| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Independent Events |
| Type | Find unknown probability given independence |
| Difficulty | Moderate -0.3 This is a straightforward application of independence (P(A∩B) = P(A)×P(B)) leading to a simple quadratic equation, followed by routine use of probability formulas. The algebraic manipulation is minimal and the concepts are standard S1 material, making it slightly easier than average but not trivial due to the multi-step nature. |
| Spec | 2.03a Mutually exclusive and independent events2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks |
|---|---|
| \(\therefore 2[P(B)]^2 = \frac{1}{8}\); \(\therefore [P(B)]^2 = \frac{1}{16}\); \(\therefore P(B) = \frac{1}{4}\) | M2 M2 A1 |
| (b) \(P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{2} + \frac{1}{4} - \frac{1}{8} = \frac{5}{8}\) | M2 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\therefore P(A \mid B') = P(A) = \frac{1}{2}\) | M1 A1 | (10 marks total) |
**(a)** $P(A \cap B) = P(A) \times P(B) = 2P(B) \times P(B) = 2[P(B)]^2$
$\therefore 2[P(B)]^2 = \frac{1}{8}$; $\therefore [P(B)]^2 = \frac{1}{16}$; $\therefore P(B) = \frac{1}{4}$ | M2 M2 A1 |
**(b)** $P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{2} + \frac{1}{4} - \frac{1}{8} = \frac{5}{8}$ | M2 A1 |
**(c)** A and B independent $\therefore$ A and $B'$ independent
$\therefore P(A \mid B') = P(A) = \frac{1}{2}$ | M1 A1 | **(10 marks total)**
---The events $A$ and $B$ are independent and such that
$$\text{P}(A) = 2\text{P}(B) \text{ and } \text{P}(A \cap B) = \frac{1}{8}.$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\text{P}(B) = \frac{1}{4}$. [5 marks]
\item Find $\text{P}(A \cup B)$. [3 marks]
\item Find $\text{P}(A | B')$. [2 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q2 [10]}}