| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Repeated trials until stopping condition |
| Difficulty | Standard +0.3 This is a straightforward tree diagram question with clear stopping conditions. Parts (i)-(ii) involve basic probability calculations with given probabilities (2/3 red, 1/3 blue). Part (iii) requires systematic case analysis but is routine. Part (iv) tests understanding of geometric distribution definition, which is standard S1 knowledge. Slightly above average due to the multi-part nature and part (iii) requiring careful enumeration of cases. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space5.02f Geometric distribution: conditions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Correct structure with no extra branches | B1 | Allow extra branches with correct 0 & 1; ignore probs and R & B |
| Probs and R and B all correct | B1dep | Ignore other probs |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{2}{3} \times \frac{2}{3} \times \frac{2}{3}\) | M1 | or \(\frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \times \frac{1}{3} + \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3}\); NOT \(\frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3}\); ft their tree e.g. "without replacement" gives \(\frac{2}{3} \times \frac{3}{5} \times \frac{2}{4}\ (=\frac{1}{5})\) M1A0 |
| \(= \frac{8}{27}\) or \(0.296\) (3 sf) | A1 | No ft from tree for A1 |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| All six cases seen or implied: \(2\&1\); \(3\&2, 3\&1\); \(4\&3, 4\&2, 4\&1\) or \(2\&1\); \(3\ \&\ (<3)\); \(4\&\ (<4)\) | M1 | |
| \(P(2\&1) = \frac{2}{3} \times \frac{1}{3} \times \frac{1}{3}\) or \(\frac{2}{27}\) | ||
| \(P(3\&2) = (\frac{2}{3})^2 \times \frac{1}{3} \times \frac{2}{3} \times \frac{1}{3}\) or \(\frac{8}{243}\) | ||
| \(P(3\&1) = (\frac{2}{3})^2 \times \frac{1}{3} \times \frac{1}{3}\) or \(\frac{4}{81}\) | ||
| \(P(4\&3) = (\frac{2}{3})^3 \times (\frac{2}{3})^2 \times \frac{1}{3}\) or \((\frac{2}{3})^4 \times (\frac{2}{3})^2 \times \frac{1}{3} + (\frac{2}{3})^3 \times \frac{1}{3} \times (\frac{2}{3})^2 \times \frac{1}{3}\) or \(\frac{32}{729}\) | ||
| \(P(4\&2) = (\frac{2}{3})^3 \times \frac{2}{3} \times \frac{1}{3}\) or \((\frac{2}{3})^4 \times \frac{2}{3} \times \frac{1}{3} + (\frac{2}{3})^3 \times \frac{1}{3} \times \frac{2}{3} \times \frac{1}{3}\) or \(\frac{16}{243}\) | ||
| \(P(4\&1) = (\frac{2}{3})^3 \times \frac{1}{3}\) or \((\frac{2}{3})^4 \times \frac{1}{3} + (\frac{2}{3})^3 \times \frac{1}{3} \times (\frac{1}{3})^2\) or \(\frac{8}{81}\) | ||
| Correct expressions (or results) for 3 of these 6 probs | M1 | |
| Correct expressions (or results) for the other 3 of these 6 probs & no extra cases, and add all 6 cases, i.e. completely correct method | M1 | |
| \(\frac{266}{729}\) or \(0.365\) (3 sf) | A1 | |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| R & B; RR & RB; RRR & RRB or R & B; RR & RB; RRRR & RRB; RRRB & RRB | M1 | i.e. 3 cases: \((\geq 2\ \&\ 1)\), \((\geq 3\ \&\ 2)\), \((4\ \&\ 3)\) |
| \(P(\text{R \& B}) = \frac{2}{3} \times \frac{1}{3}\) or \(\frac{2}{9}\) | M1 | NB Must be clearly part of 3-case method |
| \(P(\text{RR \& RB}) = (\frac{2}{3})^2 \times \frac{2}{3} \times \frac{1}{3}\) or \(\frac{8}{81}\) | ||
| \(P(\text{RRR \& RRB}) = (\frac{2}{3})^3 \times (\frac{2}{3})^2 \times \frac{1}{3}\) or \(\frac{32}{729}\) | ||
| Both these correct expressions or results and add all 3 cases, i.e. completely correct method | M1 | |
| \(\frac{266}{729}\) or \(0.365\) (3 sf) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| B&B; RB & RB; RRB & RRB; RRRX & RRRX i.e. 1&1 or 2&2 or 3&3 or 4&4 | M1 | |
