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Section A: STATISTICS

Qu 1 Scheme Marks AO
(a) c 0 1 2 3 4 5 6 7 8 B1 1.2
P(C = c) 1
9
1
9
1
9
1
9
1
9
1
9
1
9
1
9
1
9
B1ft 1.2
(2)
(b) P(C < 4) = 4
9
(accept 0.444 or better) B1 3.4
(1)
(c) Probability lower than expected suggests model is not good B1ft 3.5a
(1)
(d) e.g. Cloud cover will vary from month to month and place to place B1 3.5c
So e.g. use a non-uniform distribution (1)
( 5 marks)
Notes
(a) 1 B1 for a correct set of values for c. Allow 18 , 82 ,... 88
st

2nd B1ft for correct probs from their values for c, consistent with discrete nifo m di ib n
Maybe as a prob. function. Allow P(X = x ) = 19 for 0 x 8 provided x = 0, 1, 2, ,8 i
clearly defined somewhere.

(b) B1 for using correct model to get 94 (o.e.)

SC Sam le ace 1, , 8 If scored B0B1 in (a) for this allow P(C < 4) = 3
8 to score B1 in (b)

(c) B1ft for comment that states that the model proposed is or is not a good one based on
their model in part (a) and their probability in (b)
|(b) 0.315| > 0.05 Allo e.g. i i no i able ; i i no acc a e e c
|(b) 0.315| 0.05 Allow a comment that suggests it is suitable
No prob in (b) Allow a comparison that mentions 50% or 0.5 and rejects the model
No prob in (b) and no 50% or 0.5 or (b) > 1 scores B0
Ignore any comments about location or weather patterns.

(d) B1 for a sensible refinement considering variations in month or location

J a ing no nifo m i B0
C n e & n n- nif m Allow mention of different locations, months and non-uniform
or use more locations to form a new distribution with probabilities based on frequencies
C n e & bin mial Allow mention of different locations, months and binomial
Just refined model Model must be outlined and discrete and non-uniform
e.g. higher probabilities for more cloud cover or lower probabilities for less cloud cover
Continuous model Any model that is based on a continuous distribution. e.g. normal is B0
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Qu 3 Scheme Marks AO
(a) The probability of a dart hitting the target is constant (from child to child and B1 1.2
for each throw by each child) (o.e.)
The throws of each of the darts are independent (o.e.) B1 1.2
(2)
(b) [P(H 4) = 1 P(H 3) = 1 0.9872 = 0.012795.. =] awrt 0.0128 B1 1.1b
(1)
(c) P(F = 5) = 0.9 4
0.1 , = 0.06561 M1, 3.4
= awrt 0.0656 A1 1.1b
(2)
(d) n 1 2 10
M1 3.1b
P(F = n) 0.01 0.01 + 0.01+9
10 3.1a
Sum of probs = 1 2 0.01 9 1 M1A1
2 1.1b
[i.e. 5(0.02 + 9 ) = 1 or 0.1 + 45 = 1] so = 0.02 A1 1.1b
(4)
(e) P(F = 5 Thoma model) = 0.09 B1ft 3.4
(1)
(f) Pe a model assumes the probability of hitting target is constant (o.e.)
B1 3.5a
and Thomas model assumes this probability increases with each attempt(o.e.)
(1)
(11 marks)
Notes
st
(a) 1 B1 for stating that the probability (or possibility or chance) is constant (or fixed or same)
2nd B1 for stating that throws are independent [ ial a e independen i B0]

(b) B1 for awrt 0.0128 (found on calculator)

(c) M1 for a probability expression of the form (1 p)4 p where 0 < p < 1
A1 for awrt 0.0656
SC Allow M1A0 for answer only of 0.066

(d) 1st M1 for setting up the distribution of F with at least 3 correct values of n and P(F = n) in
terms of . (Can be implied by 2nd M1 or 1st A1)
nd
2 M1 for use of sum of probs = 1 and clear summation or use of arithmetic series formula
(allow 1 error or missing term). (Can be implied by 1st A1)
1st A1 for a correct equation for
2nd A1 for = 0.02 (must be exact and come from correct working)

