| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Definitions |
| Type | Combined event algebra |
| Difficulty | Standard +0.3 This is a standard S1 probability question requiring systematic application of probability rules (addition rule, conditional probability, mutual exclusivity) and simultaneous equations. While it has multiple parts and requires careful algebraic manipulation, all techniques are routine for A-level—no novel insight or complex problem-solving is needed, making it slightly easier than average. |
| Spec | 2.03a Mutually exclusive and independent events2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(B | A) = \dfrac{P(B\cap A)}{P(A)}\) | |
| \(0.3 = \dfrac{P(B\cap A)}{x} \Rightarrow P(B\cap A) = 0.3x\) | M1 | Use of \(P(B |
| \(P(A\cup B) = P(A)+P(B)-P(A\cap B)\) | M1 | Use of \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\) with substitution; the equation may be seen in a Venn diagram; \(0.65=x+y-0.3x\) implies M1M1 |
| \(0.65 = x+y-0.3x \Rightarrow 0.65 = 0.7x+y\) | ||
| \(14x + 20y = 13\) | A1* | Answer is given so no incorrect working can be seen |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(B\cup C) = P(B)+P(C)\) or \(P(B\cap C)=0\) | M1 | Use of \(P(B\cup C)=P(B)+P(C)\) or sight of \(P(B\cap C)=0\) |
| \(0.85 = \frac{1}{2}x + 2y\) | A1 | Any correct second equation in \(x\) and \(y\) which need not be simplified |
| Answer | Marks | Guidance |
|---|---|---|
| Attempt to solve the 2 equations simultaneously | M1 | Either a correct substitution seen or a correct method to eliminate \(x\) or \(y\) |
| \(x=0.5\), \(y=0.3\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(B | A)=0.3\) and \(P(B)=0.3\) | M1 |
| \(P(A)\times P(B) = 0.5\times 0.3\) and \(P(A\cap B)=0.3\times 0.5\) or \(P(A\cap B)=0.5+0.3-0.65\) | ||
| So \(A\) and \(B\) are statistically independent | A1ft | For correct ft conclusion for their values of \(x\) and \(y\) (must have scored M1) |
## Question 6:
### Part (a):
$P(B|A) = \dfrac{P(B\cap A)}{P(A)}$ | |
$0.3 = \dfrac{P(B\cap A)}{x} \Rightarrow P(B\cap A) = 0.3x$ | M1 | Use of $P(B|A)=\frac{P(B\cap A)}{P(A)}$; assuming independence is M0; may be implied by $P(B\cap A)=0.3x$
$P(A\cup B) = P(A)+P(B)-P(A\cap B)$ | M1 | Use of $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ with substitution; the equation may be seen in a Venn diagram; $0.65=x+y-0.3x$ implies M1M1
$0.65 = x+y-0.3x \Rightarrow 0.65 = 0.7x+y$ | |
$14x + 20y = 13$ | A1* | Answer is given so no incorrect working can be seen
### Part (b)(i):
$P(B\cup C) = P(B)+P(C)$ or $P(B\cap C)=0$ | M1 | Use of $P(B\cup C)=P(B)+P(C)$ or sight of $P(B\cap C)=0$
$0.85 = \frac{1}{2}x + 2y$ | A1 | Any correct second equation in $x$ and $y$ which need not be simplified
### Part (b)(ii):
Attempt to solve the 2 equations simultaneously | M1 | Either a correct substitution seen or a correct method to eliminate $x$ or $y$
$x=0.5$, $y=0.3$ | A1 |
### Part (c):
$P(B|A)=0.3$ and $P(B)=0.3$ | M1 | Finding all of the probabilities needed for a test for independence (probabilities must be labelled) ft their values of $x$ and $y$; $P(B|A)$ and $P(B)$ or $P(A)$, $P(B)$ and $P(A\cap B)$; for $P(A\cap B)$ we **must** see working shown
$P(A)\times P(B) = 0.5\times 0.3$ and $P(A\cap B)=0.3\times 0.5$ or $P(A\cap B)=0.5+0.3-0.65$ | |
So $A$ and $B$ are statistically independent | A1ft | For correct ft conclusion for their values of $x$ and $y$ (must have scored M1)\begin{enumerate}
\item The events $A$ and $B$ satisfy
\end{enumerate}
$$\mathrm { P } ( A ) = x \quad \mathrm { P } ( B ) = y \quad \mathrm { P } ( A \cup B ) = 0.65 \quad \mathrm { P } ( B \mid A ) = 0.3$$
(a) Show that
$$14 x + 20 y = 13$$
The events $B$ and $C$ are mutually exclusive such that
$$\mathrm { P } ( B \cup C ) = 0.85 \quad \mathrm { P } ( C ) = \frac { 1 } { 2 } x + y$$
(b) (i) Find a second equation in $x$ and $y$\\
(ii) Hence find the value of $x$ and the value of $y$\\
(c) Determine whether or not $A$ and $B$ are statistically independent. You must show your working clearly.
\hfill \mbox{\textit{Edexcel S1 2024 Q6 [9]}}