| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2010 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Principle of Inclusion/Exclusion |
| Type | Three-Set Venn Diagram Probability Calculation |
| Difficulty | Moderate -0.8 This is a straightforward Venn diagram probability question requiring only basic counting and probability calculations. All information is explicitly given in the diagram, requiring no problem-solving insight—just routine application of probability formulas (counting favorable outcomes, conditional probability, and independence test). Typical of S1 material, this is easier than average A-level maths questions. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{2+3}{\text{their total}}\) or \(\frac{5}{\text{their total}}\) or \(\frac{5}{30}\) | M1 A1cso | M1 for \(\frac{2+3}{\text{their total}}\) or \(\frac{5}{30}\). ( given answer) |
| (2) | ||
| \(\frac{4+2+5+3}{\text{total}}\) | M1 A1 | M1 for adding at least 3 of "4, 2, 5, 3" and dividing by their total to give a probability. Can be written as separate fractions substituted into the completely correct Addition Rule |
| = \(\frac{14}{30}\) or \(\frac{7}{15}\) or 0.46 | (2) | |
| \(P(A \cap C) = 0\) | B1 | B1 for 0 or 0/30 |
| (1) | ||
| \(P(C \mid \text{reads at least one magazine}) = \frac{6+3}{20} = \frac{9}{20}\) | M1 A1 | M1 for a denominator of 20 or \(\frac{20}{30}\) leading to an answer with denominator of 20 |
| (2) | ||
| \(P(B) = \frac{10}{30} = \frac{1}{3}\) | M1 | |
| \(P(C) = \frac{9}{30} = \frac{3}{10}\) | ||
| \(P(B \cap C) = \frac{3}{30} = \frac{1}{10}\) or \(P(B \mid C) = \frac{3}{9}\) | ||
| \(P(B) \times P(C) = \frac{1}{3} \times \frac{3}{10} = \frac{1}{10} = P(B \cap C)\) or \(P(B \mid C) = \frac{3}{9} = \frac{1}{3} = P(B)\) | M1 | |
| So yes they are statistically independent | A1cso | 1st M1 for attempting all the required probabilities for a suitable test. 2nd M1 for use of a correct test - must have attempted all the correct probabilities. A1 for fully correct test carried out with a comment |
| (3) | ||
| Total 10 |
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{2+3}{\text{their total}}$ or $\frac{5}{\text{their total}}$ or $\frac{5}{30}$ | M1 A1cso | M1 for $\frac{2+3}{\text{their total}}$ or $\frac{5}{30}$. **(** given answer**)** |
| | | **(2)** |
| $\frac{4+2+5+3}{\text{total}}$ | M1 A1 | M1 for adding at least 3 of "4, 2, 5, 3" and dividing by their total to give a probability. Can be written as separate fractions substituted into the completely correct Addition Rule |
| = $\frac{14}{30}$ or $\frac{7}{15}$ or 0.46 | | **(2)** |
| $P(A \cap C) = 0$ | B1 | B1 for 0 or 0/30 |
| | | **(1)** |
| $P(C \mid \text{reads at least one magazine}) = \frac{6+3}{20} = \frac{9}{20}$ | M1 A1 | M1 for a **denominator of 20** or $\frac{20}{30}$ leading to an answer with denominator of 20 |
| | | **(2)** |
| $P(B) = \frac{10}{30} = \frac{1}{3}$ | M1 | |
| $P(C) = \frac{9}{30} = \frac{3}{10}$ | | |
| $P(B \cap C) = \frac{3}{30} = \frac{1}{10}$ or $P(B \mid C) = \frac{3}{9}$ | | |
| $P(B) \times P(C) = \frac{1}{3} \times \frac{3}{10} = \frac{1}{10} = P(B \cap C)$ or $P(B \mid C) = \frac{3}{9} = \frac{1}{3} = P(B)$ | M1 | |
| So yes they are statistically independent | A1cso | 1st M1 for attempting all the required probabilities for a suitable test. 2nd M1 for use of a correct test - must have attempted all the correct probabilities. A1 for fully correct test carried out with a comment |
| | | **(3)** |
| | | **Total 10** |
---4. 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]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{039e6fcf-3222-40cc-95ea-37b8dc4a4ddb-07_397_934_374_502}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
One of these students is selected at random.
\begin{enumerate}[label=(\alph*)]
\item Show that the probability that the student reads more than one magazine is $\frac { 1 } { 6 }$.
\item Find the probability that the student reads $A$ or $B$ (or both).
\item Write down the probability that the student reads both $A$ and $C$.
Given that the student reads at least one of the magazines,
\item find the probability that the student reads $C$.
\item Determine whether or not reading magazine $B$ and reading magazine $C$ are statistically independent.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2010 Q4 [10]}}