| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | October |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Principle of Inclusion/Exclusion |
| Type | Constrained Survey to Venn Diagram |
| Difficulty | Standard +0.3 This is a standard S1 Venn diagram question with three sets and given intersections. While it requires careful bookkeeping to fill in all regions correctly and involves multiple conditional probability calculations, it follows a completely routine template with no novel problem-solving required. The inclusion-exclusion principle is applied mechanically, making this slightly easier than average for A-level. |
| 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 |
|---|---|---|
| 3(a) | Four correctly labelled Venn diagrams showing: 1st B1 for 3 labelled circles with 12, 13 & n(C ∩ G) = 0 marked or implied (e.g. RH diagram); 2nd B1 for 8 and 10 correctly placed; 3rd B1 for 23 correctly placed; 4th B1 for box and 14 | B1, B1, B1, B1 |
| 3(b)(i) | \(P(S) = \left[\frac{12 + 23 + 13}{80}\right] = \frac{48}{80}\) or \(\frac{3}{5}\) or 0.6 | B1ft |
| 3(b)(ii) | \(P(S | C) = \frac{P(S \cap C)}{P(C)} = \frac{12/80}{20/80} = \frac{12}{20}\) or 0.6 |
| 3(b)(iii) | \(P(S) = P(S | C)\) or \(P(C) = 0.25\), \(P(C \cap S) = 0.15\) and \(P(C) \times P(S) = 0.6 \times 0.25\) so S and C are independent |
| 3(c) | Need \(P(S | G) = \frac{13}{?}\). \(P(S |
**3(a)** | Four correctly labelled Venn diagrams showing: 1st B1 for 3 labelled circles with 12, 13 & n(C ∩ G) = 0 marked or implied (e.g. RH diagram); 2nd B1 for 8 and 10 correctly placed; 3rd B1 for 23 correctly placed; 4th B1 for box and 14 | B1, B1, B1, B1 |
**3(b)(i)** | $P(S) = \left[\frac{12 + 23 + 13}{80}\right] = \frac{48}{80}$ or $\frac{3}{5}$ or 0.6 | B1ft | For 0.6 or any exact equivalent (single fraction) or ft their values (ft blank as 0)
**3(b)(ii)** | $P(S | C) = \frac{P(S \cap C)}{P(C)} = \frac{12/80}{20/80} = \frac{12}{20}$ or 0.6 | M1, A1cso | M1 for correct conditional prob. Correct expression and one correct ft prob. Num < Den; A1cso for 0.6 which must come from a denominator of 20
**3(b)(iii)** | $P(S) = P(S|C)$ or $P(C) = 0.25$, $P(C \cap S) = 0.15$ and $P(C) \times P(S) = 0.6 \times 0.25$ so S and C are independent | B1ft, dB1ft | 1st B1ft for full reason. If not P(S) = P(S|C) then all values must be stated, labelled and correct or correct ft from diagram. Correct not'n required so $P(S \cup C) = 0.15$ is B0B0; 2nd dB1ft dep. on correct reason for correct conclusion for their values
**3(c)** | Need $P(S | G) = \frac{13}{?}$. $P(S | C) = 0.6 > 0.565$ so assistant selling coats has better performance | M1A1, A1 | M1 for attempt at $P(S | G)$ correct ratio of probabilities or numbers using their figs; 1st A1 for $\frac{13}{?}$ (accept awrt 0.565) [Sight of $P(S | G) = \frac{13}{23}$ is M1A1]; 2nd A1 for correct conclusion that chooses "coats" based on correct comparison. Allow incorrect $P(S|C)$ provided $> 0.565$ to score 2nd A1 and so all 3 marks. Condone poor use of notation eg $S|G$ with no P(…). Probabilities may be described in words. Condone comparison of $\frac{13}{23}$ with 0.6 even if $\frac{13}{23}$ not labelled as $P(S|G)$
---\begin{enumerate}
\item Hugo recorded the purchases of 80 customers in the ladies fashion department of a large store. His results were as follows
\end{enumerate}
20 customers bought a coat\\
12 customers bought a coat and a scarf\\
23 customers bought a pair of gloves\\
13 customers bought a pair of gloves and a scarf no customer bought a coat and a pair of gloves 14 customers did not buy a coat nor a scarf nor a pair of gloves.\\
(a) Draw a Venn diagram to represent all of this information.\\
(b) One of the 80 customers is selected at random.\\
(i) Find the probability that the customer bought a scarf.\\
(ii) Given that the customer bought a coat, find the probability that the customer also bought a scarf.\\
(iii) State, giving a reason, whether or not the event 'the customer bought a coat' and the event 'the customer bought a scarf' are statistically independent.
Hugo had asked the member of staff selling coats and the member of staff selling gloves to encourage customers also to buy a scarf.\\
(c) By considering suitable conditional probabilities, determine whether the member of staff selling coats or the member of staff selling gloves has the better performance at selling scarves to their customers. Give a reason for your answer.
\hfill \mbox{\textit{Edexcel S1 2016 Q3 [12]}}