| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2009 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Dice/random device selects population |
| Difficulty | Moderate -0.3 This is a straightforward conditional probability question using the law of total probability and Bayes' theorem with clearly presented data. Part (a) involves standard tree diagram calculations, while part (b) requires recognizing a permutation scenario (3 different locations from 3 selections). The question is slightly easier than average because the table structure makes the problem setup transparent and all required techniques are routine S3 applications with no conceptual surprises. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| \multirow{2}{*}{} | Type of location | ||||
| City | Coastal | Country | |||
| \multirow{3}{*}{Type of reservation} | Bed \ | Breakfast | 80 | 10 | 30 |
| Half Board | 15 | 65 | 50 | ||
| Full Board | 5 | 25 | 20 | ||
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(P(B\ \&\ B) = (0.30 \times 0.80) + (0.55 \times 0.10) + (0.15 \times 0.30)\) | M1 | Use of 3 possibilities each the product of 2 probabilities |
| \(= 0.24 + 0.055 + 0.045 = 0.34\) | A1 | CAO; AG |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(P(HB \cap \text{Coastal}) = 0.55 \times 0.65\) | M1 | Can be implied by correct answer |
| \(= 143/400\) or \(0.357\) to \(0.358\) | A1 | CAO/AWFW (0.3575) |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(P(\text{Coastal} \mid HB) = \frac{P(\text{Coastal} \cap HB)}{P(HB)}\) | M1, M1 | answer to (ii) / \(\sum(3\times2)\) probabilities |
| \(= \frac{0.3575}{(0.3 \times 0.15)+(0.3575)+(0.15 \times 0.5)}\) | A1F | F on (ii) |
| \(= \frac{0.3575}{0.4775} = 143/191\) or \(0.747\) to \(0.75\) | A1 | CAO/AWFW (0.74869) |
| Total | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(P(\text{City} \mid HB) = \frac{0.3 \times 0.15}{P(HB)} = \frac{0.045}{0.4775} = \frac{90}{955}\) | M1 | |
| \(P(\text{Country} \mid HB) = \frac{0.15 \times 0.5}{P(HB)} = \frac{0.075}{0.4775} = \frac{30}{191}\) | M1 | Or \(\left(1 - (a)(\text{iii}) - \frac{0.045}{0.4775}\right)\) |
| Thus Probability \(= \frac{0.045}{P(HB)} \times \frac{0.3575}{P(HB)} \times \frac{0.075}{P(HB)}\) | M1 | Multiplication of 3 different probabilities |
| Multiplied by \(3! = 6\) | B1 | CAO |
| \(= 0.09424 \times 0.74869 \times 0.15707 \times 6\) | ||
| \(= 0.063\) to \(0.068\) | A1 | AWFW (0.06649) |
| Total | 5 |
# Question 2:
## Part (a)(i):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $P(B\ \&\ B) = (0.30 \times 0.80) + (0.55 \times 0.10) + (0.15 \times 0.30)$ | M1 | Use of **3 possibilities** each the product of 2 probabilities |
| $= 0.24 + 0.055 + 0.045 = 0.34$ | A1 | CAO; **AG** |
| **Total** | **2** | |
## Part (a)(ii):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $P(HB \cap \text{Coastal}) = 0.55 \times 0.65$ | M1 | Can be implied by correct answer |
| $= 143/400$ or $0.357$ to $0.358$ | A1 | CAO/AWFW (0.3575) |
| **Total** | **2** | |
## Part (a)(iii):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $P(\text{Coastal} \mid HB) = \frac{P(\text{Coastal} \cap HB)}{P(HB)}$ | M1, M1 | answer to (ii) / $\sum(3\times2)$ probabilities |
| $= \frac{0.3575}{(0.3 \times 0.15)+(0.3575)+(0.15 \times 0.5)}$ | A1F | F on (ii) |
| $= \frac{0.3575}{0.4775} = 143/191$ or $0.747$ to $0.75$ | A1 | CAO/AWFW (0.74869) |
| **Total** | **4** | |
## Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $P(\text{City} \mid HB) = \frac{0.3 \times 0.15}{P(HB)} = \frac{0.045}{0.4775} = \frac{90}{955}$ | M1 | |
| $P(\text{Country} \mid HB) = \frac{0.15 \times 0.5}{P(HB)} = \frac{0.075}{0.4775} = \frac{30}{191}$ | M1 | Or $\left(1 - (a)(\text{iii}) - \frac{0.045}{0.4775}\right)$ |
| Thus Probability $= \frac{0.045}{P(HB)} \times \frac{0.3575}{P(HB)} \times \frac{0.075}{P(HB)}$ | M1 | Multiplication of 3 different probabilities |
| Multiplied by $3! = 6$ | B1 | CAO |
| $= 0.09424 \times 0.74869 \times 0.15707 \times 6$ | | |
| $= 0.063$ to $0.068$ | A1 | AWFW (0.06649) |
| **Total** | **5** | |
---2 A hotel chain has hotels in three types of location: city, coastal and country. The percentages of the chain's reservations for each of these locations are 30,55 and 15 respectively.
Each of the chain's hotels offers three types of reservation: Bed \& Breakfast, Half Board and Full Board.
The percentages of these types of reservation for each of the three types of location are shown in the table.
\begin{center}
\begin{tabular}{|l|l|l|l|l|}
\hline
\multicolumn{2}{|c|}{\multirow{2}{*}{}} & \multicolumn{3}{|c|}{Type of location} \\
\hline
& & City & Coastal & Country \\
\hline
\multirow{3}{*}{Type of reservation} & Bed \& Breakfast & 80 & 10 & 30 \\
\hline
& Half Board & 15 & 65 & 50 \\
\hline
& Full Board & 5 & 25 & 20 \\
\hline
\end{tabular}
\end{center}
For example, 80 per cent of reservations for hotels in city locations are for Bed \& Breakfast.
\begin{enumerate}[label=(\alph*)]
\item For a reservation selected at random:
\begin{enumerate}[label=(\roman*)]
\item show that the probability that it is for Bed \& Breakfast is 0.34 ;
\item calculate the probability that it is for Half Board in a hotel in a coastal location;
\item calculate the probability that it is for a hotel in a coastal location, given that it is for Half Board.
\end{enumerate}\item A random sample of 3 reservations for Half Board is selected.
Calculate the probability that these 3 reservations are for hotels in different types of location.
\end{enumerate}
\hfill \mbox{\textit{AQA S3 2009 Q2 [13]}}