# Question 3:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Row minima: $-5, -3, -1, -4$; max is $-1$ | M1 | Finding row minimums and column maximums; condone one error |
| Column maxima: $4, 6, 3$; min is $3$; Row(maximin) $\neq$ Col(minimax) therefore game is not stable | A1 | Row maximin $\neq$ col minimax so not stable |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct reduced table for Meera (2 rows only): Option X: $-4, -1, 1, 4$; Option Z: $5, 3, -1, -3$ | B1 | Correct reduced table for Meera (2 rows only) |

## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| If $H$ plays Option A, $M$'s gains are $-4p + 5(1-p) = 5-9p$; If $H$ plays Option B, $M$'s gains are $-p + 3(1-p) = 3-4p$; If $H$ plays Option C, $M$'s gains are $p + (-1)(1-p) = -1+2p$; If $H$ plays Option D, $M$'s gains are $4p + (-3)(1-p) = -3+7p$ | M1 A1 | Setting up four expressions in terms of $p$; all four expressions correct and fully simplified |
| Correct graph with axes correct, at least one line correctly drawn | M1 A1 | Axes correct, at least one line correctly drawn; correct graph with all lines drawn with ruler, scale clear |
| $5-9p = -1+2p \Rightarrow p = \frac{6}{11}$ | A1 | Using graph to obtain correct probability; correct value of $p$ interpreted in context |
| Meera should play Option X with probability $\frac{6}{11}$ and Option Z with probability $\frac{5}{11}$ | A1 | 3.2a |
| (ii) Value to Meera $= -1 + 2 \times \frac{6}{11} = \frac{1}{11}$; Value to Haruki $= \frac{-1}{11}$ | B1ft | Correct value of game to Haruki (ft their $p$) |

## Part (c)(iii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Haruki never plays B and D | B1 | CAO — B and D |

## Part (d):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Value of game $= 0$ | B1 | States value $= 0$; may be implied by forming equation in $p$ or $q = 0$ |
| When $H$ plays Option A, $M$'s gain: $5 - 9p = 0 \Rightarrow p = \frac{5}{9}$ | M1 | Considers Option A and obtains value for $p$ (alt: considers $H$ and obtains value for $q$) |
| When $H$ plays Option C, $M$'s gain: $-\frac{5k}{9} - \frac{4}{9} = 0$ | M1 | Considers Option C and sets up equation in $k$; dependent on previous M mark |
| $k = -\frac{4}{5}$ | A1 | CAO $\left(-\frac{4}{5}\right)$ |

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