| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Principle of Inclusion/Exclusion |
| Type | Basic Inclusion-Exclusion with Two Sets |
| Difficulty | Easy -1.3 This is a straightforward application of the inclusion-exclusion formula P(R∪S) = P(R) + P(S) - P(R∩S) with all values given, requiring only algebraic rearrangement. The Venn diagram and conditional probability are routine follow-ups using basic definitions. This is easier than average A-level content—purely procedural with no problem-solving or insight required. |
| 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 |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(R \cap S) = P(R) + P(S) - P(R \cup S)\) | M1 | For correct use of formula; Or \(0.28 - x + 0.87 - x + x = 0.94\) |
| \(= 0.28 + 0.87 - 0.94\) | ||
| \(= 0.21\) | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Venn diagram with two labelled intersecting circles with 0.07, 0.21, 0.66 and 0.06 | G1 | For two labelled intersecting circles; Allow labels such as \(P(R)\) and \(P(S)\); Allow other sensible shapes in place of circles |
| At least 2 correct probabilities, FT their \(P(R \cap S)\) | G1 | Allow their \(P(R \cap S)\) rounded to 2dp; For both G1 marks FT their 0.21 provided \(< 0.28\) |
| Remaining probabilities, FT their \(P(R \cap S)\) | G1 [3] | For FT if \(P(R \cap S) = x\) then others are \(0.28-x\), \(0.87-x\), \(x-0.15\); 0.2436 leads to 0.0364, 0.6264, 0.0936 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(R \mid S) = \frac{P(R \cap S)}{P(S)} = \frac{0.21}{0.87} = \frac{21}{87} = 0.241\) | M1 | for fraction; Allow \(\frac{7}{29}\) or \(\frac{21}{87}\) as final answer |
| Exact answer \(0.241379\ldots\) | A1 | CAO; FT their part (i) (for M1 only) but M0 if their answer to part (i) is \(P(R) \times P(S)\); Allow 0.24 with working; Condone 'if' or 'when' for 'given that' but not the words 'and' or 'because' or 'due to' for E1 |
| This is the probability that (on a randomly selected day) there is at least 1 mm of rain, given that there is at least 1 hour of sun. | E1 [3] | Need more than just probability of rain given sun; Must include 'probability' or 'chance'; Do not allow just P(at least 1 mm of rain, given that there is at least 1 hour of sun); E1 (independent of M1): the order/structure must be correct i.e. no reverse statement; Allow 'The probability that on a randomly selected day when there is at least 1 hour of sun there is at least 1 mm of rain.' oe |
# Question 5:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(R \cap S) = P(R) + P(S) - P(R \cup S)$ | M1 | For correct use of formula; Or $0.28 - x + 0.87 - x + x = 0.94$ |
| $= 0.28 + 0.87 - 0.94$ | | |
| $= 0.21$ | A1 [2] | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Venn diagram with two labelled intersecting circles with 0.07, 0.21, 0.66 and 0.06 | G1 | For two labelled intersecting circles; Allow labels such as $P(R)$ and $P(S)$; Allow other sensible shapes in place of circles |
| At least 2 correct probabilities, FT their $P(R \cap S)$ | G1 | Allow their $P(R \cap S)$ rounded to 2dp; For both G1 marks FT their 0.21 **provided $< 0.28$** |
| Remaining probabilities, FT their $P(R \cap S)$ | G1 [3] | For FT if $P(R \cap S) = x$ then others are $0.28-x$, $0.87-x$, $x-0.15$; 0.2436 leads to 0.0364, 0.6264, 0.0936 |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(R \mid S) = \frac{P(R \cap S)}{P(S)} = \frac{0.21}{0.87} = \frac{21}{87} = 0.241$ | M1 | for fraction; Allow $\frac{7}{29}$ or $\frac{21}{87}$ as final answer |
| Exact answer $0.241379\ldots$ | A1 | CAO; FT their part (i) (for M1 only) but M0 if their answer to part (i) is $P(R) \times P(S)$; Allow 0.24 with working; Condone 'if' or 'when' for 'given that' but not the words 'and' or 'because' or 'due to' for E1 |
| This is the probability that (on a randomly selected day) there is at least 1 mm of rain, given that there is at least 1 hour of sun. | E1 [3] | Need more than just probability of rain given sun; Must include 'probability' or 'chance'; Do not allow just P(at least 1 mm of rain, given that there is at least 1 hour of sun); E1 (independent of M1): the order/structure must be correct i.e. no reverse statement; Allow 'The probability that on a randomly selected day when there is at least 1 hour of sun there is at least 1 mm of rain.' oe |5 Measurements of sunshine and rainfall are made each day at a particular weather station. For a randomly chosen day,
\begin{itemize}
\item $R$ is the event that at least 1 mm of rainfall is recorded,
\item $S$ is the event that at least 1 hour of sunshine is recorded.
\end{itemize}
You are given that $\mathrm { P } ( R ) = 0.28 , \mathrm { P } ( S ) = 0.87$ and $\mathrm { P } ( R \cup S ) = 0.94$.\\
(i) Find $\mathrm { P } ( R \cap S )$.\\
(ii) Draw a Venn diagram showing the events $R$ and $S$, and fill in the probability corresponding to each of the four regions of your diagram.\\
(iii) Find $\mathrm { P } ( R \mid S )$ and state what this probability represents in this context.
\hfill \mbox{\textit{OCR MEI S1 2016 Q5 [8]}}