| Exam Board | AQA |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Definitions |
| Type | Listing outcomes and counting |
| Difficulty | Moderate -0.8 This is a straightforward probability question requiring basic counting and understanding of divisibility rules. Part (a) only needs identifying even digits in bowl C, part (b) is direct probability multiplication, and part (c) requires recognizing that divisibility by 111 means all three digits are equal, then counting matching numbers across bowls. All parts use standard AS-level probability techniques with no complex reasoning or novel insight required. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space |
| Answer | Marks | Guidance |
|---|---|---|
| 15(a) | Obtains the correct probability | |
| ACF CAO | 3.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Subtotal | 1 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 15(b) | Multiplies 3 fractions with at |
| Answer | Marks | Guidance |
|---|---|---|
| in a single calculation added | 3.1b | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Accept AWRT 0.021 | 1.1b | A1 |
| Subtotal | 2 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 15(c) | Multiplies three correct |
| Answer | Marks | Guidance |
|---|---|---|
| ACF | 3.1b | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Accept AWRT 0.014 | 1.1b | A1 |
| Subtotal | 2 | |
| Question 15 Total | 5 | |
| Q | Marking instructions | AO |
Question 15:
--- 15(a) ---
15(a) | Obtains the correct probability
ACF CAO | 3.1b | B1 | 3 1
P(Even) = =
9 3
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 15(b) ---
15(b) | Multiplies 3 fractions with at
least 2 correct
Condone the correct 3 fractions
in a single calculation added | 3.1b | M1 | 1 2 3
P(703) = × ×
4 8 9
1
=
48
Obtains correct answer
ACF
Accept AWRT 0.021 | 1.1b | A1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 15(c) ---
15(c) | Multiplies three correct
probabilities together for either
P(222) or P(333)
1 1
PI by or
288 96
ACF | 3.1b | M1 | P(divisible by 111)
= P(222) or P(333)
1 1 1 1 1 3
= × × + × ×
4 8 9 4 8 9
1
=
72
Obtains correct answer
ACF
Accept AWRT 0.014 | 1.1b | A1
Subtotal | 2
Question 15 Total | 5
Q | Marking instructions | AO | Marks | Typical solutionNumbered balls are placed in bowls A, B and C
In bowl A there are four balls numbered 1, 2, 3 and 7
In bowl B there are eight balls numbered 0, 0, 2, 3, 5, 6, 8 and 9
In bowl C there are nine balls numbered 0, 1, 1, 2, 3, 3, 3, 6 and 7
This information is shown in the diagram below.
\includegraphics{figure_15}
A three-digit number is generated using the following method:
• a ball is selected at random from each bowl
• the first digit of the number is the ball drawn from bowl A
• the second digit of the number is the ball drawn from bowl B
• the third digit of the number is the ball drawn from bowl C
\begin{enumerate}[label=(\alph*)]
\item Find the probability that the number generated is even. [1 mark]
\item Find the probability that the number generated is 703 [2 marks]
\item Find the probability that the number generated is divisible by 111 [2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 2 2023 Q15 [5]}}