(e) B1ft for value resulting from 0.01 + 4 "their " (provided and the answer are probs)
Beware If their answer is the same as their (c) (or a rounded version of their (c)) score B0

(f) B1 for a suitable comment about the probability of hitting the target
ALT Allo idea ha Pe a model gge he da ma ne e hi he a ge b Thoma a ha
it will hit at least once (in the first 10 throws).
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Question Scheme Marks AOs

1(a) 2
3
G B1
4
G 1.1b
5
1
9 G R dB1
10 3
1.1b
1
5 R
1
10 R (2)
(b) 9 4 2
M1 1.1b
10 5 3
12 (= 0.48)
A1 1.1b
25
(2)
(c) 9 1 9 4 1 or 1 9 4 2
1 M1 3.1b
10 5 10 5 3 10 10 5 3
21 (= 0.42) A1 1.1b
50
(2)
(d) 9 1 9
[P(Red from B|Red selected) ] 10 5 50
M1 3.1b
1 9 1 9 4 1 13
10 10 5 10 5 3 25

9
A1 1.1b
26
(2)
(8 marks)
Notes
Allow decimals or percentages throughout this question.
B1: for correct shape (3 pairs) and at least one label on at least two pairs
G(reen) and R(ed)
allo G and G or R and R as labels, etc.
(a)
condone e tra pairs if they are labelled with a probability of 0
dB1: (dep on previous B1) all correct i.e. for all 6 correct probabilities on the
correct branches with at least one label on each pair
M1: Multiplication of 3 correct probabilities (allow ft from their tree diagram)
(b) A1: 12 oe
25
M1: Either addition of only two correct products (product of two probs +
product of three probs) which may ft from their tree diagram
(c) or for 1 (' 101 ' '(b) ')
A1: 21 oe
50
M1: Correct ratio of probabilities
or correct ft ratio of probabilities e.g. ' 10 ' ' 5 ' or ' 10 ' ' 5 ' with num < den
9 1 9 1

(d) 1 '(b) ' ' 101 ' '(c) '

A1: 9 (allow awrt 0.346)
26
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Qu 1 Scheme Marks AO
(a) A, C or D, B or D,C B1 1.2
(1)
(b) [p = 0.4 – 0.07 – 0.24 = ] 0.09 B1 1.1b
(1)
(c) A and B independent implies 1.1b
P( A) 0.4 0.24 or q 0.16 0.24 0.4 0.24 M1

so P(A) = 0.6 and q = 0.20 A1cso 1.1b

(2)
(d)(i) r r
P B | C = 0.64 gives 0.64 or 0.64 M1 3.1a
r p r "0.09"
r = 0.64r + 0.64 “p” so 0.36r = 0.0576 so r = 0.16 A1 1.1b

(ii) Using sum of probabilities = 1 e.g. “0.6” + 0.07 + “0.25” + s =1 M1 1.1b

so s = 0.08 A1 1.1b
(4)

( 8 marks)
Notes
(a) B1 for one correct pair. If more than one pair they must all be correct.
Condone in a correct probability statement such as P A C = 0
or correct use of set notation e.g. A C
BUT e.g. “P(A) and P(C) are mutually exclusive” alone is B0

(b) B1 for p = 0.09 (Maybe stated in Venn Diagram [VD])

[ If values in VD and text conflict, take text or a value used in a later part]

(c) M1 for a correct equation in one variable for P(A) or q using independence
or for seeing both P( A B) P( A) P( B) and 0.24 0.6 0.4
A1cso for q = 0.20 or exact equivalent (dep on correct use of independence)
Beware Use of P(A) = 1 – P(B) = 0.6 leading to q = 0.2 scores M0A0

