| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2007 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Population partition tree diagram |
| Difficulty | Moderate -0.8 This is a standard S1 tree diagram question testing basic conditional probability and Bayes' theorem. Part (a) is routine diagram construction, (b)(i) is simple multiplication along branches (0.35 × 0.03), (b)(ii) requires adding three products (law of total probability), and (c) applies Bayes' theorem with given values. All techniques are textbook exercises with no problem-solving insight required, making it easier than average but not trivial due to the multi-part calculation in (b)(ii). |
| 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 |
|---|---|---|
| M1 | Tree diagram, 3 branches and then two from each. At least one probability attempted. | |
| A1 | Correct tree shape | |
| A1 | \(A\), \(B\) and \(C\) and 0.35 and 0.25 | |
| A1 (3) | \(D\) (x3) and 0.03, 0.06, 0.05 (May be implied by seeing \(P(A \cap D)\) etc at the ends) | |
| (b)(i) \(P(A \cap D) = 0.35 \times 0.03 = \mathbf{0.0105}\) or \(\frac{21}{2000}\) | M1, A1 | |
| \(P(C) = 0.4\) (anywhere) | B1 | |
| (ii) \(P(D) = (i) + 0.25 \times 0.06 + (0.4 \times 0.05)\) | M1 | |
| \(= \mathbf{0.0455}\) or \(\frac{91}{2000}\) | A1 | (5 marks) |
| (c) \(P(C | D) = \frac{P(C \cap D)}{P(D)} = \frac{0.4 \times 0.05}{(ii)}\) | M1, A1 f.t. |
| \(= 0.43956...\) or \(\frac{40}{91}\) | A1 | \(\mathbf{0.44}\) or awrt 0.440 |
**(a)**
| | M1 | Tree diagram, 3 branches and then two from each. At least one probability attempted. |
| A1 | Correct tree shape |
| A1 | $A$, $B$ and $C$ and 0.35 and 0.25 |
| A1 (3) | $D$ (x3) and 0.03, 0.06, 0.05 (May be implied by seeing $P(A \cap D)$ etc at the ends) |
**(b)(i)** $P(A \cap D) = 0.35 \times 0.03 = \mathbf{0.0105}$ or $\frac{21}{2000}$ | M1, A1 |
$P(C) = 0.4$ (anywhere) | B1 |
**(ii)** $P(D) = (i) + 0.25 \times 0.06 + (0.4 \times 0.05)$ | M1 |
$= \mathbf{0.0455}$ or $\frac{91}{2000}$ | A1 | (5 marks)
**(c)** $P(C|D) = \frac{P(C \cap D)}{P(D)} = \frac{0.4 \times 0.05}{(ii)}$ | M1, A1 f.t. |
$= 0.43956...$ or $\frac{40}{91}$ | A1 | $\mathbf{0.44}$ or awrt **0.440** | (3 marks)
**Total: 11 marks**
---\begin{enumerate}
\item In a factory, machines $A , B$ and $C$ are all producing metal rods of the same length. Machine $A$ produces $35 \%$ of the rods, machine $B$ produces $25 \%$ and the rest are produced by machine $C$. Of their production of rods, machines $A , B$ and $C$ produce $3 \% , 6 \%$ and $5 \%$ defective rods respectively.\\
(a) Draw a tree diagram to represent this information.\\
(b) Find the probability that a randomly selected rod is\\
(i) produced by machine $A$ and is defective,\\
(ii) is defective.\\
(c) Given that a randomly selected rod is defective, find the probability that it was produced by machine $C$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2007 Q2 [11]}}