprobable) routes | M1 | or $\frac{1}{3} \times \frac{1}{6} \times 3$: M2; or $\frac{1}{3}x\frac{1}{6}$ or $\frac{1}{6}x\frac{1}{6}x3$ or $\frac{1}{3}x\frac{1}{3}x3$ or $\frac{1}{4} - \frac{1}{6}$: M1; but $\frac{1}{6} \times \frac{1}{6} \times 2$: M0 |
| $\frac{(\frac{1}{6})^2 \times 3}{\frac{1}{2}}$ or $\frac{\frac{1-1}{4-6}}{\frac{1}{2}}$ or $\frac{\frac{1\times1}{2\cdot6}}{\frac{1}{2}}$ oe | M2 | or $\frac{P(4\&\text{twice})}{P(\text{twice})}$ stated or $\frac{\text{prob}}{\frac{1}{2}}$: M1; Whatever $1^{st}$, only one possibility on $2^{nd}$: M2 |
| $\frac{1}{6}$ | A1 3 | $\frac{1}{6}$, no working: M1M1A1; $\frac{1}{12}$, no working: M0 |8 A game uses an unbiased die with faces numbered 1 to 6 . The die is thrown once. If it shows 4 or 5 or 6 then this number is the final score. If it shows 1 or 2 or 3 then the die is thrown again and the final score is the sum of the numbers shown on the two throws.\\
(i) Find the probability that the final score is 4 .\\
(ii) Given that the die is thrown only once, find the probability that the final score is 4 .\\
(iii) Given that the die is thrown twice, find the probability that the final score is 4 .
\hfill \mbox{\textit{OCR S1 2009 Q8 [7]}}