| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Session | Specimen |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Venn diagram with three events |
| Difficulty | Standard +0.3 This is a standard Venn diagram probability question requiring reading probabilities from a diagram, conditional probability calculation, and using independence to solve simultaneous equations. While part (c) involves algebraic manipulation with independence conditions, the techniques are routine for A-level statistics and the multi-step reasoning is straightforward rather than requiring novel insight. |
| 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 |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Moments about \(A\) (or any other complete method) | M1 | Correct overall strategy e.g. M(\(A\)), with usual rules, to give equation in \(T\) only |
| \(T\cos 30° \times (1\sin 30°) = 20g \times 1.5\) | A1 | (A1A0 one error) Condone 1 error |
| \(T\cos 30° \times (1\sin 30°) = 20g \times 1.5\) | A1 | (A0A0 two or more errors) |
| \(T = 679\) or \(680\ \text{(N)}\) | A1 | Either 679 or 680 (since \(g = 9.8\) used) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Resolve horizontally | M1 | Using appropriate strategy to set up first equation, usual rules applying; e.g. Resolve horiz. or M(\(C\)) |
| \(X = T\cos 60°\) | A1 | Correct equation in \(X\) only |
| Resolve vertically | M1 | Using appropriate strategy to set up second equation, usual rules applying; e.g. Resolve vert. or M(\(D\)) |
| \(Y = T\cos 30° - 20g\) | A1 | Correct equation in \(Y\) only |
| Use of \(\tan\theta = \dfrac{Y}{X}\) and sub for \(T\) | M1 | Using the model and their \(X\) and \(Y\) |
| \(49°\) (or better), below horizontal, away from wall | A1 | \(49°\) or better (since \(g\) cancels); need all three parts; or any appropriate angle e.g. \(41°\) to wall, downwards and away from wall |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Tension would increase as you move from \(D\) to \(C\) | B1 | Appropriate equivalent comment |
| Since each point of the rope has to support the length of rope below it | B1 | Appropriate equivalent reason |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Take moments about \(G\), \(1.5Y = 0\) | M1 | Using the model and any other complete method e.g. the three force condition for equilibrium |
| \(Y = 0\) hence force acts horizontally. | A1* | Correct conclusion — given answer |
## Question 4(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Moments about $A$ (or any other complete method) | M1 | Correct overall strategy e.g. M($A$), with usual rules, to give equation in $T$ only |
| $T\cos 30° \times (1\sin 30°) = 20g \times 1.5$ | A1 | (A1A0 one error) Condone 1 error |
| $T\cos 30° \times (1\sin 30°) = 20g \times 1.5$ | A1 | (A0A0 two or more errors) |
| $T = 679$ or $680\ \text{(N)}$ | A1 | Either 679 or 680 (since $g = 9.8$ used) |
**Total: 4 marks**
## Question 4(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Resolve horizontally | M1 | Using appropriate strategy to set up first equation, usual rules applying; e.g. Resolve horiz. or M($C$) |
| $X = T\cos 60°$ | A1 | Correct equation in $X$ only |
| Resolve vertically | M1 | Using appropriate strategy to set up second equation, usual rules applying; e.g. Resolve vert. or M($D$) |
| $Y = T\cos 30° - 20g$ | A1 | Correct equation in $Y$ only |
| Use of $\tan\theta = \dfrac{Y}{X}$ and sub for $T$ | M1 | Using the model and their $X$ and $Y$ |
| $49°$ (or better), below horizontal, away from wall | A1 | $49°$ or better (since $g$ cancels); need all three parts; or any appropriate angle e.g. $41°$ to wall, downwards and away from wall |
**Total: 6 marks**
## Question 4(c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Tension would increase as you move from $D$ to $C$ | B1 | Appropriate equivalent comment |
| Since each point of the rope has to support the length of rope below it | B1 | Appropriate equivalent reason |
**Total: 2 marks**
## Question 4(d):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Take moments about $G$, $1.5Y = 0$ | M1 | Using the model and any other complete method e.g. the three force condition for equilibrium |
| $Y = 0$ hence force acts horizontally. | A1* | Correct conclusion — given answer |
**Total: 2 marks**4. The Venn diagram shows the probabilities of students' lunch boxes containing a drink, sandwiches and a chocolate bar.\\
\includegraphics[max width=\textwidth, alt={}, center]{565bfa73-8095-4242-80b6-cd47aaff6a31-05_655_899_392_484}\\
$D$ is the event that a lunch box contains a drink, $S$ is the event that a lunch box contains sandwiches,\\
$C$ is the event that a lunch box contains a chocolate bar, $u , v$ and $w$ are probabilities.
\begin{enumerate}[label=(\alph*)]
\item Write down $\mathrm { P } \left( S \cap D ^ { \prime } \right)$.
One day, 80 students each bring in a lunch box.\\
Given that all 80 lunch boxes contain sandwiches and a drink,
\item estimate how many of these 80 lunch boxes will contain a chocolate bar.
Given that the events $S$ and $C$ are independent and that $\mathrm { P } ( D \mid C ) = \frac { 14 } { 15 }$,
\item calculate the value of $u$, the value of $v$ and the value of $w$.\\
(7)\\
(Total 11 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 3 Q4 [11]}}