| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2011 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Independent Events |
| Type | Test independence using definition |
| Difficulty | Moderate -0.8 This is a straightforward S1 probability question testing basic concepts: Venn diagrams, independence test (checking if P(W∩F)=P(W)×P(F)), and conditional probability formula. All parts require direct application of standard formulas with no problem-solving insight or multi-step reasoning. Easier than average A-level. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables |
| Answer | Marks |
|---|---|
| Marks: | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Marks: M1 for \(0.41 \times 0.14\), A1 Condone dependent, Must have full method, www | 2 | |
| Guidance: Answer of 0.574 gets Max M1A0. Omission of 0.0574 gets M1A0 Max. Or: \(P(W | F) = 0.11/0.41 = 0.268 \neq P(W) (= 0.14)\) M1 for full working. \(P(F | W) = 0.11/0.14 = 0.786 \neq P(F) (= 0.41)\) M1 for full working. No marks without correct working |
| Answer | Marks |
|---|---|
| Answer/Working: \(P(W | F) = \frac{P(W \cap F)}{P(F)} = \frac{0.11}{0.41} = \frac{11}{41} = 0.268\). This is the probability that a randomly selected respondent works part time, given that the respondent is female. |
| Marks: M1 for correct fraction, A1, E1 | 3 |
## (i)
**Answer/Working:** G1 for two labelled intersecting circles, G1 for at least 2 correct probabilities, G1 for remaining correct probabilities
**Marks:** | 3
**Guidance:** Allow labels such as P(W) and P(F). Allow other sensible shapes in place of circles
## (ii)
**Answer/Working:** $P(W) \times P(F) = 0.14 \times 0.41 = 0.0574 \neq P(W \cap F) = 0.11$. So not independent.
**Marks:** M1 for $0.41 \times 0.14$, A1 Condone dependent, Must have full method, www | 2
**Guidance:** Answer of 0.574 gets Max M1A0. Omission of 0.0574 gets M1A0 Max. Or: $P(W|F) = 0.11/0.41 = 0.268 \neq P(W) (= 0.14)$ M1 for full working. $P(F|W) = 0.11/0.14 = 0.786 \neq P(F) (= 0.41)$ M1 for full working. No marks without correct working
## (iii)
**Answer/Working:** $P(W | F) = \frac{P(W \cap F)}{P(F)} = \frac{0.11}{0.41} = \frac{11}{41} = 0.268$. This is the probability that a randomly selected respondent works part time, given that the respondent is female.
**Marks:** M1 for correct fraction, A1, E1 | 3
**Guidance:** Allow 0.27 with working. Allow 11/41 as final answer. Condone 'if' or 'when' for 'given that' but not the words 'and' or 'because' or 'due to' for E1. E1 (independent of M1): the order/structure must be correct i.e. no reverse statement. Allow 'The probability that a randomly selected female respondent works part time' oe
---In a recent survey, a large number of working people were asked whether they worked full-time or part-time, with part-time being defined as less than 25 hours per week. One of the respondents is selected at random.
• $W$ is the event that this person works part-time.
• $F$ is the event that this person is female.
You are given that P($W$) = 0.14, P($F$) = 0.41 and P($W \cap F$) = 0.11.
\begin{enumerate}[label=(\roman*)]
\item Draw a Venn diagram showing the events $W$ and $F$, and fill in the probability corresponding to each of the four regions of your diagram. [3]
\item Determine whether the events $W$ and $F$ are independent. [2]
\item Find P($W$ | $F$) and explain what this probability represents. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 2011 Q5 [8]}}