OCR MEI S1 2007 June — Question 4 8 marks

Exam BoardOCR MEI
ModuleS1 (Statistics 1)
Year2007
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProbability Definitions
TypeVenn diagram completion
DifficultyEasy -1.2 This is a straightforward Venn diagram reading exercise requiring only basic probability calculations (counting regions, simple division, and one conditional probability). All information is given visually; no problem-solving or insight needed—just careful arithmetic with the diagram values.
Spec2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables

4 A local council has introduced a recycling scheme for aluminium, paper and kitchen waste. 50 residents are asked which of these materials they recycle. The numbers of people who recycle each type of material are shown in the Venn diagram. \includegraphics[max width=\textwidth, alt={}, center]{5e4f3310-b96e-43db-9b6d-61da3270db06-3_803_803_406_671} One of the residents is selected at random.
  1. Find the probability that this resident recycles
    (A) at least one of the materials,
    (B) exactly one of the materials.
  2. Given that the resident recycles aluminium, find the probability that this resident does not recycle paper. Two residents are selected at random.
  3. Find the probability that exactly one of them recycles kitchen waste.

Question 4:
Part (i)
AnswerMarks Guidance
\((A)\) \(P(\text{at least one}) = \frac{36}{50} = \frac{18}{25} = 0.72\)B1 aef
\((B)\) \(P(\text{exactly one}) = \frac{9+6+5}{50} = \frac{20}{50} = \frac{2}{5} = 0.4\)M1 for \((9+6+5)/50\), A1 aef 3 marks
Part (ii)
AnswerMarks Guidance
\(P(\text{not paper} \mid \text{aluminium}) = \frac{13}{24}\)M1 for denominator 24 or \(24/50\) or \(0.48\), A1 CAO 2 marks
Part (iii)
AnswerMarks Guidance
\(P(\text{one kitchen waste}) = 2\times\frac{18}{50}\times\frac{32}{49} = \frac{576}{1225} = 0.470\)M1 for both fractions, M1 for \(2\times\) product of both or sum of 2 pairs, A1 3 marks
# Question 4:

## Part (i)
$(A)$ $P(\text{at least one}) = \frac{36}{50} = \frac{18}{25} = 0.72$ | B1 aef |

$(B)$ $P(\text{exactly one}) = \frac{9+6+5}{50} = \frac{20}{50} = \frac{2}{5} = 0.4$ | M1 for $(9+6+5)/50$, A1 aef | 3 marks

## Part (ii)
$P(\text{not paper} \mid \text{aluminium}) = \frac{13}{24}$ | M1 for denominator 24 or $24/50$ or $0.48$, A1 CAO | 2 marks

## Part (iii)
$P(\text{one kitchen waste}) = 2\times\frac{18}{50}\times\frac{32}{49} = \frac{576}{1225} = 0.470$ | M1 for both fractions, M1 for $2\times$ product of both or sum of 2 pairs, A1 | 3 marks

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4 A local council has introduced a recycling scheme for aluminium, paper and kitchen waste. 50 residents are asked which of these materials they recycle. The numbers of people who recycle each type of material are shown in the Venn diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{5e4f3310-b96e-43db-9b6d-61da3270db06-3_803_803_406_671}

One of the residents is selected at random.
\begin{enumerate}[label=(\roman*)]
\item Find the probability that this resident recycles\\
(A) at least one of the materials,\\
(B) exactly one of the materials.
\item Given that the resident recycles aluminium, find the probability that this resident does not recycle paper.

Two residents are selected at random.
\item Find the probability that exactly one of them recycles kitchen waste.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI S1 2007 Q4 [8]}}