| Exam Board | SPS |
|---|---|
| Module | SPS SM Statistics (SPS SM Statistics) |
| Year | 2024 |
| Session | September |
| Marks | 11 |
| Topic | Discrete Probability Distributions |
| Type | Sum or product of two independent values |
| Difficulty | Standard +0.3 This is a straightforward A-level statistics question involving geometric probability distributions. Part (i) requires summing a geometric series (standard technique), parts (ii)-(iv) involve basic probability calculations with independence, and part (v) identifies that Y is an infinite geometric distribution. All techniques are routine for S1/S2 level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.03a Mutually exclusive and independent events5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| \(x\) | 1 | 2 | 3 | 4 | 5 |
| P\((X = x)\) | \(\frac{16}{31}\) | \(\frac{8}{31}\) | \(\frac{4}{31}\) | \(\frac{2}{31}\) | \(\frac{1}{31}\) |
The discrete random variable $X$ takes values 1, 2, 3, 4 and 5, and its probability distribution is defined as follows.
$$\mathrm{P}(X = x) = \begin{cases}
a & x = 1, \\
\frac{1}{2}\mathrm{P}(X = x - 1) & x = 2, 3, 4, 5, \\
0 & \text{otherwise,}
\end{cases}$$
where $a$ is a constant.
\begin{enumerate}[label=(\roman*)]
\item Show that $a = \frac{16}{31}$. [2]
\end{enumerate}
The discrete probability distribution for $X$ is given in the table.
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
$x$ & 1 & 2 & 3 & 4 & 5 \\
\hline
P$(X = x)$ & $\frac{16}{31}$ & $\frac{8}{31}$ & $\frac{4}{31}$ & $\frac{2}{31}$ & $\frac{1}{31}$ \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the probability that $X$ is odd. [1]
\end{enumerate}
Two independent values of $X$ are chosen, and their sum $S$ is found.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Find the probability that $S$ is odd. [2]
\item Find the probability that $S$ is greater than 8, given that $S$ is odd. [3]
\end{enumerate}
Sheila sometimes needs several attempts to start her car in the morning. She models the number of attempts she needs by the discrete random variable $Y$ defined as follows.
$$\mathrm{P}(Y = y + 1) = \frac{1}{2}\mathrm{P}(Y = y) \quad \text{for all positive integers } y.$$
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{4}
\item Find P$(Y = 1)$. [2]
\item Give a reason why one of the variables, $X$ or $Y$, might be more appropriate as a model for the number of attempts that Sheila needs to start her car. [1]
\end{enumerate}
\hfill \mbox{\textit{SPS SPS SM Statistics 2024 Q3 [11]}}