| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2011 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Independence test requiring preliminary calculations |
| Difficulty | Moderate -0.8 This is a straightforward S1 conditional probability question requiring basic probability rules (complement rule, addition rule, conditional probability formula, and independence test). All parts follow standard textbook procedures with no problem-solving insight needed—students simply apply memorized formulas to given values across multiple routine sub-questions. |
| 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 |
|---|---|
| P(\(J \cup K\)) = 1 − 0.7 or 0.1 + 0.15 + 0.05 = 0.3 | B1 |
| Answer | Marks |
|---|---|
| P(\(K\)) = 0.05 + 0.15 or "0.3" − 0.25 + 0.15 or "0.3" = 0.25 +P(\(K\)) − 0.15 | M1 |
| May be seen on Venn diagram = 0.2 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \([\text{P}(K | J)] = \frac{\text{P}(K \cap J)}{\text{P}(J)}\) | M1 |
| \(= \frac{0.15}{0.25}\) | A1 | |
| \(= \frac{3}{5}\) or 0.6 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| P(\(J\)) × P(\(K\)) = 0.25 × 0.2(= 0.05), P(\(J \cap K\)) = 0.15 or P(\(K | J\)) = 0.6, P(\(K\)) = 0.2 or may see P(J | K) = 0.75 and P(J) = 0.25 not equal therefore not independent |
| Answer | Marks |
|---|---|
| Not independent so confirms the teacher's suspicion or they are linked (This requires a statement about independence in (d) or in (e)) | B1ft |
| Answer | Marks |
|---|---|
| - 1st A1 for a correct ratio of probabilities (if this is seen the M1 is awarded by implication). Must be in (c). Condone no LHS but wrong LHS (e.g. P(K) or P(J | K)) is M0A0 |
| Answer | Marks |
|---|---|
| - (d) M1 for a correct comparison of known probabilities for an independence test - ft their values. E.g. P(J) × P(K) with P(\(J \cap K\)) or P(K | J) with P(K) [Must have expressions] |
## (a)
P($J \cup K$) = 1 − 0.7 or 0.1 + 0.15 + 0.05 = 0.3 | B1 |
## (b)
P($K$) = 0.05 + 0.15 or "0.3" − 0.25 + 0.15 or "0.3" = 0.25 +P($K$) − 0.15 | M1 |
May be seen on Venn diagram = 0.2 | A1 |
## (c)
$[\text{P}(K | J)] = \frac{\text{P}(K \cap J)}{\text{P}(J)}$ | M1 |
$= \frac{0.15}{0.25}$ | A1 |
$= \frac{3}{5}$ or 0.6 | A1 |
## (d)
P($J$) × P($K$) = 0.25 × 0.2(= 0.05), P($J \cap K$) = 0.15 or P($K | J$) = 0.6, P($K$) = 0.2 or may see P(J|K) = 0.75 and P(J) = 0.25 not equal therefore not independent | M1, A1ft |
## (e)
Not independent so confirms the teacher's suspicion or they are linked (This requires a statement about independence in (d) or in (e)) | B1ft |
**Notes:**
- **(b)** M1 for a complete method, follow through their 0.3, leading to a linear equation for P(K)
- NB You may see this Venn diagram: [diagram shown]
- A correct diagram (Venn or table) implies M1 in (b). Need not include box or 0.7. **Correct answer only is 2/2**
- In parts (c) and (d) they must have defined A and B
- **(c)** M1 for a correct expression (including ratio) in symbols.
- 1st A1 for a correct ratio of probabilities (if this is seen the M1 is awarded by implication). Must be in (c). Condone no LHS but wrong LHS (e.g. P(K) or P(J|K)) is M0A0
- 2nd A1 for correct answer as printed only. **Correct answer only 3/3**
- **Mark (d) and (e) together**
- **(d)** M1 for a correct comparison of known probabilities for an independence test - ft their values. E.g. P(J) × P(K) with P($J \cap K$) or P(K|J) with P(K) [Must have expressions]
- The values of these probabilities should be given unless they are in the question or stated elsewhere. A1ft for correct calculations and correct comment for their probabilities
- **(e)** B1ft for their conclusion on independence so not independent confirms teacher...independent contradicts teacher. **Methods leading to negative probabilities should score M0**
---Jake and Kamil are sometimes late for school.
The events $J$ and $K$ are defined as follows
$J =$ the event that Jake is late for school
$K =$ the event that Kamil is late for school
$\text{P}(J) = 0.25$, $\text{P}(J \cap K) = 0.15$ and $\text{P}(J' \cap K') = 0.7$
On a randomly selected day, find the probability that
\begin{enumerate}[label=(\alph*)]
\item at least one of Jake or Kamil are late for school, [1]
\item Kamil is late for school. [2]
\end{enumerate}
Given that Jake is late for school,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item find the probability that Kamil is late. [3]
\end{enumerate}
The teacher suspects that Jake being late for school and Kamil being late for school are linked in some way.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Determine whether or not $J$ and $K$ are statistically independent. [2]
\item Comment on the teacher's suspicion in the light of your calculation in (d). [1]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2011 Q6 [9]}}