| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Independent Events |
| Type | Venn diagram with independence constraint |
| Difficulty | Standard +0.8 This question requires understanding of Venn diagram probabilities, applying the independence condition P(B∩C) = P(B)×P(C) to form equations, solving simultaneous equations for p and q, and calculating conditional probability. While the individual concepts are standard A-level, combining independence with Venn diagrams and solving the resulting system requires solid problem-solving skills beyond routine exercises. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables |
| Answer | Marks | Guidance |
|---|---|---|
| \([0.13 + 0.25 =]\ \mathbf{0.38}\) | B1 (1) | For 0.38 or exact equivalent. If answers given on Venn Diagram and in script, script takes precedence. |
| Answer | Marks | Guidance |
|---|---|---|
| Independence implies: \(P(B \cap C) = P(B) \times P(C) \Rightarrow 0.3 = (0.3 + 0.05 + 0.25) \times (0.3 + p)\) | M1 | For a correct equation in \(p\) or \(P(C)\) only. May be implied by \(p = 0.2\) provided this does not come from incorrect working. Condone missing brackets if they get 0.2. |
| So \(p = \mathbf{0.2}\) | A1 | For \(p = 0.2\) or exact equivalent. Beware: if \(p = 0.2\) comes from incorrect working e.g. \(p = \frac{0.6}{0.3} = 0.2\), score M0A0. |
| \([\text{Sum of probabilities} = 1 \text{ gives}]\ q = \mathbf{0.07}\) | B1ft (3) | For \(q = 0.07\) ft their \(p\), i.e. \(q = 0.27 - \text{"0.2"}\) where \(0 < p < 0.27\). |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left[P(A \mid B') =\right] \dfrac{P(A \cap B')}{P(B')}\) or \(\dfrac{0.13}{(1-0.6)}\) or \(\dfrac{0.13}{(0.13 + \text{"0.2"} + \text{"0.07"})}\) | M1 | For a correct ratio of probability expressions or a correct ratio of probabilities ft their values of \(p\) and \(q\) (provided both probabilities) or letters \(p\) and \(q\). |
| \(= \dfrac{13}{40}\) or \(\mathbf{0.325}\) | A1 (2) | For 0.325 or exact equivalent. Correct answer only scores 2/2. NB on epen this is labelled M1 but treat it as A1. |
# Question 1:
## Part (a)
$[0.13 + 0.25 =]\ \mathbf{0.38}$ | B1 **(1)** | For 0.38 or exact equivalent. If answers given on Venn Diagram and in script, script takes precedence.
## Part (b)
Independence implies: $P(B \cap C) = P(B) \times P(C) \Rightarrow 0.3 = (0.3 + 0.05 + 0.25) \times (0.3 + p)$ | M1 | For a correct equation in $p$ or $P(C)$ only. May be implied by $p = 0.2$ provided this does not come from incorrect working. Condone missing brackets if they get 0.2.
So $p = \mathbf{0.2}$ | A1 | For $p = 0.2$ or exact equivalent. **Beware:** if $p = 0.2$ comes from incorrect working e.g. $p = \frac{0.6}{0.3} = 0.2$, score M0A0.
$[\text{Sum of probabilities} = 1 \text{ gives}]\ q = \mathbf{0.07}$ | B1ft **(3)** | For $q = 0.07$ ft their $p$, i.e. $q = 0.27 - \text{"0.2"}$ where $0 < p < 0.27$.
## Part (c)
$\left[P(A \mid B') =\right] \dfrac{P(A \cap B')}{P(B')}$ or $\dfrac{0.13}{(1-0.6)}$ or $\dfrac{0.13}{(0.13 + \text{"0.2"} + \text{"0.07"})}$ | M1 | For a correct ratio of probability expressions or a correct ratio of probabilities ft their values of $p$ and $q$ (provided both probabilities) or letters $p$ and $q$.
$= \dfrac{13}{40}$ or $\mathbf{0.325}$ | A1 **(2)** | For 0.325 or exact equivalent. Correct answer only scores 2/2. **NB on epen this is labelled M1 but treat it as A1.**
---\begin{enumerate}
\item The Venn diagram, where $p$ and $q$ are probabilities, shows the three events $A , B$ and $C$ and their associated probabilities.\\
\includegraphics[max width=\textwidth, alt={}, center]{a067577e-e2a6-440b-9d22-d558fade15f0-02_745_935_347_566}\\
(a) Find $\mathrm { P } ( A )$
\end{enumerate}
The events $B$ and $C$ are independent.\\
(b) Find the value of $p$ and the value of $q$\\
(c) Find $\mathrm { P } \left( A \mid B ^ { \prime } \right)$
\hfill \mbox{\textit{Edexcel Paper 3 2023 Q1 [6]}}