| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Venn diagram with three events |
| Difficulty | Moderate -0.8 This is a straightforward Venn diagram question requiring basic probability calculations: reading values from the diagram, adding regions, and applying conditional probability formula P(A|B) = P(A∩B)/P(B). All techniques are standard S1 material with no problem-solving insight needed, making it easier than average but not trivial due to multiple parts. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables2.04b Binomial distribution: as model B(n,p) |
| Answer | Marks |
|---|---|
| B1 aef isw |
| Answer | Marks |
|---|---|
| P(Reduced car use | Avoided air travel) = \(\frac{6}{7}\) = 0.857 |
| M1 for denominator 7 or \(\frac{7}{100}\) or 0.07 FT their (i)A | |
| A1 CAO |
| Answer | Marks |
|---|---|
| M1 for \(\frac{93}{100} \times\) (triple product) | |
| M1 for product of remaining fractions | |
| A1 |
| Answer | Marks |
|---|---|
| TOTAL | 8 |
# Question 1
## (i)
**P(Avoided air travel)** = $\frac{7}{100}$ = 0.07
**P(At least two)** = $\frac{11 + 2 + 1 + 4}{100}$ = $\frac{18}{100}$ = $\frac{9}{50}$ = 0.18
| B1 aef isw |
**M1** for $\frac{11 + 2 + 1 + 4}{100}$
**A1** aef isw
*For M1: terms must be added. Must be as above or better with no extra terms (added or subtracted) for M1.*
*Must simplify to 18/100 or 9/50 or 0.18 for A1.*
*SC1 for 18/58*
*Or $1 - \frac{14 + 26 + 0 + 42}{100}$ = 0.18 gets M1A1*
## (ii)
**P(Reduced car use | Avoided air travel)** = $\frac{6}{7}$ = 0.857
| M1 for denominator 7 or $\frac{7}{100}$ or 0.07 FT their (i)A |
| A1 CAO |
*Allow 0.86*
## (iii)
**P(None have avoided air travel)** = $\frac{93}{100} \times \frac{92}{99} \times \frac{91}{98}$ = 0.8025
| M1 for $\frac{93}{100} \times$ (triple product) |
| M1 for product of remaining fractions |
| A1 |
*Fuller answer 0.802511, so allow 0.803 without working, but 0.80 or 0.8 only with working.*
*$\left(\frac{93}{100}\right)^3$ scores M1M0A0 which gives answer 0.804357 so watch for this.*
*M0M0A0 for binomial probability including $0.93^{100}$ but $^3C_0 \times 0.07^0 \times 0.93^3$ still scores M1*
*$\left(\frac{k}{100}\right)^3$ for values of $k$ other than 93 scores M0M0A0*
*$\frac{k(k-1)(k-2)}{100 \times 99 \times 98}$ for values of $k$ other than 93 scores M1M0A0*
*Correct working but then multiplied or divided by some factor scores M1M0A0*
*$\frac{^{93}P_3}{^{100}P_3}$ = 0.803 or $^{93}P_3$ seen M1, divided by $^{100}P_3$ M1, 0.803 A1*
*$\frac{^{93}C_3}{^{100}C_3}$ = 0.803*
*Allow unsimplified fractional answer $\frac{778596}{970200}$ = $\frac{9269}{11550}$*
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**TOTAL** | 8 |1 A survey is being carried out into the carbon footprint of individual citizens. As part of the survey, 100 citizens are asked whether they have attempted to reduce their carbon footprint by any of the following methods.
\begin{itemize}
\item Reducing car use
\item Insulating their homes
\item Avoiding air travel
\end{itemize}
The numbers of citizens who have used each of these methods are shown in the Venn diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{e54eba7c-d862-435a-acdd-27df6ede5fab-1_699_1085_849_569}
One of the citizens is selected at random.
\begin{enumerate}[label=(\roman*)]
\item Find the probability that this citizen\\
(A) has avoided air travel,\\
(B) has used at least two of the three methods.
\item Given that the citizen has avoided air travel, find the probability that this citizen has reduced car use.
Three of the citizens are selected at random.
\item Find the probability that none of them have avoided air travel.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 Q1 [8]}}