| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2021 |
| Session | October |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Definitions |
| Type | Venn diagram completion |
| Difficulty | Moderate -0.8 This is a routine S1 Venn diagram question testing basic probability definitions (independence, conditional probability, mutual exclusivity). Parts (a)-(c) involve straightforward calculations using standard formulas, while part (d) requires understanding set relationships but no complex problem-solving. Easier than average A-level maths. |
| 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 | Mark | Guidance |
| Sum of probs \(= 1\) gives \(p + q = 0.2\), so \(P(B) = \mathbf{0.5}\) | B1 | For 0.5 or exact equivalent |
| (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| e.g. \(P(A) = 0.3\) or \(0.1 +\) "their value for \(p+q\)", \(P(A \cap B) = 0.2\) or "their value for \(p+q\)", and \([P(A) \times P(B) =] 0.3 \times \text{"0.5"}\ [=\text{"0.15"}]\) | M1 | For sight of correct probabilities for \(P(A)\) and \(P(A \cap B)\) clearly labelled, \(0.3 \times \text{"0.5"}\) seen or \(P(A) \times P(B) = 0.15\). Allow \(0.04 + 0.06 + 0.2\) for \(P(A)\) if clearly labelled. \(P(A \cap B)\) may be stated in part (a). May see \(P(B\ |
| \(0.15 \neq 0.2\) so \([A\) and \(B\) are\(]\) not independent | A1 | For all correct probabilities, calculations, a comparison and correct conclusion. Accept \(P(A \cap B) \neq P(A) \times P(B)\) instead of \(0.15 \neq 0.2\) for comparison |
| (2) | SC Allow M1A0 for \(P(A) = 0.1 + p + q\); \(P(A \cap B) = p + q\) clearly labelled and \(0.5 \times (0.1 + p + q)\) or \((p + q + 0.3)(0.1 + p + q)\) given |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([P(C \ | B) =]\ \dfrac{p}{\text{"0.5"}} = p + 0.06\) | M1 |
| \([2p = p + 0.06\) so \(]\ \mathbf{p = 0.06}\) | A1 | 1st A1 for \((p =)\ 0.06\) |
| \([\text{Use of } p + q = 0.2 \text{ gives}]\ \mathbf{q = 0.14}\) | A1 | 2nd A1 for \((q =)\ 0.14\). Ans only: \((p =)\ 0.06\) and \((q =)\ 0.14\) — 3/3 |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Suitable event \(D\) drawn [see Venn diagrams below] | B1 | For a suitable event \(D\) drawn that has an intersection with \(B\) but not with \(A\). Condone if not labelled \(D\) |
| (1) |
# Question 1:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sum of probs $= 1$ gives $p + q = 0.2$, so $P(B) = \mathbf{0.5}$ | B1 | For 0.5 or exact equivalent |
| | **(1)** | |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| e.g. $P(A) = 0.3$ or $0.1 +$ "their value for $p+q$", $P(A \cap B) = 0.2$ or "their value for $p+q$", **and** $[P(A) \times P(B) =] 0.3 \times \text{"0.5"}\ [=\text{"0.15"}]$ | M1 | For sight of correct probabilities for $P(A)$ and $P(A \cap B)$ **clearly labelled**, $0.3 \times \text{"0.5"}$ seen or $P(A) \times P(B) = 0.15$. Allow $0.04 + 0.06 + 0.2$ for $P(A)$ if clearly labelled. $P(A \cap B)$ may be stated in part (a). May see $P(B\|A) = \frac{2}{3}$ compared with $P(B)$ **or** $P(A\|B) = 0.4$ and $P(A) = 0.3$ |
| $0.15 \neq 0.2$ so $[A$ and $B$ are$]$ not independent | A1 | For all correct probabilities, calculations, a comparison and correct conclusion. Accept $P(A \cap B) \neq P(A) \times P(B)$ instead of $0.15 \neq 0.2$ for comparison |
| | **(2)** | **SC** Allow M1A0 for $P(A) = 0.1 + p + q$; $P(A \cap B) = p + q$ clearly labelled and $0.5 \times (0.1 + p + q)$ or $(p + q + 0.3)(0.1 + p + q)$ given |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[P(C \| B) =]\ \dfrac{p}{\text{"0.5"}} = p + 0.06$ | M1 | ft their $P(B)$ from part (a). For a correct equation in $p$ or $q$ based on the given statement. Allow $\dfrac{p}{0.3 + p + q} = p + 0.06$ |
| $[2p = p + 0.06$ so $]\ \mathbf{p = 0.06}$ | A1 | 1st A1 for $(p =)\ 0.06$ |
| $[\text{Use of } p + q = 0.2 \text{ gives}]\ \mathbf{q = 0.14}$ | A1 | 2nd A1 for $(q =)\ 0.14$. Ans only: $(p =)\ 0.06$ and $(q =)\ 0.14$ — 3/3 |
| | **(3)** | |
## Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Suitable event $D$ drawn [see Venn diagrams below] | B1 | For a suitable event $D$ drawn that has an intersection with $B$ but not with $A$. Condone if not labelled $D$ |
| | **(1)** | |
**Total: [7]**\begin{enumerate}
\item The Venn diagram shows the events $A$, $B$ and $C$ and their associated probabilities, where $p$ and $q$ are probabilities.\\
\includegraphics[max width=\textwidth, alt={}, center]{29ac0c0b-f963-40a1-beba-7146bbb2d021-02_579_1054_347_447}\\
(a) Find $\mathrm { P } ( B )$\\
(b) Determine whether or not $A$ and $B$ are independent.
\end{enumerate}
Given that $\mathrm { P } ( C \mid B ) = \mathrm { P } ( C )$\\
(c) find the value of $p$ and the value of $q$
The event $D$ is such that
\begin{itemize}
\item $\quad A$ and $D$ are mutually exclusive
\item $\mathrm { P } ( B \cap D ) > 0$\\
(d) On the Venn diagram show a possible position for the event $D$\\
\end{itemize}
\hfill \mbox{\textit{Edexcel S1 2021 Q1 [7]}}