| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2007 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Multi-stage with stopping condition |
| Difficulty | Standard +0.3 This is a straightforward tree diagram problem with clear stopping conditions. Students must complete a partially-given tree diagram and calculate probabilities using basic multiplication and addition rules. The conditional probabilities are simple fractions (3/10, 7/10, etc.) and the multi-stage aspect is scaffolded by the diagram structure. While it requires careful tracking of the stopping condition, this is a standard S1 exercise slightly easier than average due to the visual support and clear structure. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{2}{9}\) or \(\frac{1}{9}\) oe seen | B1 | |
| \(\frac{3}{9}\) or \(\frac{6}{9}\) oe seen | B1 | |
| \(\frac{1}{8}\) or \(\frac{8}{8}\) oe seen | B1 | |
| Correct structure | B1 | ie 8 correct branches only, ignore probs & values; but headings not req'd |
| All correct | B1 | 5 marks; including probs and values |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{10} x \frac{7}{9} + \frac{1}{10} x \frac{3}{9} + \frac{1}{10} x \frac{0}{9}\) | M2 | or \(\frac{1}{10}x \frac{7}{9} + \frac{1}{10}\) or \(1 - \frac{1}{10} x \frac{3}{9}\) M1: one correct prod or any prod + \(\frac{1}{10}\) |
| \(\frac{14}{113}\) or \(0.933\) oe | A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{21}{100}\) or \(0.525\) oe | A1 | 3 marks cao |
**i)**
$\frac{2}{9}$ or $\frac{1}{9}$ oe seen | B1 |
$\frac{3}{9}$ or $\frac{6}{9}$ oe seen | B1 |
$\frac{1}{8}$ or $\frac{8}{8}$ oe seen | B1 |
Correct structure | B1 | ie 8 correct branches only, ignore probs & values; but headings not req'd
All correct | B1 | 5 marks; including probs and values
**ii)**
$\frac{1}{10} x \frac{7}{9} + \frac{1}{10} x \frac{3}{9} + \frac{1}{10} x \frac{0}{9}$ | M2 | or $\frac{1}{10}x \frac{7}{9} + \frac{1}{10}$ or $1 - \frac{1}{10} x \frac{3}{9}$ M1: one correct prod or any prod + $\frac{1}{10}$
$\frac{14}{113}$ or $0.933$ oe | A1 | 3 marks
**iii)**
$\frac{21}{100}$ or $0.525$ oe | A1 | 3 marks cao
**Total for Question 7:** 11 marks7 A bag contains three 1 p coins and seven 2 p coins. Coins are removed at random one at a time, without replacement, until the total value of the coins removed is at least 3p. Then no more coins are removed.\\
(i) Copy and complete the probability tree diagram.
First coin\\
\includegraphics[max width=\textwidth, alt={}, center]{43f7e091-9ae7-4373-a209-e2ebdba5260f-4_350_317_1279_568}
Find the probability that\\
(ii) exactly two coins are removed,\\
(iii) the total value of the coins removed is 4p.
\hfill \mbox{\textit{OCR S1 2007 Q7 [11]}}