| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Given conditional, find joint or marginal |
| Difficulty | Moderate -0.3 This is a straightforward conditional probability question requiring direct application of standard formulas: P(A|B) = P(A∩B)/P(B) for part (a), the addition rule for part (b), and another conditional probability calculation for part (c). All three parts follow routine procedures with no problem-solving insight needed, making it slightly easier than average but still requiring careful algebraic manipulation across multiple steps. |
| Spec | 2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | M2 A1 | \(\frac{1}{4} = \mathrm{P}(B) \times \frac{3}{8}\) \(\therefore \mathrm{P}(B) = \frac{1}{4} \div \frac{3}{8} = \frac{8}{3}\) |
| (b) | M2 A1 | \(\frac{7}{12} + \frac{3}{8} - \frac{1}{4} = \frac{17}{24}\) |
| (c) | M2 A1, (9) | \(\mathrm{P}(B \mid A') = \frac{\mathrm{P}(B \cap A')}{\mathrm{P}(A')} = \frac{\frac{3}{8} - \frac{1}{4}}{1 - \frac{1}{12}} = \frac{\frac{1}{8}}{\frac{11}{12}} = \frac{3}{22}\) |
(a) | M2 A1 | $\frac{1}{4} = \mathrm{P}(B) \times \frac{3}{8}$ $\therefore \mathrm{P}(B) = \frac{1}{4} \div \frac{3}{8} = \frac{8}{3}$
(b) | M2 A1 | $\frac{7}{12} + \frac{3}{8} - \frac{1}{4} = \frac{17}{24}$
(c) | M2 A1, (9) | $\mathrm{P}(B \mid A') = \frac{\mathrm{P}(B \cap A')}{\mathrm{P}(A')} = \frac{\frac{3}{8} - \frac{1}{4}}{1 - \frac{1}{12}} = \frac{\frac{1}{8}}{\frac{11}{12}} = \frac{3}{22}$
---The events $A$ and $B$ are such that
$$\text{P}(A) = \frac{7}{12}, \quad \text{P}(A \cap B) = \frac{1}{4} \quad \text{and} \quad \text{P}(A|B) = \frac{2}{3}.$$
Find
\begin{enumerate}[label=(\alph*)]
\item P$(B)$, [3 marks]
\item P$(A \cup B)$, [3 marks]
\item P$(B|A')$. [3 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q3 [9]}}