| Exam Board | SPS |
|---|---|
| Module | SPS FM Statistics (SPS FM Statistics) |
| Year | 2025 |
| Session | April |
| Marks | 9 |
| Topic | Discrete Probability Distributions |
| Type | Conditional probability with random variables |
| Difficulty | Standard +0.8 This is a multi-part probability question requiring careful counting of outcomes with replacement, understanding of joint probability distributions, and conditional probability. Part (a) is routine verification, but parts (b) and (c) require systematic enumeration of cases where two independent selections yield the same sum, then applying conditional probability - this demands organization and careful reasoning beyond standard textbook exercises. |
| Spec | 2.03a Mutually exclusive and independent events2.03c Conditional probability: using diagrams/tables5.01a Permutations and combinations: evaluate probabilities |
| \(m\) | 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{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}
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 FM Statistics 2025 Q3 [9]}}