| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2004 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Definitions |
| Type | Combined event algebra |
| Difficulty | Moderate -0.8 This is a straightforward S1 probability question testing basic set operations and definitions. All parts follow directly from applying standard formulas (P(A∩B') = P(A) - P(A∩B), addition rule, conditional probability) with no conceptual challenges or problem-solving required—purely mechanical application of learned techniques. |
| 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/Working | Marks | Guidance |
| \(P(A \cap B') = P(A/B') \cdot P(B') = \frac{4}{5} \times \frac{1}{2} = \frac{4}{10} = \frac{2}{5}\) | M1 A1 | Use of \(P(A/B')P(B')\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(A \cap B) = P(A) - P(A \cap B')\) | M1 | |
| \(= \frac{2}{5} - \frac{2}{5}\) | ||
| \(= \underline{0}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) | M1 | |
| \(= \frac{2}{5} + \frac{1}{2} - 0\) | ||
| \(= \frac{9}{10}\) | A1 \(\checkmark\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(A/B) = P\frac{(A \cap B)}{P(B)} = 0\) | B1 \(\checkmark\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Since \(P(A \cap B) = 0\) (seen), \(A\) and \(B\) are mutually exclusive | B1 B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Since \(P(A/B) \neq P(A)\) or equivalent, \(A\) and \(B\) are NOT independent | B1 B1 |
## Question 4:
### Part (a)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(A \cap B') = P(A/B') \cdot P(B') = \frac{4}{5} \times \frac{1}{2} = \frac{4}{10} = \frac{2}{5}$ | M1 A1 | Use of $P(A/B')P(B')$ | |
### Part (a)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(A \cap B) = P(A) - P(A \cap B')$ | M1 | |
| $= \frac{2}{5} - \frac{2}{5}$ | | |
| $= \underline{0}$ | A1 | | |
### Part (a)(iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ | M1 | |
| $= \frac{2}{5} + \frac{1}{2} - 0$ | | |
| $= \frac{9}{10}$ | A1 $\checkmark$ | | |
### Part (a)(iv)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(A/B) = P\frac{(A \cap B)}{P(B)} = 0$ | B1 $\checkmark$ | | **(7)** |
### Part (b)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Since $P(A \cap B) = 0$ (seen), $A$ and $B$ are mutually exclusive | B1 B1 | | **(2)** |
### Part (b)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Since $P(A/B) \neq P(A)$ or equivalent, $A$ and $B$ are NOT independent | B1 B1 | | **(2)** |
---4. $\quad$ The events $A$ and $B$ are such that $\mathrm { P } ( A ) = \frac { 2 } { 5 } , \mathrm { P } ( B ) = \frac { 1 } { 2 }$ and $\mathrm { P } \left( A \quad B ^ { \prime } \right) = \frac { 4 } { 5 }$.
\begin{enumerate}[label=(\alph*)]
\item Find
\begin{enumerate}[label=(\roman*)]
\item $\mathrm { P } \left( A \cap B ^ { \prime } \right)$,
\item $\mathrm { P } ( A \cap B )$,
\item $\mathrm { P } ( A \cup B )$,
\item $\mathrm { P } \left( \begin{array} { l l } A & B \end{array} \right)$.
\end{enumerate}\item State, with a reason, whether or $\operatorname { not } A$ and $B$ are
\begin{enumerate}[label=(\roman*)]
\item mutually exclusive,
\item independent.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2004 Q4 [11]}}