| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2018 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Finding unknown probability from total probability |
| Difficulty | Moderate -0.3 This is a straightforward application of the law of total probability requiring one algebraic setup and solution, plus a basic independence check. The multi-step nature and conditional probability context place it slightly below average difficulty, but it's a standard textbook exercise with no novel insight required. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Tree diagram or other method to find equation for \(p\) | M1 | For selecting suitable method; sight of tree diagram with 0.1, 0.3, 0.6 and 0.09, 0.03, \(p\) suitably placed; or VD with 0.009 for \(A\cap F\) and \(B\cap F\) and \(0.6p\) suitably placed; or equation with at least one correct numerical and one "\(p\)" product on LHS |
| \(0.1\times0.09+0.3\times0.03+0.6\times p=0.06\) | A1 | For a correct equation for \(p\); or for expression \(\frac{0.06-(0.009+0.009)}{0.6}\) |
| \(p=0.07\) i.e. 7% | A1 | For 7% (accept 0.07); if no incorrect working and answer 0.07 seen, award 3/3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(B \text{ and } F)=0.3\times0.03=0.009\) but \(P(B)\times P(F)=0.3\times0.06=0.018\); these are not equal so not independent | B1 | For suitable explanation; may reference 2nd branches on tree diagram; \(0.03\neq0.06\) but needs supporting calculation/words; condone incorrect set notation provided calculations and words are correct |
## Question 2:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Tree diagram or other method to find equation for $p$ | M1 | For selecting suitable method; sight of tree diagram with 0.1, 0.3, 0.6 and 0.09, 0.03, $p$ suitably placed; or VD with 0.009 for $A\cap F$ and $B\cap F$ and $0.6p$ suitably placed; or equation with at least one correct numerical and one "$p$" product on LHS |
| $0.1\times0.09+0.3\times0.03+0.6\times p=0.06$ | A1 | For a correct equation for $p$; or for expression $\frac{0.06-(0.009+0.009)}{0.6}$ |
| $p=0.07$ i.e. **7%** | A1 | For 7% (accept 0.07); if no incorrect working and answer 0.07 seen, award 3/3 |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(B \text{ and } F)=0.3\times0.03=0.009$ but $P(B)\times P(F)=0.3\times0.06=0.018$; these are not equal so not independent | B1 | For suitable explanation; may reference 2nd branches on tree diagram; $0.03\neq0.06$ but needs supporting calculation/words; condone incorrect set notation provided calculations and words are correct |
---\begin{enumerate}
\item A factory buys $10 \%$ of its components from supplier $A , 30 \%$ from supplier $B$ and the rest from supplier $C$. It is known that $6 \%$ of the components it buys are faulty.
\end{enumerate}
Of the components bought from supplier $A , 9 \%$ are faulty and of the components bought from supplier $B , 3 \%$ are faulty.\\
(a) Find the percentage of components bought from supplier $C$ that are faulty.
A component is selected at random.\\
(b) Explain why the event "the component was bought from supplier $B$ " is not statistically independent from the event "the component is faulty".
\hfill \mbox{\textit{Edexcel AS Paper 2 2018 Q2 [4]}}