| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2020 |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Definitions |
| Type | Independent events test |
| Difficulty | Standard +0.3 This is a straightforward probability question involving independent events and basic counting. Students must calculate probabilities of matching numbers across bags, find intersections, and test independence using P(X∩Y)=P(X)P(Y). While it requires careful enumeration and multiple parts, the concepts are standard S1 material with no novel insight needed—slightly easier than average A-level. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X) = P(\text{exactly 2 balls have same number})\) \(P(2, N2, 2) = \frac{1}{4} \times 1 \times \frac{1}{4} = \frac{1}{28}\) (? see image) | M1 | Considering at least two options of 2s and 8s |
| \(P(8, 8, N8) = \frac{1}{4} \times \frac{3}{5} \times \frac{3}{7} = \frac{3}{70}\) | M1 | Considering three options for the 8s |
| \(P(8, N8, 8) = \frac{1}{4} \times \frac{3}{5} \times \frac{4}{7} = \frac{3}{35}\) | M1 | Summing their options if more than 3 in total |
| \(P(N8, 8, 8) = \frac{3}{4} \times \frac{2}{5} \times \frac{4}{7} = \frac{6}{35}\) | B1 | One option correct |
| \(P(X) = \text{sum} = \frac{47}{140}\ (0.336)\) | A1 | |
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X \cap Y) = P(4, 8, 8) = \frac{1}{4} \times \frac{2}{5} \times \frac{4}{7} = \frac{2}{35}\) | B1 | |
| \(P(Y) = \frac{1}{4}\) \(\frac{2}{35} \neq \frac{47}{140} \times \frac{1}{4}\) | M1 | Attempt to compare \(P(X \cap Y)\) with \(P(X) \times P(Y)\) or using conditional probabilities |
| Not independent | A1 | Correct answer, correct working only |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(2, 2 \text{ given same}) = \frac{1}{28} \div \frac{47}{140}\) | M1 | \(\frac{1}{28}\) in numerator of a fraction |
| \(= \frac{5}{47}\ (0.106)\) | A1 | |
| Total: 2 |
## Question 7:
**Part 7(a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X) = P(\text{exactly 2 balls have same number})$ $P(2, N2, 2) = \frac{1}{4} \times 1 \times \frac{1}{4} = \frac{1}{28}$ (? see image) | M1 | Considering at least two options of 2s and 8s |
| $P(8, 8, N8) = \frac{1}{4} \times \frac{3}{5} \times \frac{3}{7} = \frac{3}{70}$ | M1 | Considering three options for the 8s |
| $P(8, N8, 8) = \frac{1}{4} \times \frac{3}{5} \times \frac{4}{7} = \frac{3}{35}$ | M1 | Summing their options if more than 3 in total |
| $P(N8, 8, 8) = \frac{3}{4} \times \frac{2}{5} \times \frac{4}{7} = \frac{6}{35}$ | B1 | One option correct |
| $P(X) = \text{sum} = \frac{47}{140}\ (0.336)$ | A1 | |
| **Total: 5** | | |
**Part 7(b):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X \cap Y) = P(4, 8, 8) = \frac{1}{4} \times \frac{2}{5} \times \frac{4}{7} = \frac{2}{35}$ | B1 | |
| $P(Y) = \frac{1}{4}$ $\frac{2}{35} \neq \frac{47}{140} \times \frac{1}{4}$ | M1 | Attempt to compare $P(X \cap Y)$ with $P(X) \times P(Y)$ or using conditional probabilities |
| Not independent | A1 | Correct answer, correct working only |
| **Total: 3** | | |
**Part 7(c):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(2, 2 \text{ given same}) = \frac{1}{28} \div \frac{47}{140}$ | M1 | $\frac{1}{28}$ in numerator of a fraction |
| $= \frac{5}{47}\ (0.106)$ | A1 | |
| **Total: 2** | | |7 Bag $A$ contains 4 balls numbered 2, 4, 5, 8. Bag $B$ contains 5 balls numbered 1, 3, 6, 8, 8. Bag $C$ contains 7 balls numbered $2,7,8,8,8,8,9$. One ball is selected at random from each bag.
\begin{itemize}
\item Event $X$ is 'exactly two of the selected balls have the same number'.
\item Event $Y$ is 'the ball selected from bag $A$ has number 4'.\\
(a) Find $\mathrm { P } ( X )$.\\
(b) Find $\mathrm { P } ( X \cap Y )$ and hence determine whether or not events $X$ and $Y$ are independent.\\
(c) Find the probability that two balls are numbered 2, given that exactly two of the selected balls have the same number.\\
\end{itemize}
\hfill \mbox{\textit{CAIE S1 2020 Q7 [10]}}