| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2007 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Venn diagram with two events |
| Difficulty | Moderate -0.8 This is a straightforward S1 conditional probability question using basic Venn diagram setup. Part (a) requires simple arithmetic with percentages to find the intersection, part (b) is drawing a diagram, and part (c) applies the definition of conditional probability with clearly given information. All steps are routine applications of standard techniques with no problem-solving insight required. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(Q \cup T) = 0.6\) | B1 | |
| \(P(Q) + P(T) - P(Q \cap T) = 0.6\) | M1 | |
| \(P(Q \cap T) = 0.1\) | A1 (3 marks) | B1 for 0.6. M1 for use of \(P(Q) + P(T) - P(Q \cap T) = P(Q \cup T)\). 0.1 correct answer only for A1. Alternative: \((25+45+40=)110\%\) B1; \(110-100=10\%\) M1A1. 0.1 stated clearly as final answer with no working gets 3/3. |
| Answer | Marks | Guidance |
|---|---|---|
| Venn diagram with two intersecting closed curves; values \(0.15\), \(0.1\), \(0.35\), \(0.4\) and box | M1, A1, B1 (3 marks) | Two intersecting closed curves for M1, no box required. At least one label (\(Q\) or \(T\)) required for first A1. Follow through \((0.25 - \text{their } 0.1)\) and \((0.45 - \text{their } 0.1)\) for A1. 0.4 and box required, correct answer only for B1. Using %, whole numbers in correct ratio all OK. |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(Q \cap T' \mid Q \cup T) = \frac{0.15}{0.60} = \frac{1}{4}\) or \(0.25\) or \(25\%\) | M1A1, A1 (3 marks) | Require fraction with denominator 0.6 or equivalent from Venn diagram for M1. Follow through their values in fraction for A1. Final A1 is correct answer only. No working, no marks. |
## Question 4:
**Part (a):**
$P(Q \cup T) = 0.6$ | B1 |
$P(Q) + P(T) - P(Q \cap T) = 0.6$ | M1 |
$P(Q \cap T) = 0.1$ | A1 (3 marks) | B1 for 0.6. M1 for use of $P(Q) + P(T) - P(Q \cap T) = P(Q \cup T)$. 0.1 correct answer only for A1. Alternative: $(25+45+40=)110\%$ B1; $110-100=10\%$ M1A1. 0.1 stated clearly as final answer with no working gets 3/3.
**Part (b):**
Venn diagram with two intersecting closed curves; values $0.15$, $0.1$, $0.35$, $0.4$ and box | M1, A1, B1 (3 marks) | Two intersecting closed curves for M1, no box required. At least one label ($Q$ or $T$) required for first A1. Follow through $(0.25 - \text{their } 0.1)$ and $(0.45 - \text{their } 0.1)$ for A1. 0.4 and box required, correct answer only for B1. Using %, whole numbers in correct ratio all OK.
**Part (c):**
$P(Q \cap T' \mid Q \cup T) = \frac{0.15}{0.60} = \frac{1}{4}$ or $0.25$ or $25\%$ | M1A1, A1 (3 marks) | Require fraction with denominator 0.6 or equivalent from Venn diagram for M1. Follow through their values in fraction for A1. Final A1 is correct answer only. No working, no marks.
---\begin{enumerate}
\item A survey of the reading habits of some students revealed that, on a regular basis, $25 \%$ read quality newspapers, 45\% read tabloid newspapers and 40\% do not read newspapers at all.\\
(a) Find the proportion of students who read both quality and tabloid newspapers.\\
(b) In the space on page 13 draw a Venn diagram to represent this information.
\end{enumerate}
A student is selected at random. Given that this student reads newspapers on a regular basis,\\
(c) find the probability that this student only reads quality newspapers.
\hfill \mbox{\textit{Edexcel S1 2007 Q4 [9]}}