| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2021 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Independent Events |
| Type | Venn diagram with independence constraint |
| Difficulty | Challenging +1.2 This S1 question requires understanding of independence (P(A∩C) = P(A)P(C)), mutual exclusivity from Venn diagrams, and solving simultaneous equations from two independence conditions. While it involves multiple steps and careful algebraic manipulation, the concepts are standard S1 material with no novel insight required—harder than routine probability but well within typical A-level scope. |
| 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 |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(B\) and \(C\) | B1 | \(B\) and \(C\) seen. If they include \(A\) then B0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(A\) and \(C\) independent gives: \(P(C)\times 0.65 = 0.13\) or \(0.65\times(r+0.13)=0.13\) or \(0.65\times(0.48-s)=0.13\) | M1 | For a correct equation for \(P(C)\) using independence |
| \(P(C) = 0.2\) or \(r + 0.13 = 0.2\) or \(0.48 - s = 0.2\) | A1 | For \(P(C) = 0.2\), correct linear equation for \(r\) or \(s\) |
| \(r = 0.07\) or \(s = 0.28\) | A1 | For either \(r = 0.07\) or \(s = 0.28\) |
| \(P(A) + r + s = 1\) or \(0.65 +\) "0.07" \(+ s = 1\) or \(0.65 +\) "0.28" \(+ r = 1\) | M1 | For using \(\sum p = 1\). Allow letter \(r\) and \(s\) or their values provided they are probabilities |
| \(s = 0.28\) and \(r = 0.07\) | A1 | For both \(s = 0.28\) and \(r = 0.07\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P[(B\cup C)] =\) "0.2" \(+ q\) or \(0.13 +\) "0.07" \(+ q\) | B1ft | For an expression (in \(q\)) for \(P(B\cup C)\) ft their value of \(r\) or their "0.2" eg \(0.13 +\) "their \(r\)" \(+ q\) |
| \(P(A\cap C') = p + q\ \{= 0.52\}\) | B1 | For a correct expression for \(P(A\cap C')\) in terms of \(p\) and \(q\) or 0.52. Implied by 1st or 2nd M1 below |
| \(\{P[(A\cap C')\cap(B\cup C)]=q\Rightarrow\}\) "\((p+q)\)"\(\times\)"\((0.2+q)\)"\(= q\) or "\((p+q)\)"\(\times\)"\((0.13+\)"0.07"\(+q)\)"\(= q\) or "\((p+q)\)"\(\times\)"\((1-s-p)\)"\(= 0.52-p\) | M1 | For a correct use of independence (ft their probabilities), values or letters. Implied by 2nd M1 |
| \([\text{Using } p+q=0.52]\ \ 0.52\times\)"\((0.2+q)\)"\(= q\) or \(0.52(0.72-p) = 0.52-p\) | M1 | Using \(p + q = 0.52\) to gain a linear equation in one variable |
| \(q = \frac{13}{60}\) | A1 | For a correct fraction for \(q\) |
| \(p = \frac{91}{300}\) | A1 | For a correct fraction for \(p\). SC: If both \(p\) and \(q\) given as equivalent recurring decimals award A0A1 eg \(0.2\dot{1}\dot{6}\) and \(0.30\dot{3}\) |
# Question 2:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $B$ and $C$ | B1 | $B$ and $C$ seen. If they include $A$ then B0 |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $A$ and $C$ independent gives: $P(C)\times 0.65 = 0.13$ **or** $0.65\times(r+0.13)=0.13$ **or** $0.65\times(0.48-s)=0.13$ | M1 | For a correct equation for $P(C)$ using independence |
| $P(C) = 0.2$ **or** $r + 0.13 = 0.2$ **or** $0.48 - s = 0.2$ | A1 | For $P(C) = 0.2$, correct linear equation for $r$ or $s$ |
| $r = 0.07$ **or** $s = 0.28$ | A1 | For either $r = 0.07$ or $s = 0.28$ |
| $P(A) + r + s = 1$ **or** $0.65 +$ "0.07" $+ s = 1$ **or** $0.65 +$ "0.28" $+ r = 1$ | M1 | For using $\sum p = 1$. Allow letter $r$ and $s$ or their values provided they are probabilities |
| $s = 0.28$ and $r = 0.07$ | A1 | For both $s = 0.28$ and $r = 0.07$ |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P[(B\cup C)] =$ "0.2" $+ q$ or $0.13 +$ "0.07" $+ q$ | B1ft | For an expression (in $q$) for $P(B\cup C)$ ft their value of $r$ or their "0.2" eg $0.13 +$ "their $r$" $+ q$ |
| $P(A\cap C') = p + q\ \{= 0.52\}$ | B1 | For a correct expression for $P(A\cap C')$ in terms of $p$ and $q$ or 0.52. Implied by 1st or 2nd M1 below |
| $\{P[(A\cap C')\cap(B\cup C)]=q\Rightarrow\}$ "$(p+q)$"$\times$"$(0.2+q)$"$= q$ **or** "$(p+q)$"$\times$"$(0.13+$"0.07"$+q)$"$= q$ **or** "$(p+q)$"$\times$"$(1-s-p)$"$= 0.52-p$ | M1 | For a correct use of independence (ft their probabilities), values or letters. Implied by 2nd M1 |
| $[\text{Using } p+q=0.52]\ \ 0.52\times$"$(0.2+q)$"$= q$ or $0.52(0.72-p) = 0.52-p$ | M1 | Using $p + q = 0.52$ to gain a linear equation in one variable |
| $q = \frac{13}{60}$ | A1 | For a correct fraction for $q$ |
| $p = \frac{91}{300}$ | A1 | For a correct fraction for $p$. SC: If both $p$ and $q$ given as equivalent recurring decimals award A0A1 eg $0.2\dot{1}\dot{6}$ and $0.30\dot{3}$ |
---2. In the Venn diagram below, $A , B$ and $C$ are events and $p , q , r$ and $s$ are probabilities.
The events $A$ and $C$ are independent and $\mathrm { P } ( A ) = 0.65$\\
\includegraphics[max width=\textwidth, alt={}, center]{a439724e-b570-434d-bf75-de2b50915042-04_373_815_397_568}
\begin{enumerate}[label=(\alph*)]
\item State which two of the events $A$, $B$ and $C$ are mutually exclusive.
\item Find the value of $r$ and the value of $s$.
The events ( $A \cap C ^ { \prime }$ ) and ( $B \cup C$ ) are also independent.
\item Find the exact value of $p$ and the exact value of $q$. Give your answers as fractions.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2021 Q2 [12]}}