| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | Game theory with alternating players |
| Difficulty | Challenging +1.8 This is a sophisticated multi-part probability question requiring careful analysis of alternating geometric distributions with asymmetric winning conditions. While the individual calculations use standard geometric distribution formulas, students must correctly model the complex game rules, handle infinite series summation, and track multiple conditional outcomes across four interconnected parts—significantly above average difficulty but within reach for strong Further Maths students. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left(\frac{2}{3}\right)^4 \left(\frac{3}{4}\right)^4 \frac{1}{3} = \frac{1}{48}\) A.G. | M1 A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left(\frac{2}{3}\right)^{n-1} \left(\frac{3}{4}\right)^{n-1} \frac{2}{3} \cdot \frac{1}{4}\) | M1 A1 | |
| \(= \frac{1}{(3 \times 2^n)}\) A.G. | A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{3} \cdot \frac{3}{4} \left\{1 + \left(\frac{2}{3} \cdot \frac{3}{4}\right) + \left(\frac{2}{3} \cdot \frac{3}{4}\right)^2 + \ldots \right\}\) | M1 A1 | |
| \(= \frac{1}{4}\left(1 + \frac{1}{2} + \frac{1}{4} + \ldots\right) = \frac{1}{2}\) | A1 | 3 marks |
| Answer | Marks |
|---|---|
| \(\frac{1}{3} \cdot \frac{1}{4} \left\{1 + \left(\frac{2}{3} \cdot \frac{3}{4}\right) + \left(\frac{2}{3} \cdot \frac{3}{4}\right)^2 + \ldots \right\}\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 1 - \left\{\frac{1}{2} + \frac{1}{3}\left(\frac{1}{2} + \frac{1}{4} + \ldots\right)\right\}\) | (M1 A1) | |
| \(= \frac{1}{6}\) | A1 | 3 marks |
# Question 10:
## Part (i):
$\left(\frac{2}{3}\right)^4 \left(\frac{3}{4}\right)^4 \frac{1}{3} = \frac{1}{48}$ A.G. | M1 A1 | 2 marks
## Part (ii):
$\left(\frac{2}{3}\right)^{n-1} \left(\frac{3}{4}\right)^{n-1} \frac{2}{3} \cdot \frac{1}{4}$ | M1 A1 |
$= \frac{1}{(3 \times 2^n)}$ **A.G.** | A1 | 3 marks
## Part (iii):
$\frac{1}{3} \cdot \frac{3}{4} \left\{1 + \left(\frac{2}{3} \cdot \frac{3}{4}\right) + \left(\frac{2}{3} \cdot \frac{3}{4}\right)^2 + \ldots \right\}$ | M1 A1 |
$= \frac{1}{4}\left(1 + \frac{1}{2} + \frac{1}{4} + \ldots\right) = \frac{1}{2}$ | A1 | 3 marks
## Part (iv):
$\frac{1}{3} \cdot \frac{1}{4} \left\{1 + \left(\frac{2}{3} \cdot \frac{3}{4}\right) + \left(\frac{2}{3} \cdot \frac{3}{4}\right)^2 + \ldots \right\}$ | M1 A1 |
$= \frac{1}{3} \cdot \frac{1}{4}\left(1 + \frac{1}{2} + \frac{1}{4} + \ldots\right)$
*or* $1 - P(\text{Jill or Kate wins})$
$= 1 - \left\{\frac{1}{2} + \frac{1}{3}\left(\frac{1}{2} + \frac{1}{4} + \ldots\right)\right\}$ | (M1 A1) |
$= \frac{1}{6}$ | A1 | 3 marks | **Total: 11**
---10 Jill and Kate are playing a game as a practice for a penalty shoot-out. They take alternate turns at kicking a football at a goal. The probability that Jill will score a goal with any kick is $\frac { 1 } { 3 }$, independently of previous outcomes. The probability that Kate will score a goal with any kick is $\frac { 1 } { 4 }$, independently of previous outcomes. Jill begins the game. If Jill is the first to score, then Kate is allowed one more kick. If Kate scores with this kick, then the game is a draw, but if she does not score then Jill wins the game. If Kate is the first to score, then she wins the game, and no further kicks are taken.\\
(i) Show that the probability that Jill scores on her 5th kick is $\frac { 1 } { 48 }$.\\
(ii) Show that the probability that Kate wins the game on her $n$th kick is $\frac { 1 } { 3 \times 2 ^ { n } }$.\\
(iii) Find the probability that Jill wins the game.\\
(iv) Find the probability that the game is a draw.
\hfill \mbox{\textit{CAIE FP2 2013 Q10 [11]}}