(d)(i) 1st M1 for use of P B | C = 0.64 leading to a correct equation in r and possibly p.
Can ft their p provided 0 < p < 1
1st A1 for r = 0.16 or exact equivalent
(ii) 2nd M1 for use of total probability = 1 to form a linear equation in s. Allow p, q, r etc
Can follow through their values provided each of p, q, r are in [0, 1)
nd
2 A1 for s = 0.08 or exact equivalent
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Qu 4 Scheme Marks AO
(a) k k k k k 1
1 or 60k 30k 20k 15k 12k 1 M1 1.1b
10 20 30 40 50 600
600
So k = (*) A1cso 1.1b
137
(2)
(b) (Cases are:) D1 30, D2 50 and D1 50, D2 30 and D1 40, D2 40 M1 2.1
2
k k k
P D1 D2 80 2 + M1 3.4
50 30 40
= 0.0375619… awrt 0.0376 A1 1.1b
(3)
(c) Angles are: a, a d , a 2d , a 3d M1 3.1a
S4 = a + (a + d) + (a + 2d) + (a + 3d) = 360 M1 2.1
2a + 3d = 180 (o.e.) A1 2.2a
Smallest angle is a > 50 consider cases:
M1 3.1b
d = 10 so a = 75 or d = 20 so a = 60 [ d = 30 gives a = 45 no good]
3k 90
P(D = 10 or 20) = = A1 1.1b
20 137
(5)
( 10 marks)
Notes
(a) M1 for clear use of sum of probabilities = 1 (all terms seen)
A1 cso (*) M1 scored and no incorrect working seen.
Verify (Assume k = 137
600
) to score the final A1 they must have a final comment “ k 600
137 "

(b) 1st M1 for selecting at least 2 of the relevant cases (may be implied by their correct probs)
e.g. allow 30, 50 and 50,30 i.e. D1 and D2 labels not required
2nd M1 for using the model to obtain a correct expression for two different probabilities.
May use letter k or their value for k.
2 2
k k k k k k
Allow for + or 2 +
50 30 40 50 30 40
A1 for awrt 0.0376 (exact fraction is 705
18769 )

(c) 1st M1 for recognising the 4 angles and finding expressions in terms of d and their a
2nd M1 for using property of quad with these 4 angles (equation can be un-simplified)
Allow these two marks for use of a (possible) value of d
e.g. a + a + 10 + a + 20 + a + 30 = 360 (If at least 3 cases seen allow A1 for e.g. 4a = 300)
or allow M1M1 for a set of 4 angles with sum 360 and possible value of d (3 cases for A1)
e.g. (for d = 20) 60, 80, 100, 120
1st A1 for 2a + 3d = 180 condition (o.e.) [Must be in the form pa + qd = N]
3rd M1 for examining cases and getting d = 10 and d = 20 only
2nd A1 for 137 90
or exact equivalent
The correct answer and no obviously incorrect working will score 5/5
A final answer of awrt 0.657 (0.65693…) with no obviously incorrect working scores 4/5
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Question Scheme Marks AOs

4(a) P( A B ) 0.33
P( A | B ) or M1 3.1a
P( B ) 0.55
3
= or 0.6 A1 1.1b
5
(2)
(b) e.g. P( A) P(B) 7
20
9
20
63
400 P( A B) 0.13 52
400
B1 2.4
or P( A | B ) 0.6 P(A ) 0.65
(1)
(c) B1 2.5
M1 3.1a
A1 1.1b

M1 1.1b

A1 1.1b
(5)
(d) P( B C ) 0.22 0.22 or 1 [0.56]
o.e. M1 1.1b
or 1 [0.13 0.23 0.09 0.11]
= 0.44 A1 1.1b
(2)
(10 marks)
Notes:
(a)
M1: for a correct ratio of probabilities formula and at least one correct value.
A1: a correct answer
(b)
for a fully correct explanation: correct probabilities and correct comparisons.
(c)
B1: for box with B intersecting A and C but C not intersecting A.( Or accept three
intersecting circles, but with zeros entered for A C and A B C )No box is B0
M1: for method for finding P( B C )
A1: for 0.09
M1: for 0.13 and their 0.09 in correct places and method for their 0.23
A1: fully correct
(d)
M1: for a correct expression – ft their probabilities from their Venn diagram.
A1: cao

