Question 6:
6 | (a) | Delete W
Delete Z
X Y
A 0 2
B 2 -3
C 4 -4
D -1 5 | B1
B1
[2] | 2.2a
2.2a | From column headings in a reduced table
Ignore if any rows also deleted
In each row, value in col W > value in col Y
So col W is dominated by col Y or col Y dominates col W
From column headings in a reduced table
Ignore if any rows also deleted
In each row, value in col Z > value in X
So col Z is dominated by col X or col X dominates col Z
Alternative method
A | B | C | D
X | 0 | -2 | -4 | 1
Y | -2 | 3 | 4 | -5
Or transposed
For reference:
W X Y Z
A 4 0 2 1
B 0 2 -3 4
C 1 4 -4 5
D 6 -1 5 0
6 | (b) | X has probability p and Y has prob 1 – p
A: 0p + 2(1 – p) = 2 – 2p
B: 2p – 3(1 – p) = 5p – 3
C: 4p – 4(1 – p) = 8p – 4
D: –p + 5(1 – p) = 5 – 6p | M1 ft | 3.1a | Definition of p may be implied for M marks
Using their reduced table (or correct) and their (single) variable
At least two correct expressions (need not be simplified)
seen or implied from graph
Using values from original table means that these are losses for
Casey, so will need the minimax (lowest point of upper boundary)
Alternative method (change signs)
X has probability p and Y has prob 1 – p
A: 0p – 2(1 – p) = –2 + 2p
B: –2p + 3(1 – p) = –5p + 3
C: –4p + 4(1 – p) = –8p + 4
D: p – 5(1 – p) = –5 + 6p | X has probability p and Y has prob 1 – p | M1 ft | Definition of p may be implied for M marks
A: 0p – 2(1 – p) = –2 + 2p | M1 ft | Using their reduced table (or correct) and their (single) variable
B: –2p + 3(1 – p) = –5p + 3 | At least two correct expressions (need not be simplified)
C: –4p + 4(1 – p) = –8p + 4 | seen or implied from graph
D: p – 5(1 – p) = –5 + 6p | Using negatives of values from original table means that these are
gains for Casey, so will need the maximin (highest point of lower
boundary)
8p – 4 = 5 – 6p
9
p =
14
Optimal mixed strategy: choose randomly
9
between X and Y, choosing X with prob
14
5
and Y with prob
14 | M1 ft
A1
A1
[4] | 3.1a
1.1
3.2a | Solving for (their) p (graphically or algebraically or implied)
p = 0.643 (or better) (0.6428571…) (cao)
Correct probabilities described in context involving both X and Y
Or correct ratios o.e.
e.g randomly choose X 1.8 times as often as Y
6 | (b) | cont | If W and Z are chosen
X has probability p and Y has prob 1 – p
A: 4p + (1 – p) = 1 + 3p
B: 0 + 4(1 – p) = 4 –4p
C: p + 5(1 – p) = 5 –4p
D: 6p + 0(1 – p) = 6p
5 – 4p = 6p (= 1)
2 | Or the negatives of these for the alternative method
Definition of p may be implied for M marks
Using their reduced table (or correct) and their (single) variable
At least two correct expressions (need not be simplified)
seen or implied from graph
Or 4p – 5 = -6p for the alternative method
M1 ft
M1 ft
A0
A0
If W and Z are chosen
X has probability p and Y has prob 1 – p
A: 4p + (1 – p) = 1 + 3p
B: 0 + 4(1 – p) = 4 –4p
C: p + 5(1 – p) = 5 –4p
D: 6p + 0(1 – p) = 6p
5 – 4p = 6p (= 1)
2
Or the negatives of these for the alternative method
Definition of p may be implied for M marks
Using their reduced table (or correct) and their (single) variable
At least two correct expressions (need not be simplified)
seen or implied from graph
Or 4p – 5 = -6p for the alternative method
6 | 0 | 24
10.5 | 1.5 | 43.5
11.25 | 3.75 | 48.75