| Exam Board | AQA |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Direct probability from given distribution |
| Difficulty | Moderate -0.3 This is a straightforward probability question requiring students to sum probabilities from a given distribution and find an unknown parameter p. The steps are routine: use total probability = 1 to find p, then calculate and compare two probabilities. While it has multiple steps, each is standard AS-level technique with no conceptual difficulty or novel insight required, making it slightly easier than average. |
| Spec | 2.04a Discrete probability distributions |
| \(x\) | 1 | 2 | 3 | 4 | 5 | 6 |
| P(X = x) | 0.3 | 0.1 | 0.2 | 0.1 | 0.1 | 0.2 |
| \(y\) | 2 | 3 | 4 | 5 | 6 | 7 |
| P(Y = y) | 0.3 | p | 0.2 | 0.1 | p | 3p + 0.05 |
| Answer | Marks |
|---|---|
| 16 | Obtains at least one of |
| Answer | Marks | Guidance |
|---|---|---|
| PI by 0.57 | 1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| ACF | 3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| ACF | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| than P(Y ≤ 4) for ‘claim’ | 2.1 | R1 |
| Question 16 Total | 4 | |
| Q | Marking instructions | AO |
Question 16:
16 | Obtains at least one of
P(X ≥ 3) = 0.6
Or
P(Y ≤ 4) = 0.5 + p
PI by 0.57 | 1.1b | B1 | 0.3 + p + 0.2 + 0.1 + p + 3p + 0.05
= 1
0.65 + 5p = 1
5p = 0.35
p = 0.07
P(X ≥ 3) = 0.6
P(Y ≤ 4) = 0.57
As 0.6 > 0.57 the claim is correct.
Sums the y probabilities and
compares to 1
Allow one slip
Or
Sets up a correct inequality for p
P(X ≥ 3) > P(Y ≤ 4)
PI by p < 0.1
ACF | 3.1a | M1
Obtains the correct value for p
Or
Obtains p < 0.1
ACF | 1.1b | A1
Obtains 0.6 for P(X ≥ 3) and
0.57 for P(Y ≤ 4), and shows
that the claim is correct.
Or
States that if the claim is correct
then p < 0.1, and uses a value
of p = 0.1 to show that the sum
of the probabilities for Y is
greater than 1 and concludes
that the claim is correct.
Condone P(X ≥ 3) is greater
than P(Y ≤ 4) for ‘claim’ | 2.1 | R1
Question 16 Total | 4
Q | Marking instructions | AO | Marks | Typical solutionThe discrete random variable $X$ has probability distribution
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
$x$ & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
P(X = x) & 0.3 & 0.1 & 0.2 & 0.1 & 0.1 & 0.2 \\
\hline
\end{tabular}
\end{center}
The discrete random variable $Y$ has probability distribution
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
$y$ & 2 & 3 & 4 & 5 & 6 & 7 \\
\hline
P(Y = y) & 0.3 & p & 0.2 & 0.1 & p & 3p + 0.05 \\
\hline
\end{tabular}
\end{center}
It is claimed that P(X ≥ 3) is greater than P(Y ≤ 4)
Determine if this claim is correct.
Fully justify your answer.
[4 marks]
\hfill \mbox{\textit{AQA AS Paper 2 2023 Q16 [4]}}