128 Pearson Edexcel Level 3 Advanced GCE in Mathematics – Sample Assessment Materials –
Issue 1 – April 2017 © Pearson Education Limited 2017
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Question
Scheme Marks
Number
2. (a) B and C (only) B1
(1)
(b) P( A C ) 0.6 0.35 so [w =] 0.21 B1cso
(1)
(c) x = P(C) – w = 0.14 B1
y = P(A) – w – P(B) ,= 0.24 M1,A1
z = 1 P( A C ) = 0.26 B1ft
(4)
(d) [x + y = ] 0.38 B1ft
(1)
(e) [ P( B C ) = 0.15 + 0.35] = 0.5 B1cao
(1)
P( A [ B C ]) 0.15 0.21
(f) P A| B C M1A1ft
P(B C ) "0.5"
= 0.72 A1
(3)
[Total 11]
Notes
(a) B1 for just B and C [NB Just writing P( B C) 0 is B0]

(b) B1cso for 0.21 clearly from P(A) P(C) or 0.6 0.35 and no incorrect statements seen

(c) 1st B1 for x = 0.14

M1 for a correct expression for y
A1 for y = 0.24
2nd B1ft for z = 0.26 or correct ft of their values to make sum = 1 (provided all probs)
These values may be seen in correct regions in the Venn diagram

(d) B1ft for their x + y or 0.38

(e) B1 for 0.5 or exact equivalent

(f) M1 for a correct ratio of probabilities formula num of: P(B C A) or P( A [ B C ])

with brackets and some correct probability, ft their (e) May be implied by correct ratio.
1st A1ft for a numerator of 0.15 + 0.21 and a denominator of their (e)
0.15 0.21
Can award M1A1ft for even if their formula is incorrect
"their 0.5"
2nd A1 for 0.72 or exact equivalent e.g. 1825
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Question
Scheme Marks
Number
4. (a) [Let P(A) = p]
0.4 p 0.7(1 p) 0.45 M1A1
0.25 = 0.3p M1
5
p= A1
6

B1ft

B1

(6)
1
6 0.3
(b) P( A | B ) M1
0.55
1
= A1
11
(2)
[Total 8]
Notes
(a) 1st M1 for 0.4p or 0.7(1 – p ) seen in an equation for p
1st A1 for a fully correct equation for p
1st M1 for attempt at 2 sim’ eq’ns in p and q Allow one error.
ALT 0.4 p 0.7 q 209 and 0.6 p 0.3q 11 20
1st A1 for any correct equation in p or q
2nd M1 for simplifying their linear equation with at least 2 terms in p or q to a = bp or bq
2nd A1 for P(A) = 56 or exact equiv e.g. 0.83 (may be seen on their tree diagram)
1st B1ft for 1st 2 branches i.e. 5
6 and 1
6 (follow through their P(A))
2nd B1 for 2nd 4 branches i.e. 3
5 and 3
10

(b) M1 for a ratio of probabilities ft their numerator from their tree diagram but denom = 0.55
A1 for 111 or exact equivalent e.g. 0.09

5 P( A ) 103
SC [P(A) ] award M1A0 for ft their P(A) and P( A ) 1 P( A)
6 P( A) 53 P( A ) 3
10
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Question
Scheme Marks
Number
3. (a)
B1
B1
B1
B1

(4)
12 23 13 48 3
(b)(i) P(S) = or or 0.6 B1ft
80 80 5
(1)
P( S C ) 12
(ii) P(S | C) = 80
M1
P(C ) 20
80

