| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2010 |
| Session | June |
| Marks | 11 |
| Topic | Probability Generating Functions |
| Type | Find PGF from probability distribution |
| Difficulty | Challenging +1.2 This is a multi-part PGF question requiring geometric series manipulation and understanding of alternating turn probabilities. Part (i) involves straightforward probability calculations with independent events. Part (ii)(a) requires deriving a PGF by summing an infinite geometric series—a standard technique for Further Maths students. Part (ii)(b) is routine differentiation of the given PGF. While it requires multiple steps and careful algebra, the techniques are all standard for UFM Statistics students who have learned PGFs. The question is moderately above average difficulty due to the multi-step nature and need to handle the alternating structure, but doesn't require novel insight. |
| Spec | 2.03c Conditional probability: using diagrams/tables5.01a Permutations and combinations: evaluate probabilities |
**(i)(a)** $\frac{3}{4} \times \frac{1}{5} = \frac{3}{20}$ — B1 [1]
**(b)** $\frac{3}{4} \times \frac{4}{5} \times \frac{1}{4} = \frac{3}{20}$ — M1A1 [2]
**(c)** $P_A = \frac{1}{4} + \frac{3}{4} \times \frac{4}{5} \times P_A$ or $P_A = 0.25 + (0.25 \times 0.6) + (0.25 \times 0.6^2) + \ldots$ — M1
$\frac{2}{5}P_A = \frac{1}{4} \Rightarrow P_A = \frac{5}{8}$ or $P_A = \frac{0.25}{1-0.6} = \frac{5}{8}$ — M1A1 [3]
**(ii)(a)** $G(t) = \frac{1}{4}t + \frac{3}{20}t^2 + \frac{3}{20}t^3 + \frac{9}{100}t^4 + \frac{9}{100}t^5 + \ldots$ — M1
$= \frac{\frac{1}{4}t}{1-\frac{3t^2}{5}} + \frac{\frac{3}{20}t^2}{1-\frac{3t^2}{5}} = \frac{5t + 3t^2}{4(5-3t^2)}$ (AG) — M1A1 [3]
**(b)** $G'(t) = \frac{4(5-3t^2)(5+6t) - (5t+3t^2)(-24t)}{16(5-3t^2)^2}$ — M1
$E(R) = G'(1) = \frac{4 \times 2 \times 11 + 8 \times 24}{64} = \frac{35}{8}$ — A1 [2]
**[Total: 11]**12 Two players, $A$ and $B$, are taking turns to shoot at a basket with a basketball. The winner of this game is the first player to score a basket. The probability that $A$ scores a basket with any shot is $\frac { 1 } { 4 }$ and the probability that $B$ scores a basket with any shot is $\frac { 1 } { 5 }$. Each shot is independent of all other shots. $A$ shoots first.\\
(i) Find
\begin{enumerate}[label=(\alph*)]
\item the probability that $B$ wins with his first shot,
\item the probability that $A$ wins with his second shot,
\item the probability that $A$ wins the game.\\
(ii) $R$ is the total number of shots taken by $A$ and $B$ up to and including the shot that scores a basket.\\
(a) Show that the probability generating function of $R$ is given by
$$\mathrm { G } ( t ) = \frac { 5 t + 3 t ^ { 2 } } { 4 \left( 5 - 3 t ^ { 2 } \right) }$$
(b) Hence find $\mathrm { E } ( R )$.
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2010 Q12 [11]}}