| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2020 |
| Session | October |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Testing mutual exclusivity |
| Difficulty | Moderate -0.8 This is a straightforward Venn diagram question testing basic probability definitions (mutual exclusivity, independence, conditional probability) with simple algebraic manipulation. Parts (a)-(c) require only reading the diagram and applying standard formulas, while part (d) involves one additional step of conditional probability calculation. All techniques are routine for A-level statistics. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(A, C\) or \(D, B\) or \(D,C\) | B1 |
| (b) | \(p = 0.4 - 0.07 - 0.24 = \mathbf{0.09}\) | B1 |
| (c) | \(A\) and \(B\) independent implies \(\text{P}(A) \times 0.4 = 0.24\) or \((q + 0.16 + 0.24) \times 0.4 = 0.24\), so \(\text{P}(A) = 0.6\) and \(q = \mathbf{0.20}\) | M1; A1cso |
| (d)(i) | \(\text{P}(B' \vert C) = 0.64\) gives \(\frac{r}{r+p} = 0.64\) or \(\frac{r}{r + "0.09"} = 0.64\). \(r = 0.64r + 0.64 "p"\) so \(0.36r = 0.0576\) so \(r = \mathbf{0.16}\) | M1; A1 |
| (d)(ii) | Using sum of probabilities \(= 1\) e.g. "0.6" + 0.07 + "0.25" + \(s = 1\) so \(s = \mathbf{0.08}\) | M1; A1 |
| (a) | $A, C$ or $D, B$ or $D,C$ | B1 | One correct pair. If more than one pair they must all be correct. |
|---|---|---|---|
| (b) | $p = 0.4 - 0.07 - 0.24 = \mathbf{0.09}$ | B1 | For $p = 0.09$ (Maybe stated in Venn Diagram). If values in VD and text conflict, take text or a value used in a later part. |
| (c) | $A$ and $B$ independent implies $\text{P}(A) \times 0.4 = 0.24$ or $(q + 0.16 + 0.24) \times 0.4 = 0.24$, so $\text{P}(A) = 0.6$ and $q = \mathbf{0.20}$ | M1; A1cso | M1 for a correct equation in one variable for P(A) or $q$ using independence or for seeing both $\text{P}(A \cap B) = \text{P}(A) \times \text{P}(B)$ and $0.24 = 0.6 \times 0.4$. A1cso for $q = 0.20$ or exact equivalent (dep on correct use of independence). Use of $\text{P}(A) = 1 - \text{P}(B) = 0.6$ leading to $q = 0.2$ scores M0A0. |
| (d)(i) | $\text{P}(B' \vert C) = 0.64$ gives $\frac{r}{r+p} = 0.64$ or $\frac{r}{r + "0.09"} = 0.64$. $r = 0.64r + 0.64 "p"$ so $0.36r = 0.0576$ so $r = \mathbf{0.16}$ | M1; A1 | 1st M1 for use of $\text{P}(B' \vert C) = 0.64$ leading to a correct equation in $r$ and possibly $p$. Can ft their $p$ provided $0 < p < 1$. 1st A1 for $r = 0.16$ or exact equivalent. |
| (d)(ii) | Using sum of probabilities $= 1$ e.g. "0.6" + 0.07 + "0.25" + $s = 1$ so $s = \mathbf{0.08}$ | M1; A1 | M1 for use of total probability = 1 to form a linear equation in $s$. Allow $p, q, r$ etc. Can follow through their values provided each of $p, q, r$ are in $[0, 1)$. A1 for $s = 0.08$ or exact equivalent. |\begin{enumerate}
\item The Venn diagram shows the probabilities associated with four events, $A , B , C$ and $D$\\
\includegraphics[max width=\textwidth, alt={}, center]{2b63aa7f-bc50-4422-8dc0-e661b521c221-02_505_861_296_602}\\
(a) Write down any pair of mutually exclusive events from $A , B , C$ and $D$
\end{enumerate}
Given that $\mathrm { P } ( B ) = 0.4$\\
(b) find the value of $p$
Given also that $A$ and $B$ are independent\\
(c) find the value of $q$
Given further that $\mathrm { P } \left( B ^ { \prime } \mid C \right) = 0.64$\\
(d) find\\
(i) the value of $r$\\
(ii) the value of $s$\\
\hfill \mbox{\textit{Edexcel Paper 3 2020 Q1 [8]}}