| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2003 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Calculate end-outcome probability |
| Difficulty | Moderate -0.8 This is a straightforward tree diagram question requiring basic probability calculations with conditional probability in part (iii). The setup is clear, calculations are routine (simple fractions), and the tree diagram structure is standard with no conceptual challenges beyond applying basic probability rules. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03d Calculate conditional probability: from first principles |
| House number 1 | House number 2 | House number 3 |
| \(4 C , 1 P , 2 G\) | \(2 C , 2 P , 3 G\) | \(1 C , 1 G\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(G) = \text{number of grandparents}/\text{total people}\) | M1 | For appreciating total grandparents/total people, can be implied |
| \(= \frac{6}{16} = \frac{3}{8}\) | A1 (2) | For correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(H1,G) + P(H2,G) + P(H3,G)\) | B1 | For any correct 2-factor product, need not be evaluated |
| \(= \frac{1}{3}\times\frac{2}{7} + \frac{1}{3}\times\frac{3}{7} + \frac{1}{3}\times\frac{1}{2} = \frac{17}{42}\) \((= 0.405)\) | M1 | For addition of 3 relevant 2-factor products |
| A1 (3) | For correct answer or equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(H1\ | G) + P(H2\ | G)\) |
| \(= \dfrac{2/21}{17/42} + \dfrac{3/21}{17/42} = \dfrac{10}{17}\) | M1 | For dividing by answer to (ii), only if not multiplied as well, and \(p < 1\) |
| A1 | For one correct probability | |
| A1 (4) | For correct answer or equivalent | |
| OR \(P(H3\ | G) = 7/17\); Answer \(= 1 - 7/17 = 10/17\) | M1/M1/A2 |
## Question 6:
**(i)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(G) = \text{number of grandparents}/\text{total people}$ | M1 | For appreciating total grandparents/total people, can be implied |
| $= \frac{6}{16} = \frac{3}{8}$ | A1 (2) | For correct answer |
**(ii)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(H1,G) + P(H2,G) + P(H3,G)$ | B1 | For any correct 2-factor product, need not be evaluated |
| $= \frac{1}{3}\times\frac{2}{7} + \frac{1}{3}\times\frac{3}{7} + \frac{1}{3}\times\frac{1}{2} = \frac{17}{42}$ $(= 0.405)$ | M1 | For addition of 3 relevant 2-factor products |
| | A1 (3) | For correct answer or equivalent |
**(iii)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(H1\|G) + P(H2\|G)$ | M1 | For summing exactly 2 probability options |
| $= \dfrac{2/21}{17/42} + \dfrac{3/21}{17/42} = \dfrac{10}{17}$ | M1 | For dividing by answer to **(ii)**, only if not multiplied as well, and $p < 1$ |
| | A1 | For one correct probability |
| | A1 (4) | For correct answer or equivalent |
| **OR** $P(H3\|G) = 7/17$; Answer $= 1 - 7/17 = 10/17$ | M1/M1/A2 | For finding prob. options no parents / for subt. from 1 / for correct answer |
---6 The people living in 3 houses are classified as children ( $C$ ), parents ( $P$ ) or grandparents ( $G$ ). The numbers living in each house are shown in the table below.
\begin{center}
\begin{tabular}{ | c | c | c | }
\hline
House number 1 & House number 2 & House number 3 \\
\hline
$4 C , 1 P , 2 G$ & $2 C , 2 P , 3 G$ & $1 C , 1 G$ \\
\hline
\end{tabular}
\end{center}
(i) All the people in all 3 houses meet for a party. One person at the party is chosen at random. Calculate the probability of choosing a grandparent.\\
(ii) A house is chosen at random. Then a person in that house is chosen at random. Using a tree diagram, or otherwise, calculate the probability that the person chosen is a grandparent.\\
(iii) Given that the person chosen by the method in part (ii) is a grandparent, calculate the probability that there is also a parent living in the house.
\hfill \mbox{\textit{CAIE S1 2003 Q6 [9]}}