| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | March |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Standard Bayes with discrete events |
| Difficulty | Moderate -0.3 Part (i) is a straightforward Bayes' theorem application with clearly stated probabilities requiring basic tree diagram work and conditional probability formula. Part (ii) involves a simple geometric probability inequality (0.65^n < 0.002) solvable with logarithms. Both parts are routine S1 exercises with no conceptual challenges beyond standard textbook methods, making this slightly easier than average A-level difficulty. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{Abroad} \mid \text{camping}) = \dfrac{P(A \cap C)}{P(A \cap C) + P(H \cap C)} = \dfrac{0.35 \times 0.15}{0.35 \times 0.15 + 0.65 \times 0.4} = \dfrac{0.0525}{0.3125} = 0.168\) | M1, A1, M1, A1, A1 (5) | Attempt at \(P(A \cap C)\) seen alone anywhere; correct answer seen as num or denom of fraction; attempt at \(P(C)\) seen anywhere; correct unsimplified answer seen as num or denom; correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| \((0.65)^n < 0.002\); \(n > \frac{\lg(0.002)}{\lg(0.65)}\); \(n = 15\) | M1, M1, A1 (3) | Eqn with 0.65 or 0.35, power \(n\), 0.002 or 0.998; attempt to solve by logs or trial and error need a power; correct answer |
## Question 5:
**(i)**
$P(\text{Abroad} \mid \text{camping}) = \dfrac{P(A \cap C)}{P(A \cap C) + P(H \cap C)} = \dfrac{0.35 \times 0.15}{0.35 \times 0.15 + 0.65 \times 0.4} = \dfrac{0.0525}{0.3125} = 0.168$ | M1, A1, M1, A1, A1 (5) | Attempt at $P(A \cap C)$ seen alone anywhere; correct answer seen as num or denom of fraction; attempt at $P(C)$ seen anywhere; correct unsimplified answer seen as num or denom; correct answer
**(ii)**
$(0.65)^n < 0.002$; $n > \frac{\lg(0.002)}{\lg(0.65)}$; $n = 15$ | M1, M1, A1 (3) | Eqn with 0.65 or 0.35, power $n$, 0.002 or 0.998; attempt to solve by logs or trial and error need a power; correct answer
---5 In a certain town, 35\% of the people take a holiday abroad and 65\% take a holiday in their own country. Of those going abroad $80 \%$ go to the seaside, $15 \%$ go camping and $5 \%$ take a city break. Of those taking a holiday in their own country, $20 \%$ go to the seaside and the rest are divided equally between camping and a city break.\\
(i) A person is chosen at random. Given that the person chosen goes camping, find the probability that the person goes abroad.\\
(ii) A group of $n$ people is chosen randomly. The probability of all the people in the group taking a holiday in their own country is less than 0.002 . Find the smallest possible value of $n$.
\hfill \mbox{\textit{CAIE S1 2016 Q5 [8]}}