| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Calculate combined outcome probability |
| Difficulty | Moderate -0.3 This is a standard tree diagram probability question with straightforward calculations: parts (i) and (ii) involve basic probability rules (multiplication, addition, conditional probability), while part (iii) requires understanding of 'at least one' complement and solving a simple inequality. All techniques are routine for S1, though the multi-part structure and final inequality push it slightly above trivial exercises. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Probability \(= 1 - 0.79^5\) | M1 | for \(0.79^5\) or \(0.3077...\) |
| \(= 1 - 0.3077\) | M1 | for \(1 - 0.79^5\) dep |
| \(= 0.6923\) (accept awrt \(0.69\)) | A1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(1 - 0.79^n > 0.9\) | M1 | for equation/inequality in \(n\) (accept either statement opposite) |
| \(0.79^n < 0.1\) | M1 (indep) | for process of using logs i.e. \(\frac{\log a}{\log b}\) |
| \(n > \frac{\log 0.1}{\log 0.79}\), so \(n > 9.768...\) | A1 | CAO |
| Minimum \(n = 10\), Accept \(n \geq 10\) | ||
| OR (trial and improvement): \(0.79^9\) or \(0.79^{10}\) | M1 (indep) | for sight of \(0.8801\) or \(0.1198\) |
| \(1 - 0.79^9 = 0.8801 < 0.9\) or \(0.79^9 = 0.1198 > 0.1\) | M1 (indep) | for sight of \(0.9053\) or \(0.09468\) |
| \(1 - 0.79^{10} = 0.9053 > 0.9\) or \(0.79^{10} = 0.09468 < 0.1\) | A1 dep | on both M's cao |
| Minimum \(n = 10\), Accept \(n \geq 10\) | NOTE: \(n=10\) unsupported scores SC1 only |
## Question 1:
### Part (iii)(A)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Probability $= 1 - 0.79^5$ | M1 | for $0.79^5$ or $0.3077...$ |
| $= 1 - 0.3077$ | M1 | for $1 - 0.79^5$ dep |
| $= 0.6923$ (accept awrt $0.69$) | A1 | CAO |
### Part (iii)(B)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $1 - 0.79^n > 0.9$ | M1 | for equation/inequality in $n$ (accept either statement opposite) |
| $0.79^n < 0.1$ | M1 (indep) | for process of using logs i.e. $\frac{\log a}{\log b}$ |
| $n > \frac{\log 0.1}{\log 0.79}$, so $n > 9.768...$ | A1 | CAO |
| Minimum $n = 10$, Accept $n \geq 10$ | | |
| **OR** (trial and improvement): $0.79^9$ or $0.79^{10}$ | M1 (indep) | for sight of $0.8801$ **or** $0.1198$ |
| $1 - 0.79^9 = 0.8801 < 0.9$ or $0.79^9 = 0.1198 > 0.1$ | M1 (indep) | for sight of $0.9053$ **or** $0.09468$ |
| $1 - 0.79^{10} = 0.9053 > 0.9$ or $0.79^{10} = 0.09468 < 0.1$ | A1 dep | on both M's cao |
| Minimum $n = 10$, Accept $n \geq 10$ | | NOTE: $n=10$ unsupported scores SC1 only |
---1 In a large town, 79\% of the population were born in England, 20\% in the rest of the UK and the remaining $1 \%$ overseas. Two people are selected at random.
You may use the tree diagram below in answering this question.\\
\includegraphics[max width=\textwidth, alt={}, center]{b56ccabe-0e51-4555-b550-78ba347f69bb-1_944_1118_626_547}
\begin{enumerate}[label=(\roman*)]
\item Find the probability that\\
(A) both of these people were born in the rest of the UK,\\
(B) at least one of these people was born in England,\\
(C) neither of these people was born overseas.
\item Find the probability that both of these people were born in the rest of the UK given that neither was born overseas.
\item (A) Five people are selected at random. Find the probability that at least one of them was not born in England.\\
(B) An interviewer selects $n$ people at random. The interviewer wishes to ensure that the probability that at least one of them was not born in England is more than $90 \%$. Find the least possible value of $n$. You must show working to justify your answer.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 Q1 [16]}}