| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Independent Events |
| Type | Calculate probabilities using independence |
| Difficulty | Moderate -0.3 This is a straightforward S1 question testing standard independence formulas. Part (a) is definitional recall (mutually exclusive events). Parts (b)-(d) require applying P(A∪B) = P(A) + P(B) - P(A∩B) with independence, then basic set operations and conditional probability—all routine techniques with no novel insight needed. Slightly easier than average due to being mostly formula application. |
| 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 |
|---|---|---|
| (a) \((R \text{ and } S \text{ are mutually exclusive})\) | B1 | B1 for '(mutually) exclusive' or 'cannot occur at the same time' seen or equivalent. 'Intersection is zero' or 'no overlaps' without further explanation is B0. |
| (1 mark) | ||
| (b) \(\frac{2}{3} = \frac{1}{4} + P(B) - P(A \cap B)\) (use of Addition Rule) | M1 | M1 for use of Addition Formula, including an intersection, with at least one probability substituted. Intersection must be explicitly considered for this mark. Accept \(\frac{2}{3} = \frac{1}{4} + P(B) - 0\) for M1. |
| \(\frac{2}{3} = \frac{1}{4} + P(B) - \frac{1}{4} \times P(B)\) (use of independence) | M1 A1 | M1 for \(P(A \cap B) = \frac{1}{4}P(B)\); A1 for completely correct equation or equivalent. |
| \(\frac{5}{12} = \frac{3}{4} P(B)\) | ||
| \(P(B) = \frac{5}{9}\) | A1 | A1 for \(\frac{5}{9}\) or exact equivalent. |
| (4 marks) | ||
| (c) \(P(A' \cap B) = \frac{3}{4} \times \frac{5}{9} = \frac{15}{36} = \frac{5}{12}\) | M1 A1 ft | M1 for \(\frac{3}{4} \times\) 'their \(P(B)\)' or 'their \(P(B)\)'\(- P(A \cap B)\) or \(P(A \cup B)-P(B) = \frac{2}{3} - \frac{1}{4}\); Or \(P(A' \cap B) = P(A') + \) 'their \(P(B)'\) \(- P(A' \cup B) = \frac{3}{4} + \frac{5}{9} - \frac{8}{9}\); A1 for \(\frac{5}{12}\) or follow through from their method. Accept exact equivalent. Correct answer only with no working M1A1 but must be clearly labelled (c). |
| (2 marks) | ||
| (d) \(P(B' | A) = \frac{(1-(b)) \times 0.25}{0.25}\) or \(P(B')\) or \(\frac{9}{4}\) | M1 |
| \(= \frac{4}{9}\) | A1 | A1 for \(\frac{4}{9}\) or exact equivalent. For part (c) follow through their stated values; do not follow through incorrectly labelled regions on a Venn Diagram. Throughout the question we require probabilities between 0 and 1 for method marks. Venn Diagram with 2 overlapping closed curves and correct values possibly without \(\frac{1}{3}\), award M1M1A1. |
| (2 marks) |
**(a)** $(R \text{ and } S \text{ are mutually exclusive})$ | B1 | B1 for '(mutually) exclusive' or 'cannot occur at the same time' seen or equivalent. 'Intersection is zero' or 'no overlaps' without further explanation is B0.
| | (1 mark) |
**(b)** $\frac{2}{3} = \frac{1}{4} + P(B) - P(A \cap B)$ (use of Addition Rule) | M1 | M1 for use of Addition Formula, including an intersection, with at least one probability substituted. Intersection must be explicitly considered for this mark. Accept $\frac{2}{3} = \frac{1}{4} + P(B) - 0$ for M1.
| | |
| $\frac{2}{3} = \frac{1}{4} + P(B) - \frac{1}{4} \times P(B)$ (use of independence) | M1 A1 | M1 for $P(A \cap B) = \frac{1}{4}P(B)$; A1 for completely correct equation or equivalent.
| | |
| $\frac{5}{12} = \frac{3}{4} P(B)$ | | |
| | |
| $P(B) = \frac{5}{9}$ | A1 | A1 for $\frac{5}{9}$ or exact equivalent.
| | (4 marks) |
**(c)** $P(A' \cap B) = \frac{3}{4} \times \frac{5}{9} = \frac{15}{36} = \frac{5}{12}$ | M1 A1 ft | M1 for $\frac{3}{4} \times$ 'their $P(B)$' or 'their $P(B)$'$- P(A \cap B)$ or $P(A \cup B)-P(B) = \frac{2}{3} - \frac{1}{4}$; Or $P(A' \cap B) = P(A') + $ 'their $P(B)'$ $- P(A' \cup B) = \frac{3}{4} + \frac{5}{9} - \frac{8}{9}$; A1 for $\frac{5}{12}$ or follow through from their method. Accept exact equivalent. Correct answer only with no working M1A1 but must be clearly labelled (c).
| | (2 marks) |
**(d)** $P(B'|A) = \frac{(1-(b)) \times 0.25}{0.25}$ or $P(B')$ or $\frac{9}{4}$ | M1 | M1 for using $1-$'their $P(B)'$ or $(P(A \cup B) - P(A))/P(A)$ or $(P(A) - P(A \cap B))/P(A)$ with a correct attempt at the numerator and denominator. If mutually exclusive is assumed then the last option gives $\frac{1}{4}$ for M1.
| | |
| $= \frac{4}{9}$ | A1 | A1 for $\frac{4}{9}$ or exact equivalent. For part (c) follow through their stated values; do not follow through incorrectly labelled regions on a Venn Diagram. Throughout the question we require probabilities between 0 and 1 for method marks. Venn Diagram with 2 overlapping closed curves and correct values possibly without $\frac{1}{3}$, award M1M1A1.
| | (2 marks) |
**Total: 9 marks**
---\begin{enumerate}
\item (a) State in words the relationship between two events $R$ and $S$ when $\mathrm { P } ( R \cap S ) = 0$
\end{enumerate}
The events $A$ and $B$ are independent with $\mathrm { P } ( A ) = \frac { 1 } { 4 }$ and $\mathrm { P } ( A \cup B ) = \frac { 2 } { 3 }$ Find\\
(b) $\mathrm { P } ( B )$\\
(c) $\mathrm { P } \left( A ^ { \prime } \cap B \right)$\\
(d) $\mathrm { P } \left( B ^ { \prime } \mid A \right)$\\
\hfill \mbox{\textit{Edexcel S1 2012 Q2 [9]}}