| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Standard two-outcome diagnostic test |
| Difficulty | Standard +0.8 This is a multi-stage conditional probability problem requiring careful construction of a tree diagram, application of Bayes' theorem, and algebraic manipulation with percentages. Part (c) requires setting up and solving equations involving conditional probabilities in both directions, which is conceptually demanding for S1 level and goes beyond routine probability calculations. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks |
|---|---|
| (a) \(P(R) = P(P \cap R) + P(F \cap R) = 0.87 \times 0.98 + 0.13 \times 0.05 = 0.859\) | M1 A1 M1 A1 |
| (b) \(P(F \mid R) = \frac{0.13 \times 0.05}{0.859} = 0.00757\) | M1 M1 A1 |
| (c) Now \(P(F \mid R) = 0.003\), so \(P(F \cap R) = 0.003 \times P(R)\) and \(P(R \mid F) = 0.01x\), so \(P(F \cap R) = 0.0013x\) | B1 B1 B1 |
| \(0.0013x = 0.003(0.87 \times 0.98 + 0.0013x)\) | M1 A1 A1 |
| \(0.432x = 0.8526\) therefore \(x = 1.97\) | M1 A1 |
Let $P$ = pass, $F$ = fail, $R$ = roadworthy (no faults), $N$ = not roadworthy
**(a)** $P(R) = P(P \cap R) + P(F \cap R) = 0.87 \times 0.98 + 0.13 \times 0.05 = 0.859$ | M1 A1 M1 A1 |
**(b)** $P(F \mid R) = \frac{0.13 \times 0.05}{0.859} = 0.00757$ | M1 M1 A1 |
**(c)** Now $P(F \mid R) = 0.003$, so $P(F \cap R) = 0.003 \times P(R)$ and $P(R \mid F) = 0.01x$, so $P(F \cap R) = 0.0013x$ | B1 B1 B1 |
$0.0013x = 0.003(0.87 \times 0.98 + 0.0013x)$ | M1 A1 A1 |
$0.432x = 0.8526$ therefore $x = 1.97$ | M1 A1 |
**Total marks: 15**
---Of the cars that are taken to a certain garage for an M.O.T. test, 87% pass. However, 2% of these have faults for which they should have been failed. 5% of the cars which fail are in fact roadworthy and should have passed.
Using a tree diagram, or otherwise, calculate the probabilities that a car chosen at random
\begin{enumerate}[label=(\alph*)]
\item should have passed the test, regardless of whether it actually did or not, [4 marks]
\item failed the test, given that it should have passed. [3 marks]
\end{enumerate}
The garage is told to improve its procedures. When it is inspected again a year later, it is found that the pass rate is still 87% overall and 2% of the cars passed have faults as before, but now 0.3% of the cars which should have passed are failed and $x$% of the cars which are failed should have passed.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the value of $x$. [8 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q6 [15]}}