| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2015 |
| Session | June |
| Marks | 12 |
| 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 in a table. Parts (a)(i)-(iv) are standard textbook applications requiring simple multiplication and division of probabilities. Part (b) requires recognizing that given all 3 contain coupons, we need P(one of each variety | all have coupons), which involves multinomial probability but is still routine for S3 level. The question is slightly easier than average A-level due to clear structure and no conceptual surprises. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| \cline { 2 - 4 } \multicolumn{1}{c|}{} | Variety | ||
| \cline { 2 - 4 } \multicolumn{1}{c|}{} | Standard | Light | Very light |
| No coupon | 0.70 | 0.65 | 0.55 |
| £1 coupon | 0.20 | 0.25 | 0.30 |
| £2 coupon | 0.08 | 0.06 | 0.10 |
| £4 coupon | 0.02 | 0.04 | 0.05 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| (i) \(P(\text{standard} \cap £1) = 0.55 \times 0.20 = 0.110\) | B1 | |
| (ii) \(P(\text{no coupon}) = 0.55\times0.70 + 0.30\times0.65 + 0.15\times0.55\) | M1 | Correct method |
| \(= 0.385 + 0.195 + 0.0825 = 0.6625\) | A1 | |
| (iii) \(P(\text{light} \mid \text{no coupon}) = \frac{0.30\times0.65}{0.6625} = \frac{0.195}{0.6625}\) | M1 | Correct conditional probability method |
| \(= 0.2943...\) awrt \(0.294\) | A1 | |
| (iv) \(P(\text{very light} \mid \text{coupon}) = \frac{0.15\times0.45}{1-0.6625} = \frac{0.0675}{0.3375}\) | M1 | Correct conditional structure |
| \(= 0.2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(\text{coupon} \mid \text{standard}) = 0.30\), \(P(\text{coupon} \mid \text{light}) = 0.35\), \(P(\text{coupon} \mid \text{very light}) = 0.45\) | B1 | At least one conditional probability correct |
| \(P(\text{standard}\mid\text{coupon}) = \frac{0.55\times0.30}{0.3375}\), \(P(\text{light}\mid\text{coupon}) = \frac{0.30\times0.35}{0.3375}\), \(P(\text{very light}\mid\text{coupon}) = 0.2\) | M1 | Correct conditional probabilities for all three |
| \(= \frac{0.165}{0.3375}\), \(\frac{0.105}{0.3375}\), \(\frac{0.0675}{0.3375}\) | A1 | All three correct |
| \(P(\text{all different variety}) = 3! \times \frac{0.165}{0.3375} \times \frac{0.105}{0.3375} \times \frac{0.2}{1}\) | M1 | Multiplying by \(3!\) |
| \(= 6 \times 0.4889 \times 0.3111 \times 0.2 = 0.1829...\) awrt \(0.183\) | A1 |
# Question 3:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| **(i)** $P(\text{standard} \cap £1) = 0.55 \times 0.20 = 0.110$ | B1 | |
| **(ii)** $P(\text{no coupon}) = 0.55\times0.70 + 0.30\times0.65 + 0.15\times0.55$ | M1 | Correct method |
| $= 0.385 + 0.195 + 0.0825 = 0.6625$ | A1 | |
| **(iii)** $P(\text{light} \mid \text{no coupon}) = \frac{0.30\times0.65}{0.6625} = \frac{0.195}{0.6625}$ | M1 | Correct conditional probability method |
| $= 0.2943...$ awrt $0.294$ | A1 | |
| **(iv)** $P(\text{very light} \mid \text{coupon}) = \frac{0.15\times0.45}{1-0.6625} = \frac{0.0675}{0.3375}$ | M1 | Correct conditional structure |
| $= 0.2$ | A1 | |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{coupon} \mid \text{standard}) = 0.30$, $P(\text{coupon} \mid \text{light}) = 0.35$, $P(\text{coupon} \mid \text{very light}) = 0.45$ | B1 | At least one conditional probability correct |
| $P(\text{standard}\mid\text{coupon}) = \frac{0.55\times0.30}{0.3375}$, $P(\text{light}\mid\text{coupon}) = \frac{0.30\times0.35}{0.3375}$, $P(\text{very light}\mid\text{coupon}) = 0.2$ | M1 | Correct conditional probabilities for all three |
| $= \frac{0.165}{0.3375}$, $\frac{0.105}{0.3375}$, $\frac{0.0675}{0.3375}$ | A1 | All three correct |
| $P(\text{all different variety}) = 3! \times \frac{0.165}{0.3375} \times \frac{0.105}{0.3375} \times \frac{0.2}{1}$ | M1 | Multiplying by $3!$ |
| $= 6 \times 0.4889 \times 0.3111 \times 0.2 = 0.1829...$ awrt $0.183$ | A1 | |3 A particular brand of spread is produced in three varieties: standard, light and very light.
During a marketing campaign, the producer advertises that some cartons of spread contain coupons worth $\pounds 1 , \pounds 2$ or $\pounds 4$.
For each variety of spread, the proportion of cartons containing coupons of each value is shown in the table.
\begin{center}
\begin{tabular}{ | l | c | c | c | }
\cline { 2 - 4 }
\multicolumn{1}{c|}{} & \multicolumn{3}{c|}{Variety} \\
\cline { 2 - 4 }
\multicolumn{1}{c|}{} & Standard & Light & Very light \\
\hline
No coupon & 0.70 & 0.65 & 0.55 \\
\hline
£1 coupon & 0.20 & 0.25 & 0.30 \\
\hline
£2 coupon & 0.08 & 0.06 & 0.10 \\
\hline
£4 coupon & 0.02 & 0.04 & 0.05 \\
\hline
\end{tabular}
\end{center}
For example, the probability that a carton of standard spread contains a coupon worth $\pounds 2$ is 0.08 .
In a large batch of cartons, 55 per cent contain standard spread, 30 per cent contain light spread and 15 per cent contain very light spread.
\begin{enumerate}[label=(\alph*)]
\item A carton of spread is selected at random from the batch. Find the probability that the carton:
\begin{enumerate}[label=(\roman*)]
\item contains standard spread and a coupon worth $\pounds 1$;
\item does not contain a coupon;
\item contains light spread, given that it does not contain a coupon;
\item contains very light spread, given that it contains a coupon.
\end{enumerate}\item A random sample of 3 cartons is selected from the batch.
Given that all of these 3 cartons contain a coupon, find the probability that they each contain a different variety of spread.\\[0pt]
[4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA S3 2015 Q3 [12]}}