| Exam Board | AQA |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Piecewise or conditional probability function |
| Difficulty | Standard +0.3 This is a straightforward probability distribution question requiring basic algebraic manipulation. Part (a) is trivial substitution (P(X=3)=9k). Part (b) involves setting up two equations (the given condition and the sum-to-1 requirement) and solving simultaneously. While it requires careful algebra with two unknowns, it's a standard textbook exercise with no conceptual difficulty or novel insight needed—slightly easier than average for AS-level. |
| Spec | 2.04a Discrete probability distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(9k\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(c + 2c + 9k + 16k = 1\) | B1 | Uses probabilities \(c, 2c, 9k, 16k\) |
| \(3c + 25k = 1\) | M1 | Uses \(\Sigma p = 1\) to obtain an equation in \(k\) and \(c\), at least one term in \(k\) or \(c\) correct; or uses ratio \(1:3(:4)\) to obtain an equation either in \(c\) related to \(0.25\) or in \(k\) related to \(0.75\) \([c + 2c = 0.25;\ 9k + 16k = 0.75]\) |
| \(9k + 16k = 3(c + 2c)\), so \(25k = 9c\) | M1 | Uses ratio \(1:3\) to find equation in \(k\) and \(c\), at least one term correct; or obtains a second equation in \(k\) or \(c\) using ratio \(1:3(:4)\) |
| \(c = \dfrac{1}{12},\quad k = \dfrac{3}{100}\) | A1 | ACF but must be exact |
## Question 15(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $9k$ | B1 | |
## Question 15(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $c + 2c + 9k + 16k = 1$ | B1 | Uses probabilities $c, 2c, 9k, 16k$ |
| $3c + 25k = 1$ | M1 | Uses $\Sigma p = 1$ to obtain an equation in $k$ and $c$, at least one term in $k$ or $c$ correct; or uses ratio $1:3(:4)$ to obtain an equation either in $c$ related to $0.25$ or in $k$ related to $0.75$ $[c + 2c = 0.25;\ 9k + 16k = 0.75]$ |
| $9k + 16k = 3(c + 2c)$, so $25k = 9c$ | M1 | Uses ratio $1:3$ to find equation in $k$ and $c$, at least one term correct; or obtains a second equation in $k$ or $c$ using ratio $1:3(:4)$ |
| $c = \dfrac{1}{12},\quad k = \dfrac{3}{100}$ | A1 | ACF but must be exact |
---15 The discrete random variable $X$ is modelled by the probability distribution defined by:
$$\mathrm { P } ( X = x ) = \left\{ \begin{array} { c c }
c x & x = 1,2 \\
k x ^ { 2 } & x = 3,4 \\
0 & \text { otherwise }
\end{array} \right.$$
where $k$ and $c$ are constants.\\
15
\begin{enumerate}[label=(\alph*)]
\item State, in terms of $k$, the probability that $X = 3$\\
15
\item Given that $\mathrm { P } ( X \geq 3 ) = 3 \times \mathrm { P } ( X \leq 2 )$\\
Find the exact value of $k$ and the exact value of $c$.\\
\includegraphics[max width=\textwidth, alt={}, center]{11168e8f-5ba5-4d27-83ab-0327cc23d08c-21_2488_1716_219_153}
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 2 2022 Q15 [5]}}