# Question 3:

## Part (a)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Play-safe strategy for Rose is $R_1$ | B1 | |
| Reason: maximum of row minima; min of $R_1$ is $-1$, min of $R_2$ is $-3$, min of $R_3$ is $-2$; $-1$ is largest | B1 | 2 marks |

## Part (a)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Play-safe for Callum: column maxima are 5, 2, 5; minimum is 2, so Callum plays $C_2$ | M1 | |
| Saddle point would require Rose's play-safe to equal Callum's play-safe value; $-1 \neq 2$, so no stable solution | A1 | 2 marks |

## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $R_3$ is dominated by $R_1$ (every entry in $R_1 \geq$ every entry in $R_3$) | B1 | 1 mark |

## Part (c)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(C_1) = 5p + (-3)(1-p) = 8p - 3$ | B1 | |
| $E(C_2) = 2p + (-1)(1-p) = 3p - 1$ | B1 | |
| $E(C_3) = -p + 5(1-p) = 5 - 6p$ | B1 | 3 marks |

## Part (c)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Three straight lines correctly drawn on graph for $0 \leq p \leq 1$ | M1 A1 | 2 marks |

## Part (c)(iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Optimal point is intersection of $E(C_2)$ and $E(C_3)$: $3p-1 = 5-6p$ | M1 | |
| $9p = 6$, so $p = \frac{2}{3}$ | A1 | |
| Rose plays $R_1$ with probability $\frac{2}{3}$, $R_2$ with probability $\frac{1}{3}$ | A1 | 3 marks |

## Part (c)(iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Value of game $= 3(\frac{2}{3})-1 = 1$ | B1 | 1 mark |

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