# Question 7(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Row minima are $-2$, $\min(k,-4)$ and $-3$ | M1 | Attempt to calculate row minima; condone $k$ or $-4$ for $\min(k,-4)$ |
| Play safe for Player A is always option Q, since $-2$ is greater than $-4$ and $-3$; if row minimum for R is $k$ then $-4 > k$ so play safe still Q | A1 | Correct argument/conditions |

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# Question 7(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Column maxima are $4, 6, 3$ and $\max(k,-2)$ | B1 | Attempt to calculate column maxima; condone $k$ or $-2$ for $\max(k,-2)$ |
| Play safe is option Z, $k < 3$ or $k \leq 2$ | B1 | CAO; ignore lower limit figure if given |

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# Question 7(c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Option Y dominates option X | B1 | Must include word 'dominate'; also e.g. option Z dominates option X |
| Because $-1 < 3$, $-4 < 5$ and $3 < 6$ | B1 | Correct inequalities — must be clear all three hold |

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# Question 7(d)(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Augmented matrix: $\begin{pmatrix} 8 & 3 & 2 \\ 1 & 0 & k+4 \\ 3 & 7 & 1 \end{pmatrix}$ | B1 | Correct augmentation; possibly implied by later working; X column may be included |
| $V - 8p_1 - p_2 - 3p_3 + r = 0$ $V - 3p_1 - 7p_3 + s = 0$ $V - 2p_1 - (k+4)p_2 - p_3 + t = 0$ | M1 A1 | M1: at least three equations in $V, p_1, p_2, p_3$ with at least one dummy variable; A1: CAO |
| $p_1 + p_2 + p_3 + u = 1$ | B1 | Correct probability equation |

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# Question 7(d)(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct row and column labels for Simplex tableau | B1 | |
| Initial Simplex tableau with correct entries: b.v. $V$, $p_1$, $p_2$, $p_3$, $r$, $s$, $t$, $u$, Value; rows for $r,s,t,u,P$ as shown | M1 | Any one numerical row correct |
| CAO for full tableau | A1 | |

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# Question 7(e):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $p_3 = \dfrac{13}{37}$ | B1 | $p_3$ correctly stated |
| $V - 8\!\left(\dfrac{7}{37}\right) - \left(\dfrac{17}{37}\right) - 3\!\left(\dfrac{13}{37}\right) + 0 = 0$ or $V - 3\!\left(\dfrac{7}{37}\right) - 7\!\left(\dfrac{13}{37}\right) + 0 = 0$ | M1 | Attempt to calculate $V$ using either equation from (d) not involving $k$ |
| $V = \dfrac{112}{37}$, then $\dfrac{112}{37} - 2\!\left(\dfrac{7}{37}\right) - (k+4)\!\left(\dfrac{17}{37}\right) - \left(\dfrac{13}{37}\right) + 0 = 0$ | dM1 | Dependent on previous M; uses equation in $k$ with their $V$ to find $k$ |
| $k = 1$ | A1 | CAO |