| \((\frac{1}{3})^2 + (\frac{2}{3} \times \frac{1}{3})^2 + (\frac{2}{3} \times \frac{2}{3} \times \frac{1}{3})^2 + (\frac{2}{3} \times \frac{2}{3} \times \frac{2}{3})^2\) or \(\frac{1}{9} + \frac{4}{81} + \frac{16}{729} + \frac{64}{729}\) or \(\frac{197}{729}\) all correct | M1 | |
| \(\frac{1}{2}(1 - \frac{197}{729})\) | M1 | |
| \(\frac{266}{729}\) or \(0.365\) (3 sf) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| All six cases seen or implied: \(\frac{2}{27} + \frac{8}{243} + \frac{4}{81} + \frac{32}{729} + \frac{16}{243} + \frac{8}{243}\) oe | M1M0 | |
| \(= \frac{494}{2187}\) or \(0.226\) | A0 | |
| \(\frac{2}{3} \times \frac{1}{3} + (\frac{2}{3})^2 \times \frac{1}{3} + (\frac{2}{3})^3 \times \frac{1}{3} + (\frac{2}{3})^4\) or \(\frac{2}{3} \times \frac{1}{3} + (\frac{2}{3})^2 \times \frac{1}{3} + (\frac{2}{3})^3 \times \frac{1}{3}\) | M0M0M0M0 | |
| \(P(2\&1) + P(3\&2) + P(4\&3) = \frac{2}{3} \times \frac{1}{3} \times \frac{1}{3} + (\frac{2}{3})^2 \times \frac{2}{3} \times \frac{1}{3} + (\frac{2}{3})^4 \times (\frac{2}{3})^2 \times \frac{1}{3} \times (\frac{2}{3})^3 \times \frac{1}{3} \times (\frac{2}{3})^2 \times \frac{1}{3}\) | M0M1M0A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Unlimited number of throws oe; Not stop at 4 throws oe | B1 | Not fixed number of throws; Turn continues until blue obtained; Allow "Throw die until blue obtained"; NOT "Continue until 1st success"; NOT "Not stop at 4 throws or when blue obtained"; Ignore all else |
| [1] |
# Question 8:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Correct structure with no extra branches | B1 | Allow extra branches with correct 0 & 1; ignore probs and R & B |
| Probs and R and B all correct | B1dep | Ignore other probs |
| **[2]** | | |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{2}{3} \times \frac{2}{3} \times \frac{2}{3}$ | M1 | or $\frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \times \frac{1}{3} + \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3}$; NOT $\frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3}$; ft their tree e.g. "without replacement" gives $\frac{2}{3} \times \frac{3}{5} \times \frac{2}{4}\ (=\frac{1}{5})$ M1A0 |
| $= \frac{8}{27}$ or $0.296$ (3 sf) | A1 | No ft from tree for A1 |
| **[2]** | | |
## Part (iii):
> Note: There are basically 6 cases. Some group into 3 cases (middle column), others use 9 cases grouped into 6. Listing Adnan and Beryl separately gains no marks — must be combined into cases. No ft from tree, but if clearly stating cases may score 1st M1 only.
### 6-case method:
| Answer | Marks | Guidance |
|--------|-------|----------|
| All six cases seen or implied: $2\&1$; $3\&2, 3\&1$; $4\&3, 4\&2, 4\&1$ or $2\&1$; $3\ \&\ (<3)$; $4\&\ (<4)$ | M1 | |
| $P(2\&1) = \frac{2}{3} \times \frac{1}{3} \times \frac{1}{3}$ or $\frac{2}{27}$ | | |
| $P(3\&2) = (\frac{2}{3})^2 \times \frac{1}{3} \times \frac{2}{3} \times \frac{1}{3}$ or $\frac{8}{243}$ | | |
| $P(3\&1) = (\frac{2}{3})^2 \times \frac{1}{3} \times \frac{1}{3}$ or $\frac{4}{81}$ | | |
| $P(4\&3) = (\frac{2}{3})^3 \times (\frac{2}{3})^2 \times \frac{1}{3}$ or $(\frac{2}{3})^4 \times (\frac{2}{3})^2 \times \frac{1}{3} + (\frac{2}{3})^3 \times \frac{1}{3} \times (\frac{2}{3})^2 \times \frac{1}{3}$ or $\frac{32}{729}$ | | |
| $P(4\&2) = (\frac{2}{3})^3 \times \frac{2}{3} \times \frac{1}{3}$ or $(\frac{2}{3})^4 \times \frac{2}{3} \times \frac{1}{3} + (\frac{2}{3})^3 \times \frac{1}{3} \times \frac{2}{3} \times \frac{1}{3}$ or $\frac{16}{243}$ | | |
