| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2018 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Reverse conditional probability |
| Difficulty | Moderate -0.3 This is a standard tree diagram question with straightforward conditional probability calculations. Part (d) requires Bayes' theorem (reverse conditional probability), but the setup is simple with clear percentages and only three suppliers. The calculations are routine for S1 level with no conceptual challenges beyond applying the standard formula P(not S|faulty) = P(faulty and not S)/P(faulty). |
| 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 | Marks | Guidance |
| Tree diagram with first 3 branches: R labelled 0.4, S labelled 0.25, T labelled 0.35 | M1 | Attempt at tree with first 3 branches correctly labelled OR second 6 branches F labelled 0.02, 0.01 and 0.02 |
| All branches, all labels and all 9 probabilities correct | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{0.392}\) | B1 | Allow exact equivalent e.g. \(\frac{49}{125}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.4\times0.02+0.25\times0.01+0.35\times0.02 = \mathbf{0.0175}\) | M1 A1 | Allow ft from tree diagram for M1; allow exact equivalent e.g. \(\frac{7}{400}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([P(S'\mid F) = 1-P(S\mid F)]\) \(= \frac{0.4\times0.02+0.35\times0.02}{0.0175}\) or \(\frac{0.015}{0.0175}\) or \(1-\frac{0.25\times0.01}{0.0175}\) | M1 | Allow ft from tree diagram for numerator and ft from (c) for denominator; \(0 <\) their (c) \(< 1\) |
| \(= \dfrac{6}{7}\) | A1 | Allow awrt 0.857 |
# Question 3:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Tree diagram with first 3 branches: R labelled 0.4, S labelled 0.25, T labelled 0.35 | M1 | Attempt at tree with first 3 branches correctly labelled OR second 6 branches F labelled 0.02, 0.01 and 0.02 |
| All branches, all labels and all 9 probabilities correct | A1 | |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{0.392}$ | B1 | Allow exact equivalent e.g. $\frac{49}{125}$ |
## Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.4\times0.02+0.25\times0.01+0.35\times0.02 = \mathbf{0.0175}$ | M1 A1 | Allow ft from tree diagram for M1; allow exact equivalent e.g. $\frac{7}{400}$ |
## Part (d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[P(S'\mid F) = 1-P(S\mid F)]$ $= \frac{0.4\times0.02+0.35\times0.02}{0.0175}$ or $\frac{0.015}{0.0175}$ or $1-\frac{0.25\times0.01}{0.0175}$ | M1 | Allow ft from tree diagram for numerator and ft from (c) for denominator; $0 <$ their (c) $< 1$ |
| $= \dfrac{6}{7}$ | A1 | Allow awrt 0.857 |
---\begin{enumerate}
\item A manufacturer of electric generators buys engines for its generators from three companies, $R , S$ and $T$.
\end{enumerate}
Company $R$ supplies 40\% of the engines.
Company $S$ supplies $25 \%$ of the engines.
The rest of the engines are supplied by company $T$.
It is known that $2 \%$ of the engines supplied by company $R$ are faulty, $1 \%$ of the engines supplied by company $S$ are faulty and $2 \%$ of the engines supplied by company $T$ are faulty.
An engine is chosen at random.\\
(a) Draw a tree diagram to show all the possible outcomes and the associated probabilities.\\
(b) Calculate the probability that the engine is from company $R$ and is not faulty.\\
(c) Calculate the probability that the engine is faulty.
Given that the engine is faulty,\\
(d) find the probability that the engine did not come from company $S$.\\
\hfill \mbox{\textit{Edexcel S1 2018 Q3 [7]}}