| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2023 |
| Session | October |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Definitions |
| Type | Venn diagram completion |
| Difficulty | Easy -1.2 This is a routine S1 question testing basic probability definitions and rules. Part (i) involves straightforward Venn diagram completion using given probabilities (simple arithmetic), testing independence (multiply probabilities), and finding union probabilities. Part (ii) applies standard formulas for mutually exclusive events and independence. All parts require direct application of memorized rules with minimal problem-solving or insight—easier than average A-level questions. |
| 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 | Mark | Guidance |
| \(w = 0.15\) | B1 | If answer given in script and Venn diagram, mark the script |
| \(x = 0.7 - 0.15 = 0.55\) | B1 | If answer given in script and Venn diagram, mark the script |
| \(y = 0.65 - 0.55 = 0.1\) | B1 | If answer given in script and Venn diagram, mark the script |
| \(z = 1 - 0.15 - 0.55 - 0.1 = 0.2\) | B1 | If answer given in script and Venn diagram, mark the script |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(0.15 + 0.1 = 0.25\) | B1ft | For \(w + y = 0.25\), follow through their \(w\) and \(y\) provided this is a probability |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([P(C)\times P(O)] = 0.65 \times 0.7 \neq 0.55 [= P(C\cap O)]\) or \([P(C | O)] = \frac{0.55}{0.7} \neq 0.65 [= P(C)]\) | M1 |
| \(0.455 \neq 0.55\) or \(0.7857... \neq 0.65\) [So not independent] | A1* | Fully correct solution with values evaluated and no errors ft their \(w\), \(x\), \(y\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(F\cup H) = \frac{2}{7} + \frac{1}{4} = \frac{15}{28}\) | B1 | For \(\frac{15}{28}\) oe, allow awrt 0.536 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{5}{8} = \frac{2}{7} + P(G) - \frac{2}{7}P(G)\) | M1 | For use of \(P(F\cup G) = P(F) + P(G) - P(F)\times P(G)\) |
| \(P(G) = \frac{\frac{5}{8} - \frac{2}{7}}{1 - \frac{2}{7}} = \frac{19}{56} \div \frac{5}{7}\) | dM1 | Dependent on M1. For correct rearrangement to find \(P(G)\) e.g. \(\left(\frac{5}{8}-\frac{2}{7}\right)\div\left(1-\frac{2}{7}\right)\); allow \(\frac{19}{56} = \frac{5}{7}P(G)\), may be implied by \(\frac{19}{40}\) |
| \(P(G) = \frac{19}{40}\) | A1 | For \(\frac{19}{40}\) oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(F\cap G) = \frac{2}{7}\times\frac{19}{40} = \frac{19}{140}\) | B1ft | For \(\frac{19}{140}\) oe or \(\frac{2}{7}\times P(G)\) evaluated correctly where \(P(G)\) is a probability |
# Question 3:
## Part (i)(a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $w = 0.15$ | B1 | If answer given in script and Venn diagram, mark the script |
| $x = 0.7 - 0.15 = 0.55$ | B1 | If answer given in script and Venn diagram, mark the script |
| $y = 0.65 - 0.55 = 0.1$ | B1 | If answer given in script and Venn diagram, mark the script |
| $z = 1 - 0.15 - 0.55 - 0.1 = 0.2$ | B1 | If answer given in script and Venn diagram, mark the script |
## Part (i)(b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $0.15 + 0.1 = 0.25$ | B1ft | For $w + y = 0.25$, follow through their $w$ and $y$ provided this is a probability |
## Part (i)(c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $[P(C)\times P(O)] = 0.65 \times 0.7 \neq 0.55 [= P(C\cap O)]$ or $[P(C|O)] = \frac{0.55}{0.7} \neq 0.65 [= P(C)]$ | M1 | For $'(x+y)'\times'(w+x)' \neq 'x'$ or $\frac{'x'}{' w+x'} \neq 'x+y'$ ft their $w$, $x$, $y$ |
| $0.455 \neq 0.55$ or $0.7857... \neq 0.65$ [So not independent] | A1* | Fully correct solution with values evaluated and no errors ft their $w$, $x$, $y$ |
## Part (ii)(a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(F\cup H) = \frac{2}{7} + \frac{1}{4} = \frac{15}{28}$ | B1 | For $\frac{15}{28}$ oe, allow awrt 0.536 |
## Part (ii)(b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{5}{8} = \frac{2}{7} + P(G) - \frac{2}{7}P(G)$ | M1 | For use of $P(F\cup G) = P(F) + P(G) - P(F)\times P(G)$ |
| $P(G) = \frac{\frac{5}{8} - \frac{2}{7}}{1 - \frac{2}{7}} = \frac{19}{56} \div \frac{5}{7}$ | dM1 | Dependent on M1. For correct rearrangement to find $P(G)$ e.g. $\left(\frac{5}{8}-\frac{2}{7}\right)\div\left(1-\frac{2}{7}\right)$; allow $\frac{19}{56} = \frac{5}{7}P(G)$, may be implied by $\frac{19}{40}$ |
| $P(G) = \frac{19}{40}$ | A1 | For $\frac{19}{40}$ oe |
## Part (ii)(c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(F\cap G) = \frac{2}{7}\times\frac{19}{40} = \frac{19}{140}$ | B1ft | For $\frac{19}{140}$ oe or $\frac{2}{7}\times P(G)$ evaluated correctly where $P(G)$ is a probability |
---\begin{enumerate}
\item (i) Bob shops at a market each week. The event that
\end{enumerate}
Bob buys carrots is denoted by $C$\\
Bob buys onions is denoted by $O$\\
At each visit, Bob may buy neither, or one, or both of these items. The probability that
Bob buys carrots is 0.65\\
Bob does not buy onions is 0.3\\
Bob buys onions but not carrots is 0.15\\
The Venn diagram below represents the events $C$ and $O$\\
\begin{tikzpicture}
% Outer rectangle
\draw[thick] (-4,-2.5) rectangle (4,2.5);
% Left circle (O)
\draw[thick] (-1.2,0) circle (1.8);
\node[above] at (-1.2,1.8) {\textit{O}};
% Right circle (C)
\draw[thick] (1.2,0) circle (1.8);
\node[above] at (1.2,1.8) {\textit{C}};
% Region labels
\node at (-2.0,0) {\textit{w}};
\node at (0,0) {\textit{x}};
\node at (2.0,0) {\textit{y}};
\node at (3.2,-1.8) {\textit{z}};
\end{tikzpicture}\\
where $w , x , y$ and $z$ are probabilities.\\
(a) Find the value of $w$, the value of $x$, the value of $y$ and the value of $z$
For one visit to the market,\\
(b) find the probability that Bob buys either carrots or onions but not both.\\
(c) Show that the events $C$ and $O$ are not independent.\\
(ii) $F , G$ and $H$ are 3 events. $F$ and $H$ are mutually exclusive. $F$ and $G$ are independent. Given that
$$\mathrm { P } ( F ) = \frac { 2 } { 7 } \quad \mathrm { P } ( H ) = \frac { 1 } { 4 } \quad \mathrm { P } ( F \cup G ) = \frac { 5 } { 8 }$$
(a) find $\operatorname { P } ( F \cup H )$\\
(b) find $\mathrm { P } ( G )$\\
(c) find $\operatorname { P } ( F \cap G )$
\hfill \mbox{\textit{Edexcel S1 2023 Q3 [12]}}