| $P(4\&1) = (\frac{2}{3})^3 \times \frac{1}{3}$ or $(\frac{2}{3})^4 \times \frac{1}{3} + (\frac{2}{3})^3 \times \frac{1}{3} \times (\frac{1}{3})^2$ or $\frac{8}{81}$ | | |
| Correct expressions (or results) for 3 of these 6 probs | M1 | |
| Correct expressions (or results) for the other 3 of these 6 probs & no extra cases, and add all 6 cases, i.e. completely correct method | M1 | |
| $\frac{266}{729}$ or $0.365$ (3 sf) | A1 | |
| **[4]** | | |
### 3-case method:
| Answer | Marks | Guidance |
|--------|-------|----------|
| R & B; RR & RB; RRR & RRB or R & B; RR & RB; RRRR & RRB; RRRB & RRB | M1 | i.e. 3 cases: $(\geq 2\ \&\ 1)$, $(\geq 3\ \&\ 2)$, $(4\ \&\ 3)$ |
| $P(\text{R \& B}) = \frac{2}{3} \times \frac{1}{3}$ or $\frac{2}{9}$ | M1 | **NB Must be clearly part of 3-case method** |
| $P(\text{RR \& RB}) = (\frac{2}{3})^2 \times \frac{2}{3} \times \frac{1}{3}$ or $\frac{8}{81}$ | | |
| $P(\text{RRR \& RRB}) = (\frac{2}{3})^3 \times (\frac{2}{3})^2 \times \frac{1}{3}$ or $\frac{32}{729}$ | | |
| Both these correct expressions or results and add all 3 cases, i.e. completely correct method | M1 | |
| $\frac{266}{729}$ or $0.365$ (3 sf) | A1 | |
### 4-case method:
| Answer | Marks | Guidance |
|--------|-------|----------|
| B&B; RB & RB; RRB & RRB; RRRX & RRRX i.e. 1&1 or 2&2 or 3&3 or 4&4 | M1 | |
| $(\frac{1}{3})^2 + (\frac{2}{3} \times \frac{1}{3})^2 + (\frac{2}{3} \times \frac{2}{3} \times \frac{1}{3})^2 + (\frac{2}{3} \times \frac{2}{3} \times \frac{2}{3})^2$ or $\frac{1}{9} + \frac{4}{81} + \frac{16}{729} + \frac{64}{729}$ or $\frac{197}{729}$ all correct | M1 | |
| $\frac{1}{2}(1 - \frac{197}{729})$ | M1 | |
| $\frac{266}{729}$ or $0.365$ (3 sf) | A1 | |
### Common incorrect methods:
| Answer | Marks | Guidance |
|--------|-------|----------|
| All six cases seen or implied: $\frac{2}{27} + \frac{8}{243} + \frac{4}{81} + \frac{32}{729} + \frac{16}{243} + \frac{8}{243}$ oe | M1M0 | |
| $= \frac{494}{2187}$ or $0.226$ | A0 | |
| $\frac{2}{3} \times \frac{1}{3} + (\frac{2}{3})^2 \times \frac{1}{3} + (\frac{2}{3})^3 \times \frac{1}{3} + (\frac{2}{3})^4$ or $\frac{2}{3} \times \frac{1}{3} + (\frac{2}{3})^2 \times \frac{1}{3} + (\frac{2}{3})^3 \times \frac{1}{3}$ | M0M0M0M0 | |
| $P(2\&1) + P(3\&2) + P(4\&3) = \frac{2}{3} \times \frac{1}{3} \times \frac{1}{3} + (\frac{2}{3})^2 \times \frac{2}{3} \times \frac{1}{3} + (\frac{2}{3})^4 \times (\frac{2}{3})^2 \times \frac{1}{3} \times (\frac{2}{3})^3 \times \frac{1}{3} \times (\frac{2}{3})^2 \times \frac{1}{3}$ | M0M1M0A0 | |
## Part (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Unlimited number of throws oe; Not stop at 4 throws oe | B1 | Not fixed number of throws; Turn continues until blue obtained; Allow "Throw die until blue obtained"; NOT "Continue until 1st success"; NOT "Not stop at 4 throws or when blue obtained"; Ignore all else |
| **[1]** | | |
---8 A game is played with a fair, six-sided die which has 4 red faces and 2 blue faces. One turn consists of throwing the die repeatedly until a blue face is on top or until the die has been thrown 4 times.\\
(i) In the answer book, complete the probability tree diagram for one turn.\\
\includegraphics[max width=\textwidth, alt={}, center]{e5957185-5fe3-45d9-9ab3-c2aab9cbd8dd-5_314_302_1000_884}\\
(ii) Find the probability that in one particular turn the die is thrown 4 times.\\
(iii) Adnan and Beryl each have one turn. Find the probability that Adnan throws the die more times than Beryl.\\
(iv) State one change that needs to be made to the rules so that the number of throws in one turn will have a geometric distribution.
\hfill \mbox{\textit{OCR S1 2015 Q8 [9]}}