| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Given conditional, find joint or marginal |
| Difficulty | Moderate -0.8 This is a straightforward S1 probability question requiring basic manipulation of probability rules and set notation. Parts (a)-(c) involve direct application of standard formulas (complement rule, addition rule, conditional probability definition), while part (d) is a simple check of the independence condition. No novel insight or complex reasoning required—purely mechanical application of learned techniques. |
| Spec | 2.03a Mutually exclusive and independent events2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([P(A) = 1 - 0.18 - 0.22]\) \(= \mathbf{0.6}\) (or exact equivalent) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(A \cup B) = \text{"0.6"} + 0.22 = \mathbf{0.82}\) (or exact equivalent) | B1ft | For their (a) \(+ 0.22\) or \(1 - P(A' \cap B')\). Do not ft their (a) if it is \(> 0.78\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Let \(x = P(A \cap B)\). Set up correct equation e.g. \(\frac{x}{x+0.22} = 0.6\) or correctly derived equation for \(P(B)\) | M1 | 1st M1 for a correct equation for \(x\). Look out for assuming independence — if \(P(B) = 0.55\) check it is derived properly |
| \(x = 0.6x + 0.132\), so \(0.4x = 0.132\); or \(P(A \cap B) = P(A\mid B) \cdot P(B) = 0.6 \times 0.55\) | dM1 | Solving to get form \(kx = L\) or correct use of \(P(B)\) to find \(P(A \cap B)\). Dep on 1st M1 |
| \(x = \mathbf{0.33}\) (or exact equivalent) | A1cso | Dep on both Ms. No incorrect working seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(B) = 0.55\); \(P(B) \times P(A) = 0.55 \times 0.6 = 0.33\) or stating \(P(A) = P(A\mid B)\) \([= 0.6]\) | M1 | For finding \(P(B) \times P(A) = 0.33\) (values needed) or stating \(P(A) = P(A\mid B)\) (\(= 0.6\) not needed). NB: M1 using \(P(A \cap B)\) requires \(P(B) = 0.55\). No ft of incorrect \(P(B)\) |
| \(P(B) \times P(A) = P(A \cap B)\), therefore (statistically) independent or \(P(A) = P(A\mid B)\), therefore (statistically) independent | A1cso | For correct statement: \(P(B)\times P(A) = P(A\cap B)\) or \(P(A)=P(A\mid B)\) and stating independent. Full marks in (d) OK even if 0/3 in (c) |
## Question 8:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[P(A) = 1 - 0.18 - 0.22]$ $= \mathbf{0.6}$ (or exact equivalent) | B1 | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(A \cup B) = \text{"0.6"} + 0.22 = \mathbf{0.82}$ (or exact equivalent) | B1ft | For their (a) $+ 0.22$ or $1 - P(A' \cap B')$. Do not ft their (a) if it is $> 0.78$ |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Let $x = P(A \cap B)$. Set up correct equation e.g. $\frac{x}{x+0.22} = 0.6$ or correctly derived equation for $P(B)$ | M1 | 1st M1 for a correct equation for $x$. **Look out for assuming independence** — if $P(B) = 0.55$ check it is derived properly |
| $x = 0.6x + 0.132$, so $0.4x = 0.132$; or $P(A \cap B) = P(A\mid B) \cdot P(B) = 0.6 \times 0.55$ | dM1 | Solving to get form $kx = L$ or correct use of $P(B)$ to find $P(A \cap B)$. Dep on 1st M1 |
| $x = \mathbf{0.33}$ (or exact equivalent) | A1cso | Dep on both Ms. No incorrect working seen |
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(B) = 0.55$; $P(B) \times P(A) = 0.55 \times 0.6 = 0.33$ **or** stating $P(A) = P(A\mid B)$ $[= 0.6]$ | M1 | For finding $P(B) \times P(A) = 0.33$ (values needed) or stating $P(A) = P(A\mid B)$ ($= 0.6$ not needed). NB: M1 using $P(A \cap B)$ requires $P(B) = 0.55$. No ft of incorrect $P(B)$ |
| $P(B) \times P(A) = P(A \cap B)$, therefore (statistically) independent **or** $P(A) = P(A\mid B)$, therefore (statistically) independent | A1cso | For correct statement: $P(B)\times P(A) = P(A\cap B)$ or $P(A)=P(A\mid B)$ **and** stating independent. Full marks in (d) OK even if 0/3 in (c) |8. For the events $A$ and $B$,
$$\mathrm { P } \left( A ^ { \prime } \cap B \right) = 0.22 \text { and } \mathrm { P } \left( A ^ { \prime } \cap B ^ { \prime } \right) = 0.18$$
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { P } ( A )$.
\item Find $\mathrm { P } ( A \cup B )$.
Given that $\mathrm { P } ( A \mid B ) = 0.6$
\item find $\mathrm { P } ( A \cap B )$.
\item Determine whether or not $A$ and $B$ are independent.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2014 Q8 [7]}}