| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2019 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Principle of Inclusion/Exclusion |
| Type | Three Events with Independence Constraints |
| Difficulty | Standard +0.8 This question requires students to apply multiple probability concepts simultaneously (mutual exclusivity, independence, and the probability axiom that all probabilities sum to 1) to set up and solve a system of equations. While the individual concepts are AS-level standard, combining them to extract three unknowns from a Venn diagram requires careful algebraic manipulation and conceptual understanding beyond routine exercises. |
| Spec | 2.03a Mutually exclusive and independent events2.03c Conditional probability: using diagrams/tables2.03e Model with probability: critiquing assumptions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = 0\) | B1 | May be seen on Venn diagram |
| \(P(A) = 0.1 + z + y\), \(\quad P(C) = 0.39 + z[+x]\), \(\quad P(A \text{ and } C) = z\) | M1 | Identifying the probabilities required for independence and at least 2 correct. These must be labelled. If no labels, may be implied by \(z = (0.1+z+y)(0.39+z[+x])\), allow one numerical slip |
| \(P(A \text{ and } C) = P(A) \times P(C) \rightarrow z = (0.1+z+y)(0.39+z[+x])\) | M1 | Use of independence equation with labelled probabilities in terms of \(y\), \(z\) [and \(x\)]. All probabilities must be substituted into a correct formula. Sight of a correct equation e.g. \(z=(0.1+z+y)(0.39+z[+x])\) scores M1M1 |
| \(\left[\sum p = 1\right]\): \(0.06+0.3+0.39+0.1+z+y[+x]=1 \rightarrow [z+y[+x]=0.15]\) | M1 | Using \(\Sigma p = 1\). Implied by \([x+]\ y+z=0.15\), or their \(x+y+z=0.15\), or e.g. \(P(A)=0.25\) |
| Solving simultaneously: \(z = 0.13\), \(\quad y = 0.02\) | A1 | Both \(y=0.02\) and \(z=0.13\) |
| (5) |
## Question 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = 0$ | B1 | May be seen on Venn diagram |
| $P(A) = 0.1 + z + y$, $\quad P(C) = 0.39 + z[+x]$, $\quad P(A \text{ and } C) = z$ | M1 | Identifying the probabilities required for independence and at least 2 correct. These must be labelled. If no labels, may be implied by $z = (0.1+z+y)(0.39+z[+x])$, allow one numerical slip |
| $P(A \text{ and } C) = P(A) \times P(C) \rightarrow z = (0.1+z+y)(0.39+z[+x])$ | M1 | Use of independence equation with labelled probabilities in terms of $y$, $z$ [and $x$]. All probabilities must be substituted into a correct formula. Sight of a correct equation e.g. $z=(0.1+z+y)(0.39+z[+x])$ scores M1M1 |
| $\left[\sum p = 1\right]$: $0.06+0.3+0.39+0.1+z+y[+x]=1 \rightarrow [z+y[+x]=0.15]$ | M1 | Using $\Sigma p = 1$. Implied by $[x+]\ y+z=0.15$, or their $x+y+z=0.15$, or e.g. $P(A)=0.25$ |
| Solving simultaneously: $z = 0.13$, $\quad y = 0.02$ | A1 | Both $y=0.02$ and $z=0.13$ |
| | **(5)** | |
---\begin{enumerate}
\item The Venn diagram shows three events, $A$, $B$ and $C$, and their associated probabilities.\\
\includegraphics[max width=\textwidth, alt={}, center]{e73193ee-339e-48ab-811c-9ab6817f786d-04_680_780_296_644}
\end{enumerate}
Events $B$ and $C$ are mutually exclusive.\\
Events $A$ and $C$ are independent.\\
Showing your working, find the value of $x$, the value of $y$ and the value of $z$.
\hfill \mbox{\textit{Edexcel AS Paper 2 2019 Q2 [5]}}