OCR MEI S1 — Question 1 5 marks

Exam BoardOCR MEI
ModuleS1 (Statistics 1)
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicConditional Probability
TypeVenn diagram with three events
DifficultyModerate -0.8 This is a straightforward Venn diagram reading exercise requiring basic probability calculations (reading values, simple addition/subtraction) and one conditional probability application using P(A|B) = P(A∩B)/P(B). All values are given directly in the diagram with no algebraic setup needed, making it easier than average.
Spec2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables

1 A survey is being carried out into the sports viewing habits of people in a particular area. As part of the survey, 250 people are asked which of the following sports they have watched on television in the past month.
  • Football
  • Cycling
  • Rugby
The numbers of people who have watched these sports are shown in the Venn diagram. \includegraphics[max width=\textwidth, alt={}, center]{870b6ef1-60f7-42e3-95f8-0544a2a07b15-1_725_921_728_622} One of the people is selected at random.
  1. Find the probability that this person has in the past month
    (A) watched cycling but not football,
    (B) watched either one or two of the three sports.
  2. Given that this person has watched cycling, find the probability that this person has not watched football.

Question 1
(i) (A)
\(P(\text{Watched cyc but not fb}) = \frac{15}{250} = \frac{3}{50} = 0.06\)
B1 CAO (aef)
[1]
(i) (B)
\(P(\text{Watched one or two}) = \frac{33+12+21+14+3+65}{250}\)
\(= \frac{148}{250} = \frac{74}{125} = 0.592\)
M1 For M1 terms must be added with no extra terms (added or subtracted)
A1 CAO (aef)
[2]
OR: \(\frac{250-(64+38)}{250}\)
(ii)
\(P(\text{Not watched fb} \mid \text{watched cyc}) = \frac{15}{67} = 0.224\) (0.223880597...)
M1 For denominator of either 67 or 67/250 or 0.268
A1 CAO (aef)
[2]
Allow 0.22 with working
# Question 1

## (i) (A)

$P(\text{Watched cyc but not fb}) = \frac{15}{250} = \frac{3}{50} = 0.06$

B1 CAO (aef)

[1]

## (i) (B)

$P(\text{Watched one or two}) = \frac{33+12+21+14+3+65}{250}$

$= \frac{148}{250} = \frac{74}{125} = 0.592$

M1 For M1 terms must be added with no extra terms (added or subtracted)

A1 CAO (aef)

[2]

OR: $\frac{250-(64+38)}{250}$

## (ii)

$P(\text{Not watched fb} \mid \text{watched cyc}) = \frac{15}{67} = 0.224$ (0.223880597...)

M1 For denominator of either 67 or 67/250 or 0.268

A1 CAO (aef)

[2]

Allow 0.22 with working
1 A survey is being carried out into the sports viewing habits of people in a particular area. As part of the survey, 250 people are asked which of the following sports they have watched on television in the past month.

\begin{itemize}
  \item Football
  \item Cycling
  \item Rugby
\end{itemize}

The numbers of people who have watched these sports are shown in the Venn diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{870b6ef1-60f7-42e3-95f8-0544a2a07b15-1_725_921_728_622}

One of the people is selected at random.
\begin{enumerate}[label=(\roman*)]
\item Find the probability that this person has in the past month\\
(A) watched cycling but not football,\\
(B) watched either one or two of the three sports.
\item Given that this person has watched cycling, find the probability that this person has not watched football.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI S1  Q1 [5]}}