# Question 3:

## Part (a):
Row minima: $-3, -1, -3, -1$; max is $-1$

Column maxima: $4, 5$; min is $4$

$\text{Row(maximin)} \neq \text{Col(minimax)}$ therefore game is not stable | M1, A1 | 1.1b, 2.4 | M1: Finding row minimums and column maximums — condone one error; A1: Row maximin $(-1) \neq$ col minimax $(4)$ so unstable |

## Part (b):
If A plays Q, B's gains: $-4p + 3(1-p) = 3 - 7p$

If A plays R, B's gains: $-2p + 1(1-p) = 1 - 3p$

If A plays S, B's gains: $3p + (-5)(1-p) = -5 + 8p$

If A plays T, B's gains: $p + (-3)(1-p) = -3 + 4p$ | M1, A1 | 1.1b | M1: Setting up four expressions in $p$; A1: All four correct |

Correct graph with axes and at least two lines correctly drawn | M1dep, A1 | 1.1b | M1dep: Axes correct, at least two lines correctly drawn; A1: Correct graph |

$3 - 7p = -3 + 4p \Rightarrow p = \dfrac{6}{11}$ | A1 | 1.1b | Using graph to obtain correct probability |

B should play option X with probability $\dfrac{6}{11}$ and option Y with probability $\dfrac{5}{11}$ | A1ft | 3.2a | Must refer to play and name options |

## Part (c):
$V(B) = 3 - 7\left(\dfrac{6}{11}\right) = -\dfrac{9}{11} \Rightarrow V(A) = \dfrac{9}{11}$ | B1 | 2.2a | cao |

## Part (d):
If A plays Q with probability $q$ and T with probability $1-q$:

$$4q + (-1)(1-q) = \frac{9}{11}$$

$q = \dfrac{4}{11}$ | M1 | 3.1b | Setting up linear equation with V(A) and two options from (b) |

A should play Q with probability $\dfrac{4}{11}$, options R and S never, and T with probability $\dfrac{7}{11}$ | A1 | 3.2a | Must refer to play and name options |