# Question 4:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| **(i)** Row minima: $-6, -7, 3$; max is $3$. Column maxima: $5, 6, 3$; min is $3$. Row(maximin) $=$ Col(minimax) therefore game is stable. | M1, A1 | M1 for identifying row minima and column maxima; A1 for correct conclusion (AO 2.4) |
| **(ii)** Value of game to B is $-3$ | B1 | AO 2.2a |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Let A play option P with probability $p$ and option Q with probability $1-p$ | B1 | AO 3.3 |
| If B plays X, A's gains are $-6p + 5(1-p) = 5 - 11p$; If B plays Y, A's gains are $-p + 4(1-p) = 4 - 5p$; If B plays Z, A's gains are $2p + (-7)(1-p) = -7 + 9p$ | M1, A1 | M1 for at least one correct expression; A1 for all three correct |
| Graph plotted with three lines and highest intersection identified | M1dep, A1 | M1dep on previous M1; A1 for correct graph |
| $5 - 11p = -7 + 9p \Rightarrow p = \tfrac{3}{5}$ | A1 | CAO |
| A should play P with probability $0.6$ and Q with probability $0.4$ | A1ft | Follow through; AO 3.2a |

## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| As indicated by the graph, Brendan can for all values of $p$ gain more by playing X or Z; e.g. for $0 \le p \le \tfrac{3}{5}$ Brendan better off playing Z, and for $\tfrac{3}{5} < p \le 1$ Brendan better off playing X than playing Y | B1 | AO 3.2a |

## Question (d):

### Part (d)(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| If A plays option P then B can expect to gain $-(-6q + 2(1-q))$ as the values in the table are the pay-offs for A | B1 | AO 2.2a — correctly deducing the LHS of the given equation with clear reasoning for the change in sign; only allow stating $6q - 2(1-q)$ without seeing the change of sign if the game is restated for player B |
| The value of the game to B is $-(5 - 11(0.6)) = 1.6$, or equivalent calculation e.g. $-(-7 + 9(0.6))$ | B1 | AO 2.1 — correctly deriving the RHS; must indicate this is the value of the game to B; stating $\pm 1.6$ without working is B0; **candidates must explain where both parts of the given equation came from** |

### Part (d)(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $6q - 2(1-q) = 1.6 \Rightarrow q = 0.45$ | B1 | AO 1.1b — CAO; must come from solving $6q - 2(1-q) = 1.6$ |
| B should play option X with probability $0.45$ and option Z with probability $0.55$ | B1 | AO 3.2a — CAO in context; must refer to 'play' |

**Total for part (d): 4 marks**

**Total for question: 15 marks**