| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Multi-stage with stopping condition |
| Difficulty | Moderate -0.3 This is a standard S1 tree diagram question with straightforward probability calculations. Part (a) requires filling in probabilities that decrease by 0.2 each time. Parts (b) and (c) involve routine probability calculations using the multiplication and addition rules, plus conditional probability. Part (d) requires setting up an equation from a two-stage tree diagram (algebraic but guided), and part (e) is solving a quadratic. While multi-part, each step follows standard techniques with no novel insight required, making it slightly easier than average for A-level. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| Tree diagram with 0.3, 0.5 on first branches; 0.7, 0.1, 0.9 on second branches | B1 | For correctly placing 0.3 and 0.5 |
| B1 (2) | For correctly placing 0.7, 0.1 and 0.9 |
| Answer | Marks | Guidance |
|---|---|---|
| \(1 - 0.3 \times 0.5 \times 0.7 \times 0.9\) or \(0.7 + (0.3 \times 0.5) + (0.3 \times 0.5 \times 0.3) + (0.3 \times 0.5 \times 0.7 \times 0.1)\) | M1 | For correct expression ft from tree diagram |
| \(= 0.9055\) | A1 (2) | For 0.9055 or exact equivalent e.g. \(\frac{1811}{2000}\). Accept 0.906 only if correct expression seen |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(P_1 \cup P_2 \mid \text{Pass}) = \frac{0.7 + 0.3 \times 0.5}{(b)} = \frac{0.85}{0.9055} = 0.938707...\) = awrt \(0.939\) | M1, A1ft, A1 (3) | M1 for correct ratio of probs ft their 0.3 and their answer to (b) [if \(< 1\)]. Num \(>\) Den M0. A1ft for correct numerator and their part (b) on denominator. A1 for awrt 0.939 or exact fraction e.g. \(\frac{1700}{1811}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(p + (1-p)(p - 0.2) = 0.95\) or \(1-(1-p)(1.2-p)\) | M1 | For correct expression for P(pass) in terms of \(p\) |
| \(p^2 - 2.2p + 1.15 = 0\) | dM1, A1cso (3) | Dependent on 1st M1 for expanding and forming equation in \(p\). Allow one slip. A1cso for correct processing leading to printed answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(p = \frac{2.2 \pm \sqrt{2.2^2 - 4 \times 1.15}}{2}\) or complete the square: \((p-1.1)^2 - 1.1^2 + 1.15 = 0\) | M1 | For attempt to solve given equation, correct expression. Condone just \(+\) not \(\pm\) |
| \(= \frac{2.2 \pm 0.4898...}{2}\) or \(\frac{2.2 \pm \sqrt{0.24}}{2}\) or \(1.1 \pm \sqrt{0.06}\) or \((1.34...), 0.855...\) | A1 | For correct expression and simplified square root or 1.34... and 0.855... |
| \(p = 0.855\) | A1 (3) | For \(p = 0.855\) only (penalise any extra value \(> 1\)). Correct ans only scores 3/3. For \(\frac{1}{10}(11 - \sqrt{6})\) or 0.855... score M1A1A0 (not to 3dp) but for 0.855 can score M1A1A1 |
## Question 4:
### Part (a)
| Tree diagram with 0.3, 0.5 on first branches; 0.7, 0.1, 0.9 on second branches | B1 | For correctly placing 0.3 and 0.5 |
| | B1 (2) | For correctly placing 0.7, 0.1 and 0.9 |
### Part (b)
| $1 - 0.3 \times 0.5 \times 0.7 \times 0.9$ or $0.7 + (0.3 \times 0.5) + (0.3 \times 0.5 \times 0.3) + (0.3 \times 0.5 \times 0.7 \times 0.1)$ | M1 | For correct expression ft from tree diagram |
| $= 0.9055$ | A1 (2) | For 0.9055 or exact equivalent e.g. $\frac{1811}{2000}$. Accept 0.906 only if correct expression seen |
### Part (c)
| $P(P_1 \cup P_2 \mid \text{Pass}) = \frac{0.7 + 0.3 \times 0.5}{(b)} = \frac{0.85}{0.9055} = 0.938707...$ = awrt $0.939$ | M1, A1ft, A1 (3) | M1 for correct ratio of probs ft their 0.3 and their answer to (b) [if $< 1$]. Num $>$ Den M0. A1ft for correct numerator and their part (b) on denominator. A1 for awrt 0.939 or exact fraction e.g. $\frac{1700}{1811}$ |
### Part (d)
| $p + (1-p)(p - 0.2) = 0.95$ or $1-(1-p)(1.2-p)$ | M1 | For correct expression for P(pass) in terms of $p$ |
| $p^2 - 2.2p + 1.15 = 0$ | dM1, A1cso (3) | Dependent on 1st M1 for expanding and forming equation in $p$. Allow one slip. A1cso for correct processing leading to printed answer |
### Part (e)
| $p = \frac{2.2 \pm \sqrt{2.2^2 - 4 \times 1.15}}{2}$ or complete the square: $(p-1.1)^2 - 1.1^2 + 1.15 = 0$ | M1 | For attempt to solve given equation, correct expression. Condone just $+$ not $\pm$ |
| $= \frac{2.2 \pm 0.4898...}{2}$ or $\frac{2.2 \pm \sqrt{0.24}}{2}$ or $1.1 \pm \sqrt{0.06}$ or $(1.34...), 0.855...$ | A1 | For correct expression and simplified square root or 1.34... and 0.855... |
| $p = 0.855$ | A1 (3) | For $p = 0.855$ only (penalise any extra value $> 1$). Correct ans only scores 3/3. For $\frac{1}{10}(11 - \sqrt{6})$ or 0.855... score M1A1A0 (not to 3dp) but for 0.855 can score M1A1A1 |
---4. A training agency awards a certificate to each student who passes a test while completing a course.\\
Students failing the test will attempt the test again up to 3 more times, and, if they pass the test, will be awarded a certificate.\\
The probability of passing the test at the first attempt is 0.7 , but the probability of passing reduces by 0.2 at each attempt.
\begin{enumerate}[label=(\alph*)]
\item Complete the tree diagram below to show this information.\\
\includegraphics[max width=\textwidth, alt={}, center]{70137e9a-0a6b-48b5-8dd4-c436cb063351-08_545_1244_639_340}
A student who completed the course is selected at random.
\item Find the probability that the student was awarded a certificate.
\item Given that the student was awarded a certificate, find the probability that the student passed on the first or second attempt.
The training agency decides to alter the test taken by the students while completing the course, but will not allow more than 2 attempts. The agency requires the probability of passing the test at the first attempt to be $p$, and the probability of passing the test at the second attempt to be ( $p - 0.2$ ). The percentage of students who complete the course and are awarded a certificate is to be $95 \%$
\item Show that $p$ satisfies the equation
$$p ^ { 2 } - 2.2 p + 1.15 = 0$$
\item Hence find the value of $p$, giving your answer to 3 decimal places.\\
\includegraphics[max width=\textwidth, alt={}, center]{70137e9a-0a6b-48b5-8dd4-c436cb063351-09_2261_47_313_37}
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2016 Q4 [13]}}