12 A1cso
= or 0.6
20
(2)
(iii) P(S) = P(S|C) or P(C) = 0.25, P (C S ) = 0.15 and P(C) P(S) = 0.6 0.25 B1ft
so S and C are independent dB1ft
(2)
(c) Need P(S | G) = 13 M1A1
23
P(S | C) = 0.6 > 0.565 so assistant selling coats has the better performance A1
(3)

[Total 12]
Notes
(a) 1st B1 for 3 labelled circles with 12, 13 & n (C G) = 0 marked or implied (e.g.RH diagram)
2nd B1 for 8 and 10 correctly placed
May use probabilities not integers
3rd B1 for 23 correctly placed
A blank space does not imply a zero
4th B1 for box and 14

(b)(i) B1ft for 0.6 or any exact equivalent (single fraction) or ft their values (ft blank as 0)

(ii) M1 for a correct conditional prob. Correct expression and one correct ft prob. Num < Den
A1cso for 0.6 which must come from a denominator of 20

1st B1ft for a full reason. If not P(S) = P(S|C) then all values must be stated, labelled and
(iii) correct or correct ft from diagram. Co ec no n e i ed o P(S C ) 0.15 is B0B0
nd
2 dB1ft dep. on a correct reason for correct conclusion for their values

(c) M1 for attempt at P(S | G) correct ratio of probabilities or numbers using their figs
st
1 A1 for 1323 (accept awrt 0.565) [Sight of P(S | G) = 1323 is M1A1]
2nd A1 fo a co ec concl ion ha choo e coa ba ed on a co ec com a i on
Allow incorrect P(S|C) provided > 0.565 to score 2nd A1 and so all 3 marks
Condone poor use of notation eg S|G with no P(..). Probabilities may be described in words.
Condone comparison of 1323 with 0.6 even if 1323 is not labelled as P(S|G)
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Question Scheme Marks
4. (a) 0.4p + 0.15( 1 – p ) = 0.26 M1
0.25p = 0.11 dM1
p = 0.44 A1
(3)
(b) "0.56" q
= 0.175 M1A1ft
"0.56" q "0.44" 0.6
0.462q = 0.0462 dM1
q = 0.1 A1
(4)
(c) P(C) = (1 – p ) (1 – 0.15 – q ) = “0.56” "0.75" M1
= 0.42 A1
(2)
(d) P( R) (a) "0.44" M1
P( R | C )
P(C ) 1 (c) "0.58" M1
22
= = 0.75862… or awrt 0.759 A1
29
(3)
(12 marks)
Notes
(a) 1st M1 for attempt at correct equation for p (Must have at least 2 terms in p) and
must be set equal to 0.26
2nd dM1 dep on 1st M1 for solving their linear equation in p by reducing to Ap = B
with at least 1 of A or B correct
A1 for p = 0.44 (or exact equivalent e.g. 11
25 )

(b) rq
1st M1 for a probability ratio of the form
rq (1 r ) 0.6
1st A1ft for r = 1 – their p and the = 0.175
2nd dM1 dep on 1st M1 for rearranging their equation into the form Aq = B
with at least 1 of A or B correct or correct ft
2nd A1 for q = 0.1 or an exact equivalent

(c) M1 for (1 – their p) ( 1 – 0.15 – their q)

A1 for 0.42 or an exact equivalent

(d) 1st M1 for a ratio of probabilities with 0.44 or ‘their (a) on num.
2nd M1 for a ratio of probabilities with 0.58 or ‘1 – their (c) on denom.
A1 for 22
29 or awrt 0.759
Correct answer only scores 3 out of 3.
Note: If correct ft on num. and denom. leads to “num” > “denom” then
maximum score is M0M1A0)
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A B

