| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Conditional probability with algebraic expressions |
| Difficulty | Standard +0.8 This S1 question requires understanding of independence, conditional probability, and Venn diagram manipulation with algebraic expressions. Part (a) uses independence to set up an equation, parts (b-c) require manipulating conditional probability formulas with multiple unknowns. The algebraic complexity and multi-step reasoning across three interconnected parts elevates this above a standard conditional probability exercise, though it remains within typical A-level scope. |
| Spec | 2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \([P(B) = 0.4, P(A) = p + 0.1 \text{ so}] 0.4 \times (p+0.1) = 0.1\) or \(0.4 \times P(A) = 0.1\), \(p = \frac{1}{4} - 0.1\), \(\mathbf{p = 0.15}\) | M1, M1A1 | (3) |
| (b) \(\frac{5}{11} = \frac{[P(B \cap C)]}{P(C)} = \frac{0.2}{0.2 + q}\) or \(\frac{5}{11} = \frac{0.2}{P(C)}\), \(11 \times 0.2 = 5 \times (0.2 + q)\), \(q = \mathbf{0.24}\), \(r = 0.6 - (p+q)\), i.e. \(r = \mathbf{0.21}\) | M1, dM1, A1, A1ft | (4) |
| (c) \(\left[\frac{P((A \cup C) \cap B)}{P(B)}\right] = \frac{0.3}{0.4} = \mathbf{0.75}\) | M1, A1 | (2) [9] |
**(a)** $[P(B) = 0.4, P(A) = p + 0.1 \text{ so}] 0.4 \times (p+0.1) = 0.1$ or $0.4 \times P(A) = 0.1$, $p = \frac{1}{4} - 0.1$, $\mathbf{p = 0.15}$ | M1, M1A1 | (3) | 1st M1 for using independence in attempt to form equation in $p$ or P(A); 2nd M1 for correct attempt to solve their linear equation leading to $p = ...$; A1 for 0.15 or exact equivalent
**(b)** $\frac{5}{11} = \frac{[P(B \cap C)]}{P(C)} = \frac{0.2}{0.2 + q}$ or $\frac{5}{11} = \frac{0.2}{P(C)}$, $11 \times 0.2 = 5 \times (0.2 + q)$, $q = \mathbf{0.24}$, $r = 0.6 - (p+q)$, i.e. $r = \mathbf{0.21}$ | M1, dM1, A1, A1ft | (4) | 1st M1 for clear attempt to use P(B|C) to form equation for $q$ or P(C). Assuming indep M0; 2nd dM1 Dep. on 1st M1 for correctly simplifying to linear equation in $q$ or P(C). e.g. accept $11 \times 0.2 = 5 \times 0.2 + q$ or $5P(C) = 2.2$; 1st A1 for $q = 0.24$ or exact equivalent; 2nd A1ft for 0.6 – their $(p+q)$ Dependent on 1st M1 in (b) only.
**(c)** $\left[\frac{P((A \cup C) \cap B)}{P(B)}\right] = \frac{0.3}{0.4} = \mathbf{0.75}$ | M1, A1 | (2) [9] | M1 for correct ratio expression and one correct value (num < denom) or fully correct ratio. Allow $\frac{P(A \cup C) \times P(B)}{P(B)}$ with one probability correct but only if num < denom. A numerator of P(A ∪ C) × P(B) scores M0; A1 for 0.75 or exact equivalent
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4cf4f2d7-d912-4b65-a666-caa37009661a-11_606_1131_210_411}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The Venn diagram in Figure 1 shows three events $A , B$ and $C$ and the probabilities associated with each region of $B$. The constants $p , q$ and $r$ each represent probabilities associated with the three separate regions outside $B$.
The events $A$ and $B$ are independent.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $p$.
Given that $\mathrm { P } ( B \mid C ) = \frac { 5 } { 11 }$
\item find the value of $q$ and the value of $r$.
\item Find $\mathrm { P } ( A \cup C \mid B )$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2013 Q6 [9]}}