| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2017 |
| Session | October |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Tree diagram with two-stage events |
| Difficulty | Moderate -0.8 This is a standard S1 tree diagram question requiring basic probability rules (law of total probability for part a, conditional probability formula for part b). The question involves straightforward algebraic manipulation with given probabilities and no conceptual challenges beyond textbook exercises. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | [Let \(P(A) = p\)] | M1A1 |
| \(0.4p + 0.7(1-p) = 0.45\) | M1 | 1st A1 for a fully correct equation for \(p\) |
| \(0.25 = 0.3p\) | M1 | 1st M1 for attempt at 2 sim' eq'ns in p and \(q\) Allow one error. \(0.4p + 0.7q = \frac{9}{20}\) and \(0.6p + 0.3q = \frac{11}{20}\) |
| \(p = \frac{5}{6}\) | A1 | 1st A1 for any correct equation in \(p\) or \(q\) 2nd M1 for simplifying their linear equation with at least 2 terms in \(p\) or \(q\) to \(a = bp\) or \(bq\) 2nd A1 for \(P(A) = \frac{5}{6}\) or exact equiv e.g. 0.83 (may be seen on their tree diagram) |
| [Probability tree diagram with branches] | B1ft | B1ft for 1st 2 branches i.e. \(\frac{5}{6}\) and \(\frac{1}{6}\) (follow through their \(P(A)\)) |
| B1 | 2nd B1 for 2nd 4 branches i.e. \(\frac{3}{5}\) and \(\frac{10}{}\) | |
| (b) | \(\left[P(A' \mid B')\right] = \frac{\frac{1}{6} \times 0.3}{0.55}\) | M1 |
| \(= \frac{1}{11}\) | A1 | A1 for \(\frac{1}{11}\) or exact equivalent e.g. 0.09 |
| Answer | Marks |
|---|---|
| 1st M1 | for attempt at 2 sim' eq'ns in p and \(q\) Allow one error. \(0.4p + 0.7q = \frac{9}{20}\) and \(0.6p + 0.3q = \frac{11}{20}\) |
| 1st A1 | for any correct equation in \(p\) or \(q\) |
| 2nd M1 | for simplifying their linear equation with at least 2 terms in \(p\) or \(q\) to \(a = bp\) or \(bq\) |
| 2nd A1 | for \(P(A) = \frac{5}{6}\) or exact equiv e.g. 0.83 (may be seen on their tree diagram) |
| 1st B1ft | for 1st 2 branches i.e. \(\frac{5}{6}\) and \(\frac{1}{6}\) (follow through their \(P(A)\)) |
| 2nd B1 | for 2nd 4 branches i.e. \(\frac{3}{5}\) and \(\frac{10}{}\) |
| **(a)** | [Let $P(A) = p$] | M1A1 | 1st M1 for $0.4p$ or $0.7(1-p)$ seen in an equation for $p$ |
| | $0.4p + 0.7(1-p) = 0.45$ | M1 | 1st A1 for a fully correct equation for $p$ |
| | $0.25 = 0.3p$ | M1 | 1st M1 for attempt at 2 sim' eq'ns in p and $q$ Allow one error. $0.4p + 0.7q = \frac{9}{20}$ and $0.6p + 0.3q = \frac{11}{20}$ |
| | $p = \frac{5}{6}$ | A1 | 1st A1 for any correct equation in $p$ or $q$ 2nd M1 for simplifying their linear equation with at least 2 terms in $p$ or $q$ to $a = bp$ or $bq$ 2nd A1 for $P(A) = \frac{5}{6}$ or exact equiv e.g. 0.83 (may be seen on their tree diagram) |
| | [Probability tree diagram with branches] | B1ft | B1ft for 1st 2 branches i.e. $\frac{5}{6}$ and $\frac{1}{6}$ (follow through their $P(A)$) |
| | | B1 | 2nd B1 for 2nd 4 branches i.e. $\frac{3}{5}$ and $\frac{10}{}$ |
| **(b)** | $\left[P(A' \mid B')\right] = \frac{\frac{1}{6} \times 0.3}{0.55}$ | M1 | M1 for a ratio of probabilities ft their numerator from their tree diagram but denom = 0.55 |
| | $= \frac{1}{11}$ | A1 | A1 for $\frac{1}{11}$ or exact equivalent e.g. 0.09 |
**[Total 8]**
**ALT** For alternate approach:
| 1st M1 | for attempt at 2 sim' eq'ns in p and $q$ Allow one error. $0.4p + 0.7q = \frac{9}{20}$ and $0.6p + 0.3q = \frac{11}{20}$ |
|---|---|
| 1st A1 | for any correct equation in $p$ or $q$ |
| 2nd M1 | for simplifying their linear equation with at least 2 terms in $p$ or $q$ to $a = bp$ or $bq$ |
| 2nd A1 | for $P(A) = \frac{5}{6}$ or exact equiv e.g. 0.83 (may be seen on their tree diagram) |
| 1st B1ft | for 1st 2 branches i.e. $\frac{5}{6}$ and $\frac{1}{6}$ (follow through their $P(A)$) |
| 2nd B1 | for 2nd 4 branches i.e. $\frac{3}{5}$ and $\frac{10}{}$ |
**SC** $[P(A) = \frac{5}{6}]$ award M1A0 for $\frac{P(A') \times \frac{3}{10}}{P(A) \times \frac{3}{5} + P(A') \times \frac{3}{10}}$ ft their $P(A)$ and $P(A') = 1 - P(A)$
---\begin{enumerate}
\item The following incomplete tree diagram shows the relationships between the event $A$ and the event $B$.\\
\includegraphics[max width=\textwidth, alt={}, center]{77ae01cd-2b58-48ab-889f-272e27ecf99d-14_799_839_351_548}
\end{enumerate}
Given that $\mathrm { P } ( B ) = \frac { 9 } { 20 }$\\
(a) find $\mathrm { P } ( A )$ and complete the tree diagram,\\
(b) find $\mathrm { P } \left( A ^ { \prime } \mid B ^ { \prime } \right)$.
\hfill \mbox{\textit{Edexcel S1 2017 Q4 [8]}}