Edexcel S1 Specimen — Question 4 10 marks

Exam BoardEdexcel
ModuleS1 (Statistics 1)
SessionSpecimen
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPrinciple of Inclusion/Exclusion
TypeThree-Set Venn Diagram Probability Calculation
DifficultyModerate -0.8 This is a straightforward Venn diagram reading exercise requiring only basic probability calculations (counting regions and dividing by totals). All information is explicitly given in the diagram, requiring no problem-solving insight—just careful arithmetic and recall of independence definition. Significantly easier than average A-level questions.
Spec2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space

  1. The Venn diagram in Figure 1 shows the number of students in a class who read any of 3 popular magazines \(A , B\) and \(C\).
\begin{figure}[h]
\captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} One of these students is selected at random.
  1. Show that the probability that the student reads more than one magazine is \(\frac { 1 } { 6 }\).
  2. Find the probability that the student reads \(A\) or \(B\) (or both).
  3. Write down the probability that the student reads both \(A\) and \(C\). Given that the student reads at least one of the magazines,
  4. find the probability that the student reads \(C\).
  5. Determine whether or not reading magazine \(B\) and reading magazine \(C\) are statistically independent.

Question 4:
Part (a)
AnswerMarks Guidance
\(\frac{2+3}{\text{their total}} = \frac{5}{\text{their total}} = \frac{1}{6}\) (given answer)M1 A1cso M1 for \(\frac{2+3}{\text{their total}}\) or \(\frac{5}{30}\)
Part (b)
AnswerMarks Guidance
\(\frac{4+2+5+3}{\text{total}} = \frac{14}{30}\) or \(\frac{7}{15}\) or 0.4\(\dot{6}\)M1 A1 M1 for adding at least 3 of "4,2,5,3" and dividing by total
Part (c)
AnswerMarks Guidance
\(P(A \cap C) = 0\)B1 B1 for 0 or 0/30
Part (d)
AnswerMarks Guidance
\(P(C \text{ reads at least one magazine}) = \frac{6+3}{20} = \frac{9}{20}\)M1 A1 M1 for denominator of 20 or \(\frac{20}{30}\) leading to denominator 20
Part (e)
AnswerMarks Guidance
\(P(B) = \frac{10}{30} = \frac{1}{3}\), \(P(C) = \frac{9}{30} = \frac{3}{10}\), \(P(B \cap C) = \frac{3}{30} = \frac{1}{10}\)M1 1st M1 for attempting all required probabilities
\(P(B)\times P(C) = \frac{1}{3}\times\frac{3}{10} = \frac{1}{10} = P(B \cap C)\) or \(P(BC) = \frac{3}{9} = \frac{1}{3} = P(B)\) M1
So yes they are statistically independentA1cso A1 for fully correct test with comment
# Question 4:

## Part (a)
| $\frac{2+3}{\text{their total}} = \frac{5}{\text{their total}} = \frac{1}{6}$ (**given answer**) | M1 A1cso | M1 for $\frac{2+3}{\text{their total}}$ or $\frac{5}{30}$ |

## Part (b)
| $\frac{4+2+5+3}{\text{total}} = \frac{14}{30}$ or $\frac{7}{15}$ or 0.4$\dot{6}$ | M1 A1 | M1 for adding at least 3 of "4,2,5,3" and dividing by total |

## Part (c)
| $P(A \cap C) = 0$ | B1 | B1 for 0 or 0/30 |

## Part (d)
| $P(C \text{ reads at least one magazine}) = \frac{6+3}{20} = \frac{9}{20}$ | M1 A1 | M1 for denominator of 20 or $\frac{20}{30}$ leading to denominator 20 |

## Part (e)
| $P(B) = \frac{10}{30} = \frac{1}{3}$, $P(C) = \frac{9}{30} = \frac{3}{10}$, $P(B \cap C) = \frac{3}{30} = \frac{1}{10}$ | M1 | 1st M1 for attempting all required probabilities |
|---|---|---|
| $P(B)\times P(C) = \frac{1}{3}\times\frac{3}{10} = \frac{1}{10} = P(B \cap C)$ or $P(B|C) = \frac{3}{9} = \frac{1}{3} = P(B)$ | M1 | 2nd M1 for correct test with all correct probabilities |
| So yes they are statistically independent | A1cso | A1 for fully correct test with comment |

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\begin{enumerate}
  \item The Venn diagram in Figure 1 shows the number of students in a class who read any of 3 popular magazines $A , B$ and $C$.
\end{enumerate}

\begin{figure}[h]
\begin{center}
  \begin{tikzpicture}
  % Outer rectangle
  \draw[thick] (-4,-2.5) rectangle (4.5,2.8);

  % Circle A (upper left)
  \draw[thick] (-1.8,0.5) ellipse (2 and 1);
  \node[above left] at (-2.2,1.7) {\textit{A}};

  % Circle B (upper center-right, larger)
  \draw[thick] (1.2,0.4) ellipse (2 and 1);
  \node[above] at (1.2,1.6) {\textit{B}};

  % Ellipse C (lower right)
  \draw[thick] (2.2,-0.5) ellipse (2 and 1);
  \node[right] at (3.8,-0.0) {\textit{C}};

  % Region labels
  \node at (-2.2,0.5) {4};
  \node at (-0.5,0.7) {2};
  \node at (1.2,1.2) {5};
  \node at (1.8,-0.2) {3};
  \node at (3.0,-0.7) {6};
  \node at (-2.8,-1.8) {10};
\end{tikzpicture}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

One of these students is selected at random.\\
(a) Show that the probability that the student reads more than one magazine is $\frac { 1 } { 6 }$.\\
(b) Find the probability that the student reads $A$ or $B$ (or both).\\
(c) Write down the probability that the student reads both $A$ and $C$.

Given that the student reads at least one of the magazines,\\
(d) find the probability that the student reads $C$.\\
(e) Determine whether or not reading magazine $B$ and reading magazine $C$ are statistically independent.\\

\hfill \mbox{\textit{Edexcel S1  Q4 [10]}}