| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2020 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Two independent categorical choices |
| Difficulty | Moderate -0.8 This is a standard S1 tree diagram question requiring completion of probabilities, use of independence, conditional probability, and comparison. All techniques are routine for this module with clear scaffolding through parts (a)-(e). The independence condition and conditional probability calculation are textbook applications requiring no novel insight. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Correct tree diagram with \(0.25\) for \(P(V)\) | B1 | \(0.25\) for \(P(V)\) |
| Correct probabilities on 2nd branches: \((1-p)\), \((1-q)\) and \(0.6\) | B1 | For correct probabilities on 2nd branches \([allow their values]\) and \(0.6\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(W) = 0.4p + 0.35q + \text{"0.25"}\times0.4 \;[= 0.4p + 0.35q + 0.1]\) | B1ft | For a correct expression using their values from tree diagram |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Correct expression: \(P(W\cap V) = \text{"0.1"} = \text{"0.25"}\times P(W)\) or \(P(W) = P(W\mid V) = 0.4\) | M1 | For sight or use of a correct expression in \(V\) and \(W\) or correct equation in \(p\) and \(q\) (ft their part (b)) |
| \(0.1 = 0.25(0.4p + 0.35q + 0.25\times0.4)\) or \(0.4p + 0.35q + 0.25\times0.4 = 0.4\) | A1 | For a fully correct equation (needn't be simplified) [may see \(0.4p+0.35q=0.3\) or \(8p+7q=6\)] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{7}{30} = \frac{0.35(1-q)}{\text{"P}(J)\text{"}}\) | M1 | For using given conditional probability to form an equation in \(q\) and \(P(J)\) using \(\frac{7}{30}\) |
| Since \(V\) and \(W\) are independent so are \(V\) and \(W' = J\), so \(P(J) = 0.6\); or sub \(P(J) = 1 -\) their (b) to get equation in \(p\) and \(q\) | dM1 | dep on 1st M1; for getting \(P(J)=0.6\) or sub \(1 -\) their (b) to get 2nd equation in \(p\) and \(q\) |
| \([1 - q = \frac{2}{3}P(J)\) therefore\(]\) \(\mathbf{q = 0.6}\) | A1 | For \(q=0.6\) [NB must be \(q=0.6\) not just \(P(J)=0.6\)] |
| \(8p + 7\times\text{"0.6"} = 6\), so \(\mathbf{p = 0.225}\) or \(\frac{9}{40}\) | ddM1, A1 | dep on both Ms; for substitution to find 2nd value. Allow ft of their \(p\) or \(q\) provided both lie in \((0,1)\). \(2^{\text{nd}}\) A1 for \(p=0.225\) or exact equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\{P(V\mid W) = P(V) = 0.25\) (since independent) and \(P(M\mid W) = 0.225\,(= p)\}\) | — | — |
| \(P(F\mid W) = \frac{0.35\times\text{"0.6"}}{\text{"0.4"}}\) or \(\frac{0.35q}{(b)}\); \(= \frac{21}{40}\) or \(0.525\) | M1; A1 | M1 for a method for finding \(P(F\mid W)\); A1 for correct value \(\frac{21}{40}\) or exact equivalent |
| [Since this prob \(> 0.5\) therefore it must be the largest] so conclusion is correct | B1ft | For a correct conclusion based on enough probs found; ft their probabilities. Allow B1ft for comparing 3 calculated probs of the form \(P(M\cap W)\), needn't be correct ft |
# Question 2(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Correct tree diagram with $0.25$ for $P(V)$ | B1 | $0.25$ for $P(V)$ |
| Correct probabilities on 2nd branches: $(1-p)$, $(1-q)$ and $0.6$ | B1 | For correct probabilities on 2nd branches $[allow their values]$ and $0.6$ |
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# Question 2(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(W) = 0.4p + 0.35q + \text{"0.25"}\times0.4 \;[= 0.4p + 0.35q + 0.1]$ | B1ft | For a correct expression using their values from tree diagram |
