| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2010 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Conditional probability with given score/outcome |
| Difficulty | Standard +0.8 This is a multi-step conditional probability problem requiring Bayes' theorem. Students must enumerate all ways to score 12 under two different rules (multiplication vs addition), calculate P(score=12|even) and P(score=12|odd) by counting outcomes, then apply the conditional probability formula. The enumeration and careful probability calculation elevate this above routine exercises. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(E \text{ and } 12) = \frac{2}{5} \times \frac{4}{36} = \frac{8}{180}\) \((2/45)\) | M1 | 2/5 or 3/5 mult by dice-related probability seen anywhere |
| A1 | \(\frac{2}{5} \times \frac{4}{36}\) seen oe | |
| \(P(12) = \frac{3}{5} \times \frac{1}{36} + \frac{8}{180} = \frac{11}{180}\) \((0.0611)\) | M1 | Summing two 2-factor probs involving 2/5 and 3/5 |
| A1ft | 3/5 × 1/36 + their P(E and 12), ft their P(E 12) | |
| \(P(E \mid 12) = \frac{P(E \text{ and } 12)}{P(12)} = \frac{8}{11}\) \((0.727)\) | M1dep | Subst in condit prob formula, must have a fraction |
| A1 [6] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Even: 2 and (4,3) or (3,4) or (2,6) or (6,2). Gives 8 options | M1 | List attempt evens |
| A1 | 8 options | |
| Odd: 1 and (6,6) or 3 and (6,6) or 5 and (6,6). Gives 3 options | M1 | List attempt odds |
| A1 | 3 options | |
| \(\text{Prob}(E \mid 12) = 8/11\) | M1 | (Their even)/(their total) |
| A1 | Correct answer |
$P(E \text{ and } 12) = \frac{2}{5} \times \frac{4}{36} = \frac{8}{180}$ $(2/45)$ | M1 | 2/5 or 3/5 mult by dice-related probability seen anywhere
| A1 | $\frac{2}{5} \times \frac{4}{36}$ seen oe
$P(12) = \frac{3}{5} \times \frac{1}{36} + \frac{8}{180} = \frac{11}{180}$ $(0.0611)$ | M1 | Summing two 2-factor probs involving 2/5 and 3/5
| A1ft | 3/5 × 1/36 + their P(E and 12), ft their P(E 12)
$P(E \mid 12) = \frac{P(E \text{ and } 12)}{P(12)} = \frac{8}{11}$ $(0.727)$ | M1dep | Subst in condit prob formula, must have a fraction
| A1 [6] | Correct answer
OR list:
Even: 2 and (4,3) or (3,4) or (2,6) or (6,2). Gives 8 options | M1 | List attempt evens
| A1 | 8 options
Odd: 1 and (6,6) or 3 and (6,6) or 5 and (6,6). Gives 3 options | M1 | List attempt odds
| A1 | 3 options
$\text{Prob}(E \mid 12) = 8/11$ | M1 | (Their even)/(their total)
| A1 | Correct answer3 A fair five-sided spinner has sides numbered 1,2,3,4,5. Raj spins the spinner and throws two fair dice. He calculates his score as follows.
\begin{itemize}
\item If the spinner lands on an even-numbered side, Raj multiplies the two numbers showing on the dice to get his score.
\item If the spinner lands on an odd-numbered side, Raj adds the numbers showing on the dice to get his score.
\end{itemize}
Given that Raj's score is 12, find the probability that the spinner landed on an even-numbered side.
\hfill \mbox{\textit{CAIE S1 2010 Q3 [6]}}