A B C
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Question Scheme Marks

3(a) 30
[P(Female) ] oe B1
90
(1)
(b) P(Male 4 years) 16
( 90 ) 16
[P(Male | 4 years) ] oe M1A1
P( 4 years) 16 9
( 90 ) 25
(2)
(c) P(Male 10 years) 20 16
( 90 ) 36
[P(Male | 10 years) ] M1A1
P( 10 years) 9 16 14 20
( 90 ) 59
(2)
(d) 16 60
P(Male | 4 years) ' ' , P(Male) or
25 90
16 25
P( 4 years | Male) , P( 4 years) or M1
60 90
16 60 25
P(Male 4 years) , P(Male) , P( 4 years)
90 90 90
P(M | < 4) P(M) or P(< 4 | M) P(<4) or M1
P(Male 4 years) P(M)×P(< 4)
so not independent. A1
(3)
Total 8
Notes
(a) 30
B1 for or exact equivalent
90
(b) M1 for a correct ratio expression with at least one correct probability
substituted or for a correct ratio of probabilities. num>denom is M0
16
A1 or 0.64 (Correct answer scores 2 out of 2).
25
(c) M1 for a correct ratio expression with at least one correct probability
substituted or for a correct ratio of probabilities. num>denom is M0.
36
A1 or condone awrt 0.610 (must be 3sf) (Correct answer gets 2 out of 2).
59

(d) 1st M1 for stating all of the required numerical probabilities for a correct test
which must be labelled. The probabilities must be correct or correct ft from (b)
16
(If attempting the first test, P(Male | 4 years) ' ' was found in part(b) and
25
need not be fully restated here).
2nd M1 for use of a correct test. Must see the product if attempting the 3rd test.
A1 for correct test with all probabilities correct and a correct conclusion.

NB Use of A and B throughout scores M0M0A0 unless A and B are explicitly

defined.

9
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Question Scheme Marks

4.(a) 1 1 1
[P(both blue) ] oe B1
20 20 400
(1)
(b) 1 19 19
P(exactly 1 red) 2 ,= oe M1, A1
20 20 200
(2)
(c) 4 5 4 10 B1 M1
P(2 yellow and 1 green)=3 oe
9 8 7 21 A1
(3)
(d) 5 4 3 2 M1
P(All beads are yellow) =
9 8 7 6
P(At least 1 bead is green) = 1 – P(All beads are yellow)
5 4 3 2 121
1 M1A1
9 8 7 6 126
(3)
Total 9
Notes
(a) 1
B1 or 0.0025
400
(b) 1 19 19 1
M1 for a correct equivalent expression +
20 20 20 20
19
A1 or 0.095
200
(c) B1 for 3 × … or for the sum of exactly 3 identical products attempted
M1 for any one product correct
10
A1 (allow awrt 0.476 from correct working)
21
(d) st 5 4 3 2
1 M1
9 8 7 6
2nd M1 Use of 1 p (where p is a product of 4 probabilities)
121
A1 (condone awrt 0.960 must be at least 3sf from correct working)
126
OR
1st M1 List all 15 favourable outcomes and at least one correct product
(YYYG)×4 [(YYGY), (YGYY), (GYYY)]
(YYGG)×6 [(YGYG), (YGGY), (GYYG), (GYGY), (GGYY)]
(GGYG)×4 [(GGGY), (YGGG), (GYGG)]
(GGGG)

2nd M1 Sum all 15 correct probabilities

121
A1 (condone awrt 0.960 must be at least 3sf from correct working)
126

10
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Question Scheme Marks

6. (a) P(S) = 0.31+ p , P(D) = 0.35 , P( S D) 0.14 M1
( 0.31 + p )(0.35) = 0.14 oe M1
P(S) = 0.4 or 0.31 + p = 0.4 or 0.35p = 0.0315 A1
p = 0.09 A1
(4)
(b) P( S M D) 1 so q = 1 – ( 0.17+0.10+0.15+0.06+0.04) – p or 0.48 – p M1
q = 0.39 A1ft
(2)

