| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2021 |
| Session | October |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Principle of Inclusion/Exclusion |
| Type | Three-Set Venn Diagram Probability Calculation |
| Difficulty | Standard +0.3 This is a standard A-level statistics question on Venn diagrams and probability that requires systematic application of basic probability rules (sum of probabilities, conditional probability, independence). While it has multiple parts, each step follows directly from definitions with no novel insight required—slightly easier than average due to its routine nature. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables |
| Answer | Marks | Guidance |
|---|---|---|
| Part | Answer/Working | Marks |
| (a) | \(0.08 + 0.09 + 0.36 = \mathbf{0.53}\) | B1 |
| (b)(i) | \([P(G \cap E \cap S) = 0] \Rightarrow \mathbf{p = 0}\) | B1 |
| (b)(ii) | \([P(G) = 0.25 \Rightarrow] 0.08 + 0.05 + q + " p" = 0.25\) | M1 |
| \(q = \mathbf{0.12}\) | A1 | A1 for \(q = 0.12\) (may be placed in Venn diagram) |
| (c)(i) | \(P(S \mid E) = \frac{5}{12} \Rightarrow \frac{r + " p"}{r + " p" + 0.09 + 0.05} = \frac{5}{12}\) | M1 |
| \([12r = 5r + 5 \times 0.14 \Rightarrow] r = \mathbf{0.10}\) | A1ft, A1 | 1st A1ft for a correct ratio of probabilities (on LHS) allowing ft of their \(p\) where \(0 \leq p < 0.86\). 2nd A1 for \(r = 0.1(0)\) or exact equivalent (may be in Venn diagram) Ans only 3/3 |
| (c)(ii) | \([0.08 + 0.05 + \text{"0.12"} + \text{"0"} + 0.09 + \text{"0.10"} + 0.36 + t = 1] \Rightarrow t = \mathbf{0.20}\) | B1ft |
| (d) | \(P(S \cap E') = 0.36 + " q" = [0.48]\) | B1ft |
| \(P\left(\left[\left(S \cap E'\right) \cap G\right] = \text{"q"} = [0.12]\right)\) and \(P(G) = 0.25\) and | M1 | M1 for attempting all required probs (labelled) and using them in a correct test (allow ft of \(q\)) |
| \(P(S \cap E') \times P(G) = \text{"0.48"} \times \frac{1}{4}\) or \(0.12\) | A1 | A1 for all probs correct and a correct deduction (no ft deduction here) |
| \(P(S \cap E') \times P(G) = 0.12 = P\left[\left(S \cap E'\right) \cap G\right]\) so are independent | ||
| *Notes* | ||
| SC: No "\(P\)" If correct argument seen apart from \(P\) for probability for all 3 marks, award (B0M1A1). If unsure about an attempt using conditional probabilities, please send to review. |
| Part | Answer/Working | Marks | Guidance |
|------|---|---|---|
| (a) | $0.08 + 0.09 + 0.36 = \mathbf{0.53}$ | B1 | B1 for 0.53 (or exact equivalent) [Allow 53%] |
| (b)(i) | $[P(G \cap E \cap S) = 0] \Rightarrow \mathbf{p = 0}$ | B1 | B1 for $p = 0$ (may be placed in Venn diagram) |
| (b)(ii) | $[P(G) = 0.25 \Rightarrow] 0.08 + 0.05 + q + " p" = 0.25$ | M1 | M1 for a linear equation for $q$ (ft letter "$p$" or their value if $0 \leq p \leq 0.12$) $\Rightarrow by p + q = 0.12$ |
| | $q = \mathbf{0.12}$ | A1 | A1 for $q = 0.12$ (may be placed in Venn diagram) |
| (c)(i) | $P(S \mid E) = \frac{5}{12} \Rightarrow \frac{r + " p"}{r + " p" + 0.09 + 0.05} = \frac{5}{12}$ | M1 | M1 for a ratio of probabilities ($r$ on num and den) (on LHS) with num < den and num or den correct ft. Allow ft of letter "$p$" or their $p$ where $0 \leq p < 0.86$ but "$ + 0$" is not required. |
| | $[12r = 5r + 5 \times 0.14 \Rightarrow] r = \mathbf{0.10}$ | A1ft, A1 | 1st A1ft for a correct ratio of probabilities (on LHS) allowing ft of their $p$ where $0 \leq p < 0.86$. 2nd A1 for $r = 0.1(0)$ or exact equivalent (may be in Venn diagram) **Ans only 3/3** |
| (c)(ii) | $[0.08 + 0.05 + \text{"0.12"} + \text{"0"} + 0.09 + \text{"0.10"} + 0.36 + t = 1] \Rightarrow t = \mathbf{0.20}$ | B1ft | B1ft for $t = 0.2(0)$ (o.e.) or correct ft i.e. $0.42 - (p + q + r)$ where $p, q, r$ and $t$ are all probs |
| (d) | $P(S \cap E') = 0.36 + " q" = [0.48]$ | B1ft | B1ft for $P(S \cap E') = 0.36 + " q"$ (ft letter "$q$" or their value if $0 \leq q \leq 0.12$) |
| | $P\left(\left[\left(S \cap E'\right) \cap G\right] = \text{"q"} = [0.12]\right)$ and $P(G) = 0.25$ and | M1 | M1 for attempting all required probs (labelled) and using them in a correct test (allow ft of $q$) |
| | $P(S \cap E') \times P(G) = \text{"0.48"} \times \frac{1}{4}$ or $0.12$ | A1 | A1 for **all probs correct and a correct deduction** (no ft deduction here) |
| | $P(S \cap E') \times P(G) = 0.12 = P\left[\left(S \cap E'\right) \cap G\right]$ so **are independent** | | |
| | *Notes* | | |
| | **SC:** No "$P$" If correct argument seen apart from $P$ for probability for all 3 marks, award (B0M1A1). **If unsure about an attempt using conditional probabilities, please send to review.** | | |
**Venn Diagram Provided:**
```
G E
0.08 0.05 0.09
0.12 0.10
0.36 0.20
S
```\begin{enumerate}
\item A large college produces three magazines.
\end{enumerate}
One magazine is about green issues, one is about equality and one is about sports.\\
A student at the college is selected at random and the events $G , E$ and $S$ are defined as follows\\
$G$ is the event that the student reads the magazine about green issues\\
$E$ is the event that the student reads the magazine about equality\\
$S$ is the event that the student reads the magazine about sports\\
The Venn diagram, where $p , q , r$ and $t$ are probabilities, gives the probability for each subset.\\
\includegraphics[max width=\textwidth, alt={}, center]{10736735-3050-43eb-9e76-011ca6fa48b8-10_508_862_756_603}\\
(a) Find the proportion of students in the college who read exactly one of these magazines.
No students read all three magazines and $\mathrm { P } ( G ) = 0.25$\\
(b) Find\\
(i) the value of $p$\\
(ii) the value of $q$
Given that $\mathrm { P } ( S \mid E ) = \frac { 5 } { 12 }$\\
(c) find\\
(i) the value of $r$\\
(ii) the value of $t$\\
(d) Determine whether or not the events ( $S \cap E ^ { \prime }$ ) and $G$ are independent. Show your working clearly.
\section*{Question 4 continued.}
\section*{Question 4 continued.}
\section*{Question 4 continued.}
\hfill \mbox{\textit{Edexcel Paper 3 2021 Q4 [11]}}