| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Describe or suggest sampling method |
| Difficulty | Moderate -0.8 This is a straightforward data representation and basic probability question requiring table completion using simple arithmetic, followed by routine probability calculations (simple fractions, conditional probability, independence test, and binomial probability). All techniques are standard S1 content with no problem-solving insight needed. |
| Spec | 2.03a Mutually exclusive and independent events2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Designer labels | No designer labels | Total | |
| High-heeled shoes | |||
| Low-heeled shoes | |||
| Sports shoes | |||
| Total | 20 |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | Designer | Not designer |
| H-h shoes | 2 | 4 |
| L-h shoes | 1 | 3 |
| Sports | 5 | 5 |
| Total | 8 | 12 |
| (ii) 1/20 (0.05) | B1ft [1] | correct answer, ft their table |
| (iii) 10/20 (1/2, 0.5) | B1ft [1] | correct answer, ft their table |
| (iv) 2/8 (1/4, 0.25) | B1ft [1] | correct final answer, ft their table |
| (v) \(P(D) = 8/20\) (0.4); \(P(S) = 10/20\) (0.5); \(P(D \cap S) = 5/20\) (0.25); Not independent as \(P(D) \times P(S) \neq P(D \cap S)\) | M1 A1ft [2] | finding \(P(D \cap S)\) and comparing with their \(P(D) \times P(S)\); correct conclusion, ft their table |
| [OR1] \(P(D | S) = \frac{P(D \cap S)}{P(S)} = \frac{5}{10}\); \(P(D) = \frac{8}{20}\); Not independent as \(P(D | S) \neq P(D)\) |
| [OR2] \(P(S | D) = \frac{P(S \cap D)}{P(D)} = \frac{5}{8}\); \(P(S) = \frac{10}{20}\); Not independent as \(P(S | D) \neq P(S)\) |
| (vi) \(P(\text{at most 4}) = 1 - {^7C_5}(0.4)^5(0.6)^2 - {^7C_6}(0.4)^6(0.6)^1 - (0.4)^7 = 0.904\) | M1 M1 A1 [3] | bin probability of form \(^nC_r p^r(1-p)^{n-r}\), \(r \neq 0\) or 7; bin expression for \(1 - P(5, 6, 7)\) or \(P(0, 1, 2, 3, 4)\), any \(p\); correct answer |
(i) | Designer | Not designer | Total |
| :---: | :---: | :---: | :---: |
| H-h shoes | 2 | 4 | 6 |
| L-h shoes | 1 | 3 | 4 |
| Sports | 5 | 5 | 10 |
| Total | 8 | 12 | 20 | | B1 B1 [2] | one row or column correct; all correct
(ii) 1/20 (0.05) | B1ft [1] | correct answer, ft their table
(iii) 10/20 (1/2, 0.5) | B1ft [1] | correct answer, ft their table
(iv) 2/8 (1/4, 0.25) | B1ft [1] | correct final answer, ft their table
(v) $P(D) = 8/20$ (0.4); $P(S) = 10/20$ (0.5); $P(D \cap S) = 5/20$ (0.25); Not independent as $P(D) \times P(S) \neq P(D \cap S)$ | M1 A1ft [2] | finding $P(D \cap S)$ and comparing with their $P(D) \times P(S)$; correct conclusion, ft their table
[OR1] $P(D|S) = \frac{P(D \cap S)}{P(S)} = \frac{5}{10}$; $P(D) = \frac{8}{20}$; Not independent as $P(D|S) \neq P(D)$ | | finding $P(D|S)$ and comparing with their $P(D)$; correct conclusion, ft their table
[OR2] $P(S|D) = \frac{P(S \cap D)}{P(D)} = \frac{5}{8}$; $P(S) = \frac{10}{20}$; Not independent as $P(S|D) \neq P(S)$ | | finding $P(D|S)$ and comparing with their $P(D)$; correct conclusion, ft their table
(vi) $P(\text{at most 4}) = 1 - {^7C_5}(0.4)^5(0.6)^2 - {^7C_6}(0.4)^6(0.6)^1 - (0.4)^7 = 0.904$ | M1 M1 A1 [3] | bin probability of form $^nC_r p^r(1-p)^{n-r}$, $r \neq 0$ or 7; bin expression for $1 - P(5, 6, 7)$ or $P(0, 1, 2, 3, 4)$, any $p$; correct answer5 Suzanne has 20 pairs of shoes, some of which have designer labels. She has 6 pairs of high-heeled shoes, of which 2 pairs have designer labels. She has 4 pairs of low-heeled shoes, of which 1 pair has designer labels. The rest of her shoes are pairs of sports shoes. Suzanne has 8 pairs of shoes with designer labels in total.\\
(i) Copy and complete the table below to show the number of pairs in each category.
\begin{center}
\begin{tabular}{|l|l|l|l|}
\hline
& Designer labels & No designer labels & Total \\
\hline
High-heeled shoes & & & \\
\hline
Low-heeled shoes & & & \\
\hline
Sports shoes & & & \\
\hline
Total & & & 20 \\
\hline
\end{tabular}
\end{center}
Suzanne chooses 1 pair of shoes at random to wear.\\
(ii) Find the probability that she wears the pair of low-heeled shoes with designer labels.\\
(iii) Find the probability that she wears a pair of sports shoes.\\
(iv) Find the probability that she wears a pair of high-heeled shoes, given that she wears a pair of shoes with designer labels.\\
(v) State with a reason whether the events 'Suzanne wears a pair of shoes with designer labels' and 'Suzanne wears a pair of sports shoes' are independent.
Suzanne chooses 1 pair of shoes at random each day.\\
(vi) Find the probability that Suzanne wears a pair of shoes with designer labels on at most 4 days out of the next 7 days.
\hfill \mbox{\textit{CAIE S1 2012 Q5 [10]}}