| Exam Board | WJEC |
|---|---|
| Module | Unit 4 (Unit 4) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | Game theory with alternating players |
| Difficulty | Challenging +1.2 This is a multi-part geometric distribution problem requiring understanding of infinite geometric series and algebraic manipulation. Part (a) involves straightforward probability calculations with independent events, part (b) requires summing an infinite GP (a standard A-level technique but requiring careful setup), and part (c) is routine inequality solving. The context is slightly novel but the mathematical techniques are all standard A-level material with clear structure guiding students through the solution. |
| 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 principles2.03e Model with probability: critiquing assumptions5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1) |
| Answer | Marks | Guidance |
|---|---|---|
| (a)(i) \(P(J \text{ wins with } 1^{\text{st}} \text{ shot}) = P(M \text{ misses}) \times P(J \text{ hits}) = 0.75p\) | M1, A1 | AO1, AO1 |
| (a)(ii) \(J \text{ wins with his second shot if the first three shots miss and then } J \text{ hits the target with his second shot.}\) | M1 | AO3 |
| \(P(J \text{ wins with } 2^{\text{nd}} \text{ shot}) = 0.75 \times (1-p) \times 0.75 \times p\) | A1 | AO2 |
| (b) \(P(J \text{ wins game}) = 0.75p + 0.75^2(1-p)p + 0.75^3(1-p)^2p + \ldots\) | M1 | AO3 |
| \(\text{Attempting to sum an infinite geometric series}\) | M1 | AO3 |
| \(= \frac{0.75p}{1 - 0.75(1-p)} = \frac{3p}{1+3p}\) | A1 | AO2 |
| (c) \(\text{Mary is more likely to win if } \frac{3p}{1+3p} < 0.5\) | M1 | AO3 |
| \(\text{leading to } p < \frac{1}{3}\) | A1 | AO1 |
**(a)(i)** $P(J \text{ wins with } 1^{\text{st}} \text{ shot}) = P(M \text{ misses}) \times P(J \text{ hits}) = 0.75p$ | M1, A1 | AO1, AO1 | |
**(a)(ii)** $J \text{ wins with his second shot if the first three shots miss and then } J \text{ hits the target with his second shot.}$ | M1 | AO3 | |
| $P(J \text{ wins with } 2^{\text{nd}} \text{ shot}) = 0.75 \times (1-p) \times 0.75 \times p$ | A1 | AO2 | |
**(b)** $P(J \text{ wins game}) = 0.75p + 0.75^2(1-p)p + 0.75^3(1-p)^2p + \ldots$ | M1 | AO3 | |
| $\text{Attempting to sum an infinite geometric series}$ | M1 | AO3 | |
| $= \frac{0.75p}{1 - 0.75(1-p)} = \frac{3p}{1+3p}$ | A1 | AO2 | |
**(c)** $\text{Mary is more likely to win if } \frac{3p}{1+3p} < 0.5$ | M1 | AO3 | |
| $\text{leading to } p < \frac{1}{3}$ | A1 | AO1 | |2. Mary and Jeff are archers and one morning they play the following game. They shoot an arrow at a target alternately, starting with Mary. The winner is the first to hit the target. You may assume that, with each shot, Mary has a probability 0.25 of hitting the target and Jeff has a probability $p$ of hitting the target. Successive shots are independent.
\begin{enumerate}[label=(\alph*)]
\item Determine the probability that Jeff wins the game\\
i) with his first shot,\\
ii) with his second shot.
\item Show that the probability that Jeff wins the game is
$$\frac { 3 p } { 1 + 3 p }$$
\item Find the range of values of $p$ for which Mary is more likely to win the game than Jeff.
\end{enumerate}
\hfill \mbox{\textit{WJEC Unit 4 Q2 [9]}}