Edexcel S1 2023 January — Question 4 13 marks

Exam BoardEdexcel
ModuleS1 (Statistics 1)
Year2023
SessionJanuary
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPrinciple of Inclusion/Exclusion
TypeFinding Unknown Probabilities in Venn Diagrams
DifficultyStandard +0.3 This is a straightforward S1 question requiring basic Venn diagram algebra and probability inequalities. Part (i) involves setting up simple linear equations from given probabilities (standard textbook exercise), and part (ii) uses AM-GM inequality or simple algebra to show P(C)+P(D)>1. All steps are routine with no novel insight required, making it slightly easier than average.
Spec2.03a 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

    1. In the Venn diagram below, \(A\) and \(B\) represent events and \(p , q , r\) and \(s\) are probabilities. \includegraphics[max width=\textwidth, alt={}, center]{c316fa29-dedc-4890-bd82-31eb0bb819f9-12_400_789_347_639}
$$\mathrm { P } ( A ) = \frac { 7 } { 25 } \quad \mathrm { P } ( B ) = \frac { 1 } { 5 } \quad \mathrm { P } \left[ \left( A \cap B ^ { \prime } \right) \cup \left( A ^ { \prime } \cap B \right) \right] = \frac { 8 } { 25 }$$
  1. Use algebra to show that \(2 p + 2 q + 2 r = \frac { 4 } { 5 }\)
  2. Find the value of \(p\), the value of \(q\), the value of \(r\) and the value of \(s\) (ii) Two events, \(C\) and \(D\), are such that $$\mathrm { P } ( C ) = \frac { x } { x + 5 } \quad \mathrm { P } ( D ) = \frac { 5 } { x }$$ where \(x\) is a positive constant.
    By considering \(\mathrm { P } ( C ) + \mathrm { P } ( D )\) show that \(C\) and \(D\) cannot be mutually exclusive.

Question 4:
Part (i)(a)
AnswerMarks Guidance
AnswerMark Guidance
\(p + q = \frac{7}{25}\) oeM1 For \(p+q = \frac{7}{25}\) oe or \(p+q = P(A)\)
\(q + r = \frac{1}{5}\) oeM1 For \(q+r = \frac{1}{5}\) oe or \(q+r = P(B)\)
\(p + r = \frac{8}{25}\) oeM1 For \(p+r = \frac{8}{25}\) oe
\(2p+2q+2r = \frac{7}{25}+\frac{1}{5}+\frac{8}{25} = \frac{4}{5}\)A1* Must see \(2p+2q+2r = \frac{7}{25}+\frac{1}{5}+\frac{8}{25}\) with no errors
Part (i)(b)
AnswerMarks Guidance
AnswerMark Guidance
\(p+q+r+s = 1\)M1 Any correct equation involving at least two of \(p,q,r,s\)
\(p = \frac{1}{5}\)A1 or 0.2
\(q = \frac{2}{25}\)A1 or 0.08
\(r = \frac{3}{25}\)A1 or 0.12
\(s = \frac{3}{5}\)A1 oe
Part (ii)
AnswerMarks Guidance
AnswerMark Guidance
\(\frac{x}{x+5}+\frac{5}{x} = \frac{x^2+5(x+5)}{x(x+5)}\) or \(\frac{x}{x+5}+\frac{5}{x} = \frac{x+5-5}{x+5}+\frac{5}{x}\)M1 Attempt to add \(P(C)\) and \(P(D)\)
\(= \frac{x^2+5x+25}{x^2+5x}\) or \(= 1 - \frac{5}{x+5}+\frac{5}{x}\)M1 For \(\frac{x^2+5x+25}{x^2+5x}\) oe or \(1-\frac{5}{x+5}+\frac{5}{x}\)
\(= 1 + \frac{25}{x^2+5x}\) or as \(x^2+5x+25 > x^2+5x\) so \(P(C)+P(D)>1\)A1 For recognising \(P(C)+P(D) > 1\)
\(P(C \cup D) > 1\) or \(P(C \cap D) > 0\)A1 cso Fully correct solution showing \(C\) and \(D\) cannot be mutually exclusive
# Question 4:

## Part (i)(a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $p + q = \frac{7}{25}$ oe | M1 | For $p+q = \frac{7}{25}$ oe or $p+q = P(A)$ |
| $q + r = \frac{1}{5}$ oe | M1 | For $q+r = \frac{1}{5}$ oe or $q+r = P(B)$ |
| $p + r = \frac{8}{25}$ oe | M1 | For $p+r = \frac{8}{25}$ oe |
| $2p+2q+2r = \frac{7}{25}+\frac{1}{5}+\frac{8}{25} = \frac{4}{5}$ | A1* | Must see $2p+2q+2r = \frac{7}{25}+\frac{1}{5}+\frac{8}{25}$ with no errors |

## Part (i)(b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $p+q+r+s = 1$ | M1 | Any correct equation involving at least two of $p,q,r,s$ |
| $p = \frac{1}{5}$ | A1 | or 0.2 |
| $q = \frac{2}{25}$ | A1 | or 0.08 |
| $r = \frac{3}{25}$ | A1 | or 0.12 |
| $s = \frac{3}{5}$ | A1 | oe |

## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{x}{x+5}+\frac{5}{x} = \frac{x^2+5(x+5)}{x(x+5)}$ or $\frac{x}{x+5}+\frac{5}{x} = \frac{x+5-5}{x+5}+\frac{5}{x}$ | M1 | Attempt to add $P(C)$ and $P(D)$ |
| $= \frac{x^2+5x+25}{x^2+5x}$ or $= 1 - \frac{5}{x+5}+\frac{5}{x}$ | M1 | For $\frac{x^2+5x+25}{x^2+5x}$ oe or $1-\frac{5}{x+5}+\frac{5}{x}$ |
| $= 1 + \frac{25}{x^2+5x}$ or as $x^2+5x+25 > x^2+5x$ so $P(C)+P(D)>1$ | A1 | For recognising $P(C)+P(D) > 1$ |
| $P(C \cup D) > 1$ or $P(C \cap D) > 0$ | A1 cso | Fully correct solution showing $C$ and $D$ cannot be mutually exclusive |

---
\begin{enumerate}
  \item (i) In the Venn diagram below, $A$ and $B$ represent events and $p , q , r$ and $s$ are probabilities.\\
\includegraphics[max width=\textwidth, alt={}, center]{c316fa29-dedc-4890-bd82-31eb0bb819f9-12_400_789_347_639}
\end{enumerate}

$$\mathrm { P } ( A ) = \frac { 7 } { 25 } \quad \mathrm { P } ( B ) = \frac { 1 } { 5 } \quad \mathrm { P } \left[ \left( A \cap B ^ { \prime } \right) \cup \left( A ^ { \prime } \cap B \right) \right] = \frac { 8 } { 25 }$$

(a) Use algebra to show that $2 p + 2 q + 2 r = \frac { 4 } { 5 }$\\
(b) Find the value of $p$, the value of $q$, the value of $r$ and the value of $s$\\
(ii) Two events, $C$ and $D$, are such that

$$\mathrm { P } ( C ) = \frac { x } { x + 5 } \quad \mathrm { P } ( D ) = \frac { 5 } { x }$$

where $x$ is a positive constant.\\
By considering $\mathrm { P } ( C ) + \mathrm { P } ( D )$ show that $C$ and $D$ cannot be mutually exclusive.

\hfill \mbox{\textit{Edexcel S1 2023 Q4 [13]}}