| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2018 |
| Session | October |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Principle of Inclusion/Exclusion |
| Type | Three-Set Venn Diagram Probability Calculation |
| Difficulty | Moderate -0.3 This is a standard S1 Venn diagram probability question requiring straightforward reading of probabilities from the diagram and basic probability calculations (addition, conditional probability, independence). Part (b) uses independence to find a missing probability, and part (d) involves a simple binomial-type calculation. All techniques are routine for S1 with no novel problem-solving required, making it slightly easier than average. |
| 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 |
|---|---|---|
| (a) 0.13 | B1 (1) | |
| (b) \(P(A) \times P(C) = P(A \cap C)\) | M1 | Using \(P(A) \times P(C) = P(A \cap C)\) allow one addition error in P(A) e.g. P(A) = 0.11 |
| \(0.2 \times (0.08 + p) = 0.05\) or \(P(C) = \frac{0.05}{0.10 + 0.05 + 0.01 + 0.04}\) or \(\frac{0.05}{0.2}\) or \(0.25\) | M1 | 0.17 only |
| \(p = 0.17\) | A1 | |
| \(P(\text{no faults}) = 1 - (0.1 + 0.05 + 0.01 + 0.04 + 0.08 + 0.03 + "0.17")\) or \(1 - ["P(C)' + 0.10 + 0.05 + 0.08]\) | M1 | Allow letter p for 0.17 |
| 0.52 only (correct answer of 0.52 with no incorrect working is 4/4) | ||
| (4) | ||
| Ans only | They can get q without finding p so a correct answer to q scores 4/4 | |
| (c) \(P(\text{Fault B but not fault C} | \text{Has fault A}) = \frac{0.05}{0.2} = 0.25\) | M1, A1 (2) |
| (d) \(P(\text{exactly 2 defects}) = 0.12\) or \(\frac{3}{25}\) | B1 | Sight of 0.12 or (0.05 + 0.03 + 0.04) only. NB e.g. \(0.12 \times 2\) is B1M0A0 |
| \(P(\text{both have 2 defects}) = 0.12^2\) | M1 | |
| \(= 0.0144\) or \(\frac{9}{625}\) | A1 | |
| (3) | ||
| Total 10 |
**(a)** 0.13 | B1 (1) |
**(b)** $P(A) \times P(C) = P(A \cap C)$ | M1 | Using $P(A) \times P(C) = P(A \cap C)$ allow one addition error in P(A) e.g. P(A) = 0.11
| $0.2 \times (0.08 + p) = 0.05$ or $P(C) = \frac{0.05}{0.10 + 0.05 + 0.01 + 0.04}$ or $\frac{0.05}{0.2}$ or $0.25$ | M1 | 0.17 only
| $p = 0.17$ | A1 |
| $P(\text{no faults}) = 1 - (0.1 + 0.05 + 0.01 + 0.04 + 0.08 + 0.03 + "0.17")$ or $1 - ["P(C)' + 0.10 + 0.05 + 0.08]$ | M1 | Allow letter p for 0.17
| | | 0.52 only (correct answer of 0.52 with no incorrect working is 4/4)
| | (4) |
| Ans only | They can get q without finding p so a correct answer to q scores 4/4 |
**(c)** $P(\text{Fault B but not fault C} | \text{Has fault A}) = \frac{0.05}{0.2} = 0.25$ | M1, A1 (2) | M1: for attempt at $P\left(B \cap C' | A\right)$ allow for $\frac{0.06}{0.2}$ or $\frac{0.05}{0.2}$ allow it of their P(A) used in part(b). A1: 0.25
**(d)** $P(\text{exactly 2 defects}) = 0.12$ or $\frac{3}{25}$ | B1 | Sight of 0.12 or (0.05 + 0.03 + 0.04) only. NB e.g. $0.12 \times 2$ is B1M0A0
| $P(\text{both have 2 defects}) = 0.12^2$ | M1 |
| $= 0.0144$ or $\frac{9}{625}$ | A1 |
| | (3) |
| | **Total 10** |
---4. Pieces of wood cladding are produced by a timber merchant. There are three types of fault, $A , B$ and $C$, that can appear in each piece of wood cladding.
The Venn diagram shows the probabilities of a piece of wood cladding having the various types of fault.\\
\includegraphics[max width=\textwidth, alt={}, center]{0377c6e9-ab4f-477d-9236-0732fe81f25e-14_602_1120_497_413}
A piece of wood cladding is chosen at random.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that the piece of wood cladding has more than one type of fault.
Fault types $A$ and $C$ occur independently.
\item Find the probability that the piece of wood cladding has no faults.
Given that the piece of wood cladding has fault $A$,
\item find the probability that it also has fault $B$ but not fault $C$.
Two pieces of the wood cladding are selected at random.
\item Find the probability that both have exactly 2 types of fault.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2018 Q4 [10]}}