| Exam Board | SPS |
|---|---|
| Module | SPS SM Statistics (SPS SM Statistics) |
| Year | 2025 |
| Session | April |
| Marks | 9 |
| Topic | Discrete Probability Distributions |
| Type | Conditional probability with random variables |
| Difficulty | Standard +0.3 This is a straightforward probability question involving sampling without replacement and conditional probability. Part (a) requires basic counting of favorable outcomes. Part (b) involves calculating P(M=S) by summing products of probabilities, which is routine. Part (c) applies Bayes' theorem in a standard way. The question requires careful enumeration but no novel insight or complex problem-solving beyond standard S1/S2 techniques. |
| Spec | 2.03c Conditional probability: using diagrams/tables2.04a Discrete probability distributions |
| \(m\) | 4 | 5 | 6 | 7 | 8 |
| \(P(M = m)\) | \(\frac{1}{15}\) | \(\frac{4}{15}\) | \(\frac{1}{3}\) | \(\frac{4}{15}\) | \(\frac{1}{15}\) |
Miguel has six numbered tiles, labelled 2, 2, 3, 3, 4, 4. He selects two tiles at random, without replacement. The variable $M$ denotes the sum of the numbers on the two tiles.
\begin{enumerate}[label=(\alph*)]
\item Show that $P(M = 6) = \frac{1}{3}$ [2]
\end{enumerate}
The table shows the probability distribution of $M$
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
$m$ & 4 & 5 & 6 & 7 & 8 \\
\hline
$P(M = m)$ & $\frac{1}{15}$ & $\frac{4}{15}$ & $\frac{1}{3}$ & $\frac{4}{15}$ & $\frac{1}{15}$ \\
\hline
\end{tabular}
\end{center}
Miguel returns the two tiles to the collection. Now Sofia selects two tiles at random from the six tiles, without replacement. The variable $S$ denotes the sum of the numbers on the two tiles that Sofia selects.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find $P(M = S)$ [3]
\item Find $P(S = 7 | M = S)$ [4]
\end{enumerate}
\hfill \mbox{\textit{SPS SPS SM Statistics 2025 Q7 [9]}}