Moderate -0.5 This is a straightforward conditional probability question requiring application of the law of total probability and Bayes' theorem with clearly structured data. While it involves multiple steps and careful bookkeeping across three room types, the mathematical techniques are standard S3 fare with no conceptual surprises or novel problem-solving required.
3 A hotel has three types of room: double, twin and suite. The percentage of rooms in the hotel of each type is 40,45 and 15 respectively.
Each room in the hotel may be occupied by \(0,1,2\), or 3 or more people.
The proportional occupancy of each type of room is shown in the table.
I can see these are answer space pages (blank lined pages for students to write their answers) for Questions 3 and 4 of what appears to be a statistics exam paper (P47644/Jun12/MS03).
However, these images only show blank answer spaces — they do not contain any mark scheme content. The pages shown are:
- Pages 7, 8, 9: "Answer space for question 3" (blank)
- Pages 10, 11: Question 4 text + "Answer space for question 4" (blank)
# Question 3:
## Part (a)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{unoccupied suite}) = 0.15 \times 0.10 = 0.015$ | B1 | |
## Part (a)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{2 or more}) = 0.40(0.45+0.05)+0.45(0.30+0.10)+0.15(0.55+0.15)$ | M1 | Correct method using room proportions |
| $= 0.40(0.50)+0.45(0.40)+0.15(0.70) = 0.20+0.18+0.105 = 0.485$ | A1 | |
## Part (a)(iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{unoccupied}) = 0.40(0.15)+0.45(0.05)+0.15(0.10)$ | M1 | |
| $= 0.06+0.0225+0.015 = 0.0975$ | A1 | |
## Part (a)(iv)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{double} \mid \text{unoccupied}) = \frac{0.40 \times 0.15}{0.0975} = \frac{0.06}{0.0975} = \frac{4}{6.5} \approx 0.615$ | M1 A1 | Correct conditional probability |
## Part (a)(v)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{occupied}) = 1 - 0.0975 = 0.9025$ | M1 | |
| $P(\text{suite} \mid \text{occupied}) = \frac{0.15(0.90)}{0.9025} = \frac{0.135}{0.9025} \approx 0.1496$ | M1 A1 | |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{occupied double})=\frac{0.40\times0.85}{0.9025}$, $P(\text{occupied twin})=\frac{0.45\times0.95}{0.9025}$, $P(\text{occupied suite})=\frac{0.15\times0.90}{0.9025}$ | M1 | Finding conditional probabilities of room type given occupied |
| $P(\text{all different}) = 3! \times \frac{0.34}{0.9025}\times\frac{0.4275}{0.9025}\times\frac{0.135}{0.9025}$ | M1 A1 | Multiply by $3!$ for arrangements |
| $\approx 6 \times 0.3768 \times 0.4737 \times 0.1496 \approx 0.1602$ | A1 | awrt 0.160 |
I can see these are answer space pages (blank lined pages for students to write their answers) for Questions 3 and 4 of what appears to be a statistics exam paper (P47644/Jun12/MS03).
However, these images only show **blank answer spaces** — they do not contain any mark scheme content. The pages shown are:
- Pages 7, 8, 9: "Answer space for question 3" (blank)
- Pages 10, 11: Question 4 text + "Answer space for question 4" (blank)
3 A hotel has three types of room: double, twin and suite. The percentage of rooms in the hotel of each type is 40,45 and 15 respectively.
Each room in the hotel may be occupied by $0,1,2$, or 3 or more people.
The proportional occupancy of each type of room is shown in the table.
\hfill \mbox{\textit{AQA S3 2012 Q3 [14]}}