| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Sequential selection without replacement |
| Difficulty | Easy -1.2 This is a straightforward application of tree diagrams and conditional probability with small numbers. Part (i) requires completing a partially-drawn tree diagram with simple fractions (5/6, 1/6, 4/5, 1/5). Part (ii) involves multiplying along two branches and adding - a standard textbook exercise. Part (iii) is slightly more involved as it requires conditional probability over three draws, but the calculation is still mechanical with no conceptual difficulty. The small numbers (5 red, 1 black) make arithmetic trivial throughout. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Top: 2 branches \(\frac{4}{5}\), \(\frac{1}{5}\) & R, B shown | B1 | consistent |
| Bottom: \(1^{\text{st}}\) branch: prob \(= 1\) or \(\frac{5}{5}\), & R shown | B1 | allow eg \(\frac{4}{4}\). Any missing label(s) on first three branches, subtr B1 once |
| no \(2^{\text{nd}}\) branch OR branch with prob \(= 0\) or \(\frac{0}{5}\) | ignore any \(3^{\text{rd}}\) layer branches. No label needed on zero branch, if drawn |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{5}{6} \times \frac{1}{5}\) or \(\frac{1}{6}(\times 1)\) or \(\frac{1}{6}\) seen | M1 | or \(1 - \frac{5}{6} \times \frac{4}{5}\) or \(1 - \frac{2}{3}\): M2 |
| \(\frac{5}{6} \times \frac{1}{5} + \frac{1}{6}(\times 1)\) | M1 | all correct. ft incorrect tree dep probs \(\leq 1\); if \(3^{\text{rd}}\) tree prob \(= 1\): (ii)M1M1A0; if \(3^{\text{rd}}\) tree prob \(\neq 1\): (ii)M1M0A0. NB!! \(2 \times \frac{5}{6} \times \frac{1}{5} = \frac{1}{3}\): M1M0A0 |
| \(= \frac{1}{3}\) oe | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{4}{5} \times \frac{3}{4} + \frac{1}{5}(\times 1)\) or \(1 - \frac{4}{5} \times \frac{1}{4}\) or \(1 - 0.2\) all correct | M1 | or \((\frac{4}{6} \times \frac{4}{5} \times \frac{3}{4} + \frac{4}{6} \times \frac{5}{5} \times \frac{1}{5}) \div \frac{4}{6}\) all correct. but \(\frac{5}{6} \times (\frac{4}{5} \times \frac{3}{4} + \frac{1}{5})\): M0 |
| \(= \frac{4}{5}\) or 0.8 oe | A1 | May be seen without working M1A1 cao. ft incorrect tree: (iii) M1A0 |
## Question 4:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Top: 2 branches $\frac{4}{5}$, $\frac{1}{5}$ & R, B shown | B1 | consistent |
| Bottom: $1^{\text{st}}$ branch: prob $= 1$ or $\frac{5}{5}$, & R shown | B1 | allow eg $\frac{4}{4}$. Any missing label(s) on first three branches, subtr B1 once |
| no $2^{\text{nd}}$ branch OR branch with prob $= 0$ or $\frac{0}{5}$ | | ignore any $3^{\text{rd}}$ layer branches. No label needed on zero branch, if drawn |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{5}{6} \times \frac{1}{5}$ or $\frac{1}{6}(\times 1)$ or $\frac{1}{6}$ seen | M1 | or $1 - \frac{5}{6} \times \frac{4}{5}$ or $1 - \frac{2}{3}$: M2 |
| $\frac{5}{6} \times \frac{1}{5} + \frac{1}{6}(\times 1)$ | M1 | all correct. ft incorrect tree dep probs $\leq 1$; if $3^{\text{rd}}$ tree prob $= 1$: (ii)M1M1A0; if $3^{\text{rd}}$ tree prob $\neq 1$: (ii)M1M0A0. NB!! $2 \times \frac{5}{6} \times \frac{1}{5} = \frac{1}{3}$: M1M0A0 |
| $= \frac{1}{3}$ oe | A1 | cao |
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{4}{5} \times \frac{3}{4} + \frac{1}{5}(\times 1)$ or $1 - \frac{4}{5} \times \frac{1}{4}$ or $1 - 0.2$ all correct | M1 | or $(\frac{4}{6} \times \frac{4}{5} \times \frac{3}{4} + \frac{4}{6} \times \frac{5}{5} \times \frac{1}{5}) \div \frac{4}{6}$ all correct. but $\frac{5}{6} \times (\frac{4}{5} \times \frac{3}{4} + \frac{1}{5})$: M0 |
| $= \frac{4}{5}$ or 0.8 oe | A1 | May be seen without working M1A1 cao. ft incorrect tree: (iii) M1A0 |
---4 A bag contains 5 red discs and 1 black disc. Tina takes two discs from the bag at random without replacement.\\
(i) The diagram shows part of a tree diagram to illustrate this situation.
\section*{First disc}
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{e23cb28b-49e5-436a-942d-e6320029c634-3_264_494_479_550}
\end{center}
Complete the tree diagram in your Answer Book showing all the probabilities.
\section*{Second disc}
(ii) Find the probability that exactly one of the two discs is red.
All the discs are replaced in the bag. Tony now takes three discs from the bag at random without replacement.\\
(iii) Given that the first disc Tony takes is red, find the probability that the third disc Tony takes is also red.
\hfill \mbox{\textit{OCR S1 2012 Q4 [7]}}