| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Conditional probability with given score/outcome |
| Difficulty | Standard +0.3 This is a straightforward conditional probability question using Bayes' theorem. Students must enumerate outcomes for score=8 under both head (sum=8) and tail (product=8) scenarios, calculate P(score=8), then find P(head|score=8). While it requires systematic case-work and careful probability calculations, the structure is standard and the arithmetic is manageable, making it slightly easier than average. |
| Spec | 2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(8) = P(H 4 4) + P(T 2 4) + P(T 4 2)\) | M1 | \(\frac{1}{3}\) or \(\frac{2}{3}\) mult by dice related prob, seen anywhere |
| \(= \frac{1}{3} \times \frac{1}{16} + \frac{2}{3} \times \frac{1}{16} + \frac{2}{3} \times \frac{1}{16}\) | M1 | Summing two or three 2-factor probs involving \(\frac{1}{3}\) and \(\frac{2}{3}\) |
| \(= \frac{5}{48}\) | A1 | \(\frac{5}{48}\) oe seen as num or denom of a fraction |
| 5 marks | ||
| \(P(H \mid 8) = \frac{P(H \cap 8)}{P(8)}\) | B1 | \(\frac{1}{48}\) oe seen as num or denom of a fraction |
| \(= \frac{\frac{1}{48}}{\frac{5}{48}} = \frac{1}{5}\) | A1 | 5 marks |
| $P(8) = P(H 4 4) + P(T 2 4) + P(T 4 2)$ | M1 | $\frac{1}{3}$ or $\frac{2}{3}$ mult by dice related prob, seen anywhere |
| $= \frac{1}{3} \times \frac{1}{16} + \frac{2}{3} \times \frac{1}{16} + \frac{2}{3} \times \frac{1}{16}$ | M1 | Summing two or three 2-factor probs involving $\frac{1}{3}$ and $\frac{2}{3}$ |
| $= \frac{5}{48}$ | A1 | $\frac{5}{48}$ oe seen as num or denom of a fraction |
| | | 5 marks |
| $P(H \mid 8) = \frac{P(H \cap 8)}{P(8)}$ | B1 | $\frac{1}{48}$ oe seen as num or denom of a fraction |
| $= \frac{\frac{1}{48}}{\frac{5}{48}} = \frac{1}{5}$ | A1 | 5 marks |
---3 Jodie tosses a biased coin and throws two fair tetrahedral dice. The probability that the coin shows a head is $\frac { 1 } { 3 }$. Each of the dice has four faces, numbered $1,2,3$ and 4 . Jodie's score is calculated from the numbers on the faces that the dice land on, as follows:
\begin{itemize}
\item if the coin shows a head, the two numbers from the dice are added together;
\item if the coin shows a tail, the two numbers from the dice are multiplied together.
\end{itemize}
Find the probability that the coin shows a head given that Jodie's score is 8 .
\hfill \mbox{\textit{CAIE S1 2014 Q3 [5]}}