| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2022 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Independent Events |
| Type | Find unknown probability given independence |
| Difficulty | Moderate -0.3 This is a straightforward application of standard probability formulas (independence, union rule, conditional probability, mutual exclusivity) with no conceptual tricks. Part (a) requires solving a simple equation using P(H∪W) = P(H) + P(W) - P(H)P(W), part (b) is direct conditional probability, and part (c) is routine Venn diagram construction. Slightly easier than average due to being purely procedural with clear signposting. |
| 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 |
| \(P(H \cup W) = P(H) + P(W) - P(H \cap W)\) or \(P(H' \cap W) = P(H \cup W) - P(H)\) | M1 | Use of addition formula with at least one value correctly substituted |
| \(P(H \cap W) = \dfrac{3}{8} \times P(W)\) or \(P(H' \cap W) = P(H') \times P(W)\) | M1 | Use of independence |
| \(\dfrac{3}{4} = \dfrac{3}{8} + P(W) - \dfrac{3}{8}P(W)\) or \(\dfrac{3}{8} = \dfrac{5}{8}P(W)\) | A1 | Correct equation in \(P(W)\) |
| \(P(W) = \dfrac{3}{5}\) | A1cso* | Correct solution ending with \(P(W) = \frac{3}{5}\) with no wrong working. Dep on all previous marks. NB: method using \(\frac{3}{5}\) or \(\frac{3}{5} \times \frac{3}{8} \left[= \frac{9}{40}\right]\) scores max M1M1A0A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(N' \mid H) = \dfrac{\frac{3}{8} - \frac{1}{15}}{\frac{3}{8}}\) or \(\dfrac{\frac{9}{40} + \frac{1}{12}}{\frac{3}{8}}\) or \(1 - \dfrac{\frac{1}{15}}{\frac{3}{8}}\) | M1 | For \(\dfrac{p}{\frac{3}{8}}\) where \(0 < p < \frac{3}{8}\). Use of independence is M0 e.g. \(\dfrac{x \times \frac{3}{8}}{\frac{3}{8}}\) |
| \(= \dfrac{37}{45}\) = awrt 0.822 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| 3 circles labelled with \(N\) inside \(H\) | B1 | Either \(N\) inside \(H\) or intersecting circles with \(P(N \cap H') = 0\). Condone missing box. Allow all 3 circles overlapping with all zeros correctly labelled |
| \(P(H \cap W) = \dfrac{3}{8} \times \dfrac{3}{5} \left[= \dfrac{9}{40}\right]\) seen or correctly placed | M1 | |
| \(\dfrac{3}{5} - \dfrac{9}{40} = \dfrac{3}{8}\) (regions in \(P(W)\) adding to 0.6) | M1 | |
| \(\dfrac{3}{8} - \dfrac{9}{40} - \dfrac{1}{15} = \dfrac{1}{12}\) | M1 | |
| Fully correct diagram with \(\frac{1}{4}\) and box and correct probabilities | A1 | Allow exact decimal equivalents |
# Question 4:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(H \cup W) = P(H) + P(W) - P(H \cap W)$ **or** $P(H' \cap W) = P(H \cup W) - P(H)$ | M1 | Use of addition formula with at least one value correctly substituted |
| $P(H \cap W) = \dfrac{3}{8} \times P(W)$ **or** $P(H' \cap W) = P(H') \times P(W)$ | M1 | Use of independence |
| $\dfrac{3}{4} = \dfrac{3}{8} + P(W) - \dfrac{3}{8}P(W)$ **or** $\dfrac{3}{8} = \dfrac{5}{8}P(W)$ | A1 | Correct equation in $P(W)$ |
| $P(W) = \dfrac{3}{5}$ | A1cso* | Correct solution ending with $P(W) = \frac{3}{5}$ with no wrong working. Dep on all previous marks. NB: method using $\frac{3}{5}$ or $\frac{3}{5} \times \frac{3}{8} \left[= \frac{9}{40}\right]$ scores max M1M1A0A0 |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(N' \mid H) = \dfrac{\frac{3}{8} - \frac{1}{15}}{\frac{3}{8}}$ or $\dfrac{\frac{9}{40} + \frac{1}{12}}{\frac{3}{8}}$ or $1 - \dfrac{\frac{1}{15}}{\frac{3}{8}}$ | M1 | For $\dfrac{p}{\frac{3}{8}}$ where $0 < p < \frac{3}{8}$. Use of independence is M0 e.g. $\dfrac{x \times \frac{3}{8}}{\frac{3}{8}}$ |
| $= \dfrac{37}{45}$ = awrt 0.822 | A1 | |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| 3 circles labelled with $N$ inside $H$ | B1 | Either $N$ inside $H$ or intersecting circles with $P(N \cap H') = 0$. Condone missing box. Allow all 3 circles overlapping with all zeros correctly labelled |
| $P(H \cap W) = \dfrac{3}{8} \times \dfrac{3}{5} \left[= \dfrac{9}{40}\right]$ seen or correctly placed | M1 | |
| $\dfrac{3}{5} - \dfrac{9}{40} = \dfrac{3}{8}$ (regions in $P(W)$ adding to 0.6) | M1 | |
| $\dfrac{3}{8} - \dfrac{9}{40} - \dfrac{1}{15} = \dfrac{1}{12}$ | M1 | |
| Fully correct diagram with $\frac{1}{4}$ and box and correct probabilities | A1 | Allow exact decimal equivalents |
---\begin{enumerate}
\item The events $H$ and $W$ are such that
\end{enumerate}
$$\mathrm { P } ( H ) = \frac { 3 } { 8 } \quad \mathrm { P } ( H \cup W ) = \frac { 3 } { 4 }$$
Given that $H$ and $W$ are independent,\\
(a) show that $\mathrm { P } ( W ) = \frac { 3 } { 5 }$
The event $N$ is such that
$$\mathrm { P } ( N ) = \frac { 1 } { 15 } \quad \mathrm { P } ( H \cap N ) = \mathrm { P } ( N )$$
(b) Find $\mathrm { P } \left( N ^ { \prime } \mid H \right)$
Given that $W$ and $N$ are mutually exclusive,\\
(c) draw a Venn diagram to represent the events $H , W$ and $N$ giving the exact probabilities of each region in the Venn diagram.
\hfill \mbox{\textit{Edexcel S1 2022 Q4 [11]}}