| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Dice/random device selects population |
| Difficulty | Moderate -0.3 This is a straightforward application of the law of total probability and Bayes' theorem with clearly given probabilities and simple counting. The setup is explicit, requiring only basic probability tree calculations with no conceptual subtlety—slightly easier than average for A-level. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Playground \(X\) | Playground \(Y\) | Playground \(Z\) |
| \(3 S , 2 R , 4 P\) | \(6 S , 3 R , 1 C , 2 P\) | \(8 S , 3 R , 4 C , 1 P\) |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(P(X \text{ and } P) = \frac{1}{4} \times \frac{4}{9} = \frac{1}{9}\) | M1 | Mult a playground prob with a P prob |
| \(P(Y \text{ and } P) = \frac{1}{4} \times \frac{2}{12} = \frac{1}{24}\) | A1 | One correct prob |
| \(P(Z \text{ and } P) = \frac{1}{2} \times \frac{1}{16} = \frac{1}{32}\) | M1 | Summing at least two 2-factor probs |
| \(P(P) = \frac{53}{288} = 0.184\) | A1 | 4 |
| (ii) \(P(Y \mid C) = \frac{P(Y \cap C)}{P(C)}\) | M1 | Attempt at \(P(Y \cap C)\) as numerator of a fraction |
| M1 | Attempt at P(C) in form of summing two 2-factor products, seen anywhere | |
| \(\frac{\frac{1}{4} \times \frac{1}{12}}{\frac{1}{4} \times \frac{1}{12} + \frac{1}{2} \times \frac{1}{16}}\) | A1 | Correct unsimplified P(C) seen anywhere |
| \(= \frac{\frac{1}{48}}{\frac{1}{48}} = \frac{1}{7}\) | A1 | 4 |
(i) $P(X \text{ and } P) = \frac{1}{4} \times \frac{4}{9} = \frac{1}{9}$ | M1 | Mult a playground prob with a P prob
$P(Y \text{ and } P) = \frac{1}{4} \times \frac{2}{12} = \frac{1}{24}$ | A1 | One correct prob
$P(Z \text{ and } P) = \frac{1}{2} \times \frac{1}{16} = \frac{1}{32}$ | M1 | Summing at least two 2-factor probs
$P(P) = \frac{53}{288} = 0.184$ | A1 | 4 | Correct answer
(ii) $P(Y \mid C) = \frac{P(Y \cap C)}{P(C)}$ | M1 | Attempt at $P(Y \cap C)$ as numerator of a fraction
| M1 | Attempt at P(C) in form of summing two 2-factor products, seen anywhere
$\frac{\frac{1}{4} \times \frac{1}{12}}{\frac{1}{4} \times \frac{1}{12} + \frac{1}{2} \times \frac{1}{16}}$ | A1 | Correct unsimplified P(C) seen anywhere
$= \frac{\frac{1}{48}}{\frac{1}{48}} = \frac{1}{7}$ | A1 | 4 | Correct answer5 Playground equipment consists of swings ( $S$ ), roundabouts ( $R$ ), climbing frames ( $C$ ) and play-houses $( P )$. The numbers of pieces of equipment in each of 3 playgrounds are as follows.
\begin{center}
\begin{tabular}{ | c | c | c | }
\hline
Playground $X$ & Playground $Y$ & Playground $Z$ \\
\hline
$3 S , 2 R , 4 P$ & $6 S , 3 R , 1 C , 2 P$ & $8 S , 3 R , 4 C , 1 P$ \\
\hline
\end{tabular}
\end{center}
Each day Nur takes her child to one of the playgrounds. The probability that she chooses playground $X$ is $\frac { 1 } { 4 }$. The probability that she chooses playground $Y$ is $\frac { 1 } { 4 }$. The probability that she chooses playground $Z$ is $\frac { 1 } { 2 }$. When she arrives at the playground, she chooses one piece of equipment at random.\\
(i) Find the probability that Nur chooses a play-house.\\
(ii) Given that Nur chooses a climbing frame, find the probability that she chose playground $Y$.
\hfill \mbox{\textit{CAIE S1 2014 Q5 [8]}}