# Question 3:

## Part (a)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| Row minima: $-2, -1, 3, 0$; max is $3$. Column maxima: $3, 5$; min is $3$ | M1 | Finding row minimums and column maximums; condone one error |
| Row(maximin) $=$ Col(minimax) $= 3$, therefore game is stable | A1 | Dependent on all 6 correct values; must see two 3's explicitly compared; cao $(-3)$ is A0 |

## Part (a)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Value of game to player $B$ is $-3$ | A1 | cao |

## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\begin{pmatrix} -2 & 1 & 0 \\ 2 & -5 & -2 \end{pmatrix}$ | B1 | cao; must be reduced game with option S removed |

## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| If $A$ plays Q: $B$'s gains $= -2p + 2(1-p) = 2 - 4p$ | M1 | Setting up three expressions in terms of $p$ |
| If $A$ plays R: $B$'s gains $= p + (-5)(1-p) = -5 + 6p$ | A1 | All three expressions correctly simplified |
| If $A$ plays T: $B$'s gains $= (-2)(1-p) = -2 + 2p$ | — | — |
| At least two lines correctly drawn | M1 | If values on at least one vertical axis not given, lines must be in right position relative to each other |
| Completely correct graph with correct intersection points at ends where $p=0$ and $p=1$ | A1 | Lines must not extend past $p<0$ and/or $p>1$ |
| $2 - 4p = -5 + 6p \Rightarrow p = \frac{7}{10}$ | A1 | Using graph with 3 lines to obtain correct probability |
| $B$ plays option X with probability $\frac{7}{10}$ and option Y with probability $\frac{3}{10}$ | A1ft | Must refer to 'play' and associated probabilities |

## Part (d)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $6 \times \frac{7}{10} - 5 = -\frac{4}{5}$; value of game to player $A$ is $\frac{4}{5}$ | B1 | cao ($\frac{4}{5}$ or $0.8$) |

## Part (d)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Player $A$ should never play option T | B1 | cao |

## Part (d)(iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| If $A$ plays Q with probability $q$ and R with probability $1-q$: $2q + (-1)(1-q) = \frac{4}{5}$ | M1 | Setting up linear equation with $V(A)$ from (d)(i) using either $2q+(-1)(1-q)$ or $-2q+5(1-q)$ |
| $q = \frac{3}{5}$; $A$ plays Q with probability $\frac{3}{5}$ and R with probability $\frac{2}{5}$ (play S and T never) | A1 | Must refer to 'play' and associated probabilities |