Edexcel S1 — Question 2 11 marks

Exam BoardEdexcel
ModuleS1 (Statistics 1)
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicConditional Probability
TypeIndependence test requiring preliminary calculations
DifficultyModerate -0.3 This is a standard S1 probability question testing independence and basic probability rules. Part (a) requires checking if P(A∩B) = P(A)×P(B), parts (b-c) use complement/intersection rules, and part (d) requires Bayes' theorem. All are routine applications of formulas with straightforward arithmetic—slightly easier than average due to being purely procedural with no problem-solving insight required.
Spec2.03a Mutually exclusive and independent events2.03c Conditional probability: using diagrams/tables

2. Given that \(\mathrm { P } ( A ) = \frac { 3 } { 5 } , \mathrm { P } ( B ) = \frac { 5 } { 8 } , \mathrm { P } ( A \cap B ) = \frac { 7 } { 20 } , \mathrm { P } ( A \cup C ) = \frac { 7 } { 10 }\) and \(\mathrm { P } ( C \mid A ) = \frac { 1 } { 3 }\),
  1. determine whether \(A\) and \(B\) are independent events.
  2. Find \(\mathrm { P } \left( A \cap B ^ { \prime } \right)\).
  3. Find \(\mathrm { P } \left( ( A \cap C ) ^ { \prime } \right)\).
  4. Find \(\mathrm { P } ( A \mid C )\).

AnswerMarks Guidance
(a) \(P(A) \times P(B) = \frac{3}{8} \times \frac{7}{20}\), so not independentM1 A1
(b) \(P(A \cap B') = \frac{3}{5} - \frac{7}{20} = \frac{1}{4}\)M1 A1
(c) \(P(CA) = P(A \cap C)/P(A)\) where \(P(A \cap C) = \frac{1}{5}\) and \(P((A \cap C)') = \frac{4}{5}\) M1 A1 A1
(d) \(P(A \cup C) = \frac{2}{5} + P(C)\) where \(P(C) = \frac{7}{10} - \frac{2}{5} = \frac{3}{10}\)M1 A1
\(P(AC) = \frac{1}{5} ÷ \frac{3}{10} = \frac{2}{3}\) M1 A1
(a) $P(A) \times P(B) = \frac{3}{8} \times \frac{7}{20}$, so not independent | M1 A1 |

(b) $P(A \cap B') = \frac{3}{5} - \frac{7}{20} = \frac{1}{4}$ | M1 A1 |

(c) $P(C|A) = P(A \cap C)/P(A)$ where $P(A \cap C) = \frac{1}{5}$ and $P((A \cap C)') = \frac{4}{5}$ | M1 A1 A1 |

(d) $P(A \cup C) = \frac{2}{5} + P(C)$ where $P(C) = \frac{7}{10} - \frac{2}{5} = \frac{3}{10}$ | M1 A1 |

$P(A|C) = \frac{1}{5} ÷ \frac{3}{10} = \frac{2}{3}$ | M1 A1 | 11 marks
2. Given that $\mathrm { P } ( A ) = \frac { 3 } { 5 } , \mathrm { P } ( B ) = \frac { 5 } { 8 } , \mathrm { P } ( A \cap B ) = \frac { 7 } { 20 } , \mathrm { P } ( A \cup C ) = \frac { 7 } { 10 }$ and $\mathrm { P } ( C \mid A ) = \frac { 1 } { 3 }$,
\begin{enumerate}[label=(\alph*)]
\item determine whether $A$ and $B$ are independent events.
\item Find $\mathrm { P } \left( A \cap B ^ { \prime } \right)$.
\item Find $\mathrm { P } \left( ( A \cap C ) ^ { \prime } \right)$.
\item Find $\mathrm { P } ( A \mid C )$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel S1  Q2 [11]}}