(c)(i) P( D S M ) 0.10
[ P( D | S M ] M1
P( S M ) 0.27
10
= or awrt 0.370 A1
27
(ii) P( D S ' M ) 0.15
[ P( D | S ' M ] M1
P( S ' M ) 0.54
5
= or awrt 0.278 A1
18
(4)
(d) 27 order S M so expect 27 10
27
D or 36 order S M so expect 36 5
18
D M1
So expect 20 (desserts) A1cao
(2)
[12]
Notes
st
(a) 1 M1 for attempting P(S), P(D) and P ( S D) with at least 2 correct.
These may be seen in a conditional probability.
0.14 0.14
NB P(S | D) = and P(D| S) =
0.35 0.31 p
nd
2 M1 using the independence condit’ and their values to form a suitable equation for p or P(S)
1st A1 for P(S) = 0.4 or 0.31 + p = 0.4 or 0.35p = 0.0315 (i.e. one move from p = …)

(b) M1 for using sum of probabilities = 1 and ft their p

A1ft for 0.48 – “their p” (provided 0 < their p < 0.48 )

(c) 1st M1 for a correct ratio of probabilities or a correct ratio expression with at least one correct
probability substituted. (M0 if numerator is P(D)×P( S M ) or numerator>denominator)
10
1st A1 for or awrt 0.370
27
2nd M1 for a correct ratio of probabilities or a correct ratio expression with at least one correct
probability substituted. (M0 if numerator is P(D)×P( S M ) or numerator>denominator)
5
2nd A1 for or awrt 0.278
18

(d) M1 for at least one correct calculation ft their probabilities from (c).
i.e. either 27×their (c)(i) or 36×their (c)(ii)
 www.yesterdaysmathsexam.com

Question
Scheme Marks
Number

2. (a) B1

(1)

(b) [ P(C D) 0 ] B1
(i) 1 B1
P(C D) P(C ) P( D)
2
(ii) B1
P(C D)
P(C | D)[ ] 0
P( D ) (3)

(c)(i) P( F G ) P( F ) P(G ) P( F G)
3 1 1
P(G ) P(G ) M1 M1
8 6 6
1 A1
P(G )
4
(ii) 1 B1
P( F | G )[ P( F )]
6
(4)

[8 marks]
Notes
nd
(a) If a 2 diagram is drawn then award B0 unless the incorrect diagram is crossed out

(b) 1st B1 for writing or using P(C D) 0 anywhere in (b)(may be implied by correct P(C | D))
st
P(C D) P(C ) P(D) P(C D) 15 103 2 does imply 1
1
B1
2nd B1 for P(C D) 0.5 (o.e.) (may just be labelled (b)(i) 1
2 ) This does not imply 1st B1
3rd B1 for P(C | D) = 0…this will imply 1st B1 too

(c)(i) 1st M1 for use of addition formula (3 terms) with correct substitution of at least one term
Assuming or stating P( F G) 0 scores M0
2nd M1 for use of independence P( F G) P( F ) P(G) = 16 P(G) (i.e. must be used)
Use of e.g. x for P(G) is fine. NB 83 16 P(G) is M0M0
ALT Let y = P( F G) then P(G) = y 83 16 scores 1st M1 and y 16 ( y 83 16 ) o.e. gets 2nd M1
A1 for 14 o.e.
(ii) B1 for P( F | G ) 16 (may be labelled (c)(ii) 16 ) Accept exact equivalents.
 www.yesterdaysmathsexam.com

Question
Scheme Marks
Number
(The event that) the integer selected is prime and ends in a 3 (and is between 1
2. (a) B1
and 50 inclusive)
(1)

(b) 15 B1
(or equivalent e.g. 0.30) [condone 30%]
50 (1)

12 B1
(c) (or equivalent e.g. 0.24) [condone 24%]
50 (1)

P(A C ) 7
50 7 M1, A1
(d) P( A| C ) , (2)
P(C ) 30
50 30

15 7 M1, A1
(e) , so not independent.
50 30 (2)