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# Question 2(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Correct expression: $P(W\cap V) = \text{"0.1"} = \text{"0.25"}\times P(W)$ or $P(W) = P(W\mid V) = 0.4$ | M1 | For sight or use of a correct expression in $V$ and $W$ or correct equation in $p$ and $q$ (ft their part (b)) |
| $0.1 = 0.25(0.4p + 0.35q + 0.25\times0.4)$ or $0.4p + 0.35q + 0.25\times0.4 = 0.4$ | A1 | For a fully correct equation (needn't be simplified) [may see $0.4p+0.35q=0.3$ or $8p+7q=6$] |
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# Question 2(d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{7}{30} = \frac{0.35(1-q)}{\text{"P}(J)\text{"}}$ | M1 | For using given conditional probability to form an equation in $q$ and $P(J)$ using $\frac{7}{30}$ |
| Since $V$ and $W$ are independent so are $V$ and $W' = J$, so $P(J) = 0.6$; or sub $P(J) = 1 -$ their (b) to get equation in $p$ and $q$ | dM1 | dep on 1st M1; for getting $P(J)=0.6$ or sub $1 -$ their (b) to get 2nd equation in $p$ and $q$ |
| $[1 - q = \frac{2}{3}P(J)$ therefore$]$ $\mathbf{q = 0.6}$ | A1 | For $q=0.6$ [NB must be $q=0.6$ not just $P(J)=0.6$] |
| $8p + 7\times\text{"0.6"} = 6$, so $\mathbf{p = 0.225}$ or $\frac{9}{40}$ | ddM1, A1 | dep on both Ms; for substitution to find 2nd value. Allow ft of their $p$ or $q$ provided both lie in $(0,1)$. $2^{\text{nd}}$ A1 for $p=0.225$ or exact equivalent |
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# Question 2(e):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\{P(V\mid W) = P(V) = 0.25$ (since independent) and $P(M\mid W) = 0.225\,(= p)\}$ | — | — |
| $P(F\mid W) = \frac{0.35\times\text{"0.6"}}{\text{"0.4"}}$ or $\frac{0.35q}{(b)}$; $= \frac{21}{40}$ or $0.525$ | M1; A1 | M1 for a method for finding $P(F\mid W)$; A1 for correct value $\frac{21}{40}$ or exact equivalent |
| [Since this prob $> 0.5$ therefore it must be the largest] so conclusion is correct | B1ft | For a correct conclusion based on enough probs found; ft their probabilities. Allow B1ft for comparing 3 calculated probs of the form $P(M\cap W)$, needn't be correct ft |\begin{enumerate}
\item In a school canteen, students can choose from a main course of meat ( $M$ ), fish ( $F$ ) or vegetarian ( $V$ ). They can then choose a drink of either water ( $W$ ) or juice ( $J$ ).
\end{enumerate}
The partially completed tree diagram, where $p$ and $q$ are probabilities, shows the probabilities of these choices for a randomly selected student.
\section*{Drink}
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Main course}
\includegraphics[alt={},max width=\textwidth]{81d5e460-9559-4d25-aa08-6440559aec83-04_783_1013_593_463}
\end{center}
\end{figure}
(a) Complete the tree diagram, giving your answers in terms of $p$ and $q$ where appropriate.\\
(b) Find an expression, in terms of $p$ and $q$, for the probability that a randomly selected student chooses water to drink.
The events "choosing a vegetarian main course" and "choosing water to drink" are independent.\\
(c) Find a linear equation in terms of $p$ and $q$.
A student who has chosen juice to drink is selected at random. The probability that they chose fish for their main course is $\frac { 7 } { 30 }$\\
(d) Find the value of $p$ and the value of $q$.
The canteen manager claims that students who choose water to drink are most likely to choose a fish main course.\\
(e) State, showing your working clearly, whether or not the manager's claim is correct.
\hfill \mbox{\textit{Edexcel S1 2020 Q2 [13]}}