| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2009 |
| Session | January |
| Marks | 8 |
| 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 conditional probability question requiring basic probability rules (multiplication rule, complement rule) and independence testing. The calculations are mechanical with clearly stated probabilities, typical of S1 material but simpler than average A-level questions since it only requires applying standard formulas without problem-solving insight. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(E \cap B) = P(E | B) \times P(B) = \frac{9}{25} \times \frac{2}{3} = 0.24\) or \(\frac{6}{25}\) or \(\frac{18}{75}\) | M1, A1 (2) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(E \cup B) = \frac{2}{3} + \frac{2}{5} - \frac{6}{25} = \frac{62}{75}\) | M1, A1 | 1st M1 for use of addition rule; must have 3 terms and some values |
| or \(P(E' | B')= \frac{13}{25}\), or \(P(B' \cap E) = \frac{12}{75}\), or \(P(B \cap E') = \frac{32}{75}\) | |
| \(P(E' \cap B') = 1 - P(E \cup B) = \frac{13}{75}\) or \(0.17\dot{3}\) | M1 A1 (4) | 2nd M1 for method leading to answer e.g. \(1 - P(E \cup B)\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(E | B) = 0.36 \neq 0.40 = P(E)\) or \(P(E \cap B) = \frac{6}{25} \neq \frac{2}{5} \times \frac{2}{3} = P(E) \times P(B)\) | M1, A1 |
| So \(E\) and \(B\) are not statistically independent | (2) | A1 for correct values and correct comment |
## Question 2:
**(a)**
$E$ = take regular exercise, $B$ = always eat breakfast
$P(E \cap B) = P(E|B) \times P(B) = \frac{9}{25} \times \frac{2}{3} = 0.24$ or $\frac{6}{25}$ or $\frac{18}{75}$ | M1, A1 (2) | M1 for $\frac{9}{25} \times \frac{2}{3}$ or $P(E|B) \times P(B)$ and at least one correct value. NB $\frac{2}{5} \times \frac{2}{3}$ alone or $\frac{2}{5} \times \frac{9}{25}$ alone scores M0A0
**(b)**
$P(E \cup B) = \frac{2}{3} + \frac{2}{5} - \frac{6}{25} = \frac{62}{75}$ | M1, A1 | 1st M1 for use of addition rule; must have 3 terms and some values
or $P(E'|B')= \frac{13}{25}$, or $P(B' \cap E) = \frac{12}{75}$, or $P(B \cap E') = \frac{32}{75}$ | | Alternative valid methods accepted
$P(E' \cap B') = 1 - P(E \cup B) = \frac{13}{75}$ or $0.17\dot{3}$ | M1 A1 (4) | 2nd M1 for method leading to answer e.g. $1 - P(E \cup B)$
**(c)**
$P(E|B) = 0.36 \neq 0.40 = P(E)$ or $P(E \cap B) = \frac{6}{25} \neq \frac{2}{5} \times \frac{2}{3} = P(E) \times P(B)$ | M1, A1 | M1 for identifying suitable values to test for independence
So $E$ and $B$ are not statistically independent | (2) | A1 for correct values and correct comment
---2. A group of office workers were questioned for a health magazine and $\frac { 2 } { 5 }$ were found to take regular exercise. When questioned about their eating habits $\frac { 2 } { 3 }$ said they always eat breakfast and, of those who always eat breakfast $\frac { 9 } { 25 }$ also took regular exercise.
Find the probability that a randomly selected member of the group
\begin{enumerate}[label=(\alph*)]
\item always eats breakfast and takes regular exercise,
\item does not always eat breakfast and does not take regular exercise.
\item Determine, giving your reason, whether or not always eating breakfast and taking regular exercise are statistically independent.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2009 Q2 [8]}}