P(B A C ) 2
50 2 M1, A1
(f) P( B|(A C )) , (2)
P( A C ) 7
50 7
[9 marks]

(d) M1 for a correct ratio expression (may be in words) with at least one correct probability
7 m
substituted or correct ratio expression and or where 7 < n or m < 30
n 30
or fully correct ratio of probabilities.
7
A1 for or any exact equivalent e.g. 0.23 but 0.233 is M1A0 (Correct ans only = M1A1)
30

(e) M1 for correctly comparing their (b) ith their (d) , can be in words or symbols
e.g. P(A) P(A | C) in symbols.
A1 dependent on a correct (b) and (d) (or awrt 0.233 in (d)) and for concluding
not independent

15 30 7
For a correct test using correctly labelled P( A) , P(C ) and P( A C )
50 50 50
SC 15 30 9 7
with all correct probabilities and (o.e.) seen leading to
50 50 50 50
not independent score M0A1

(f) M1 for a correct ratio expression (may be in words) with at least one correct probability
r 2
substituted or correct ratio expression and or where r < 7 or 2 < t
7 t
or fully correct ratio of probabilities
2
A1 for or an exact equivalent. Allow awrt 0.286 here as well.(Correct ans. only = M1A1)
7
 www.yesterdaysmathsexam.com

Question
Scheme Marks
Number
5. (a) Age Computer use
0.80 Use computer every day

< 50
B1ing
p Does not use computer
(0.20)
every day
B1
(1 – p) 0.55 Use computer every day Sp
50 .98)
(2)
(0.45) Does not use computer
every day

(b) p × 0.80 + (1 – p) × 0.55 = 0.70 M1

p = 0.6 A1
(2)

(c) P( 50 use computer daily) '0.6' 0.80 M1

P( 50| use computer daily)
P(use computer daily) 0.70
48
A1oe
70
(2)
[6 marks]
Notes
(a) Allow undefined letters for labels e.g. U(use) and U or N and NE
Allow labels on branches and probabilities at the ends
Condone 80% and 55% etc on tree diagram and in (b)
st
1 B1 for correct shape (2 branches then 4 branches) and correct labels on first set of
branches (p , < 50 and 50 but condone > 50 )
nd
2 B1 for correct labels on second set of branches (0.80, 0.55, daily and not daily)
Allow 0.8p and 0.55(1 – p ) on or at the end of the appropriate branches.
NB they do not require the probabilities in brackets for either of these two marks.

(b) M1 for a correct equation to find p using their tree diagram.

A1 for 0.6 [ condone 60%] (Correct answer only will score M1A1)

M1 for a correct expression with 0.70 substituted correctly and numerator < denominator
(c) or correct ratio of probabilities f.t. their p provided 0 p 78
48 24
A1 for or an exact equivalent e.g. (Correct answer only is M1A1)
70 35
Allow awrt 0.686 following a correct expression. [68.6% is A0]
 www.yesterdaysmathsexam.com

Question
Scheme Marks
Number

B1
4.(a)
B1

(2)
(b) 1 0.3 0.5 0.7 0.9 or 0.7 0.3 0.5 0.3 0.5 0.3 0.3 0.5 0.7 0.1 M1
= 0.9055 A1
(2)
0.7 "0.3" 0.5 0.85
(c) P( P1 P2 | Pass) , = M1, A1ft
(b) "0.9055"
= 0.938707 =a 0.939 A1
(3)
(d) p (1 p)( p 0.2) or 1 (1 p)(1.2 p) (o.e.) M1
e.g. p p p 2 0.2 p 0.2 0.95 p 2 2.2 p 1.15 0 (*) dM1A1cso
(3)
2.2 2.22 4 1.15 2
(e) p or Complete the sq: p 1.1 1.12 1.15 0 M1
2
2.2 0.4898... 2.2 0.24
= or or 1.1 0.06 or (1.34 ), 0.855 A1
2 2
p = 0.85505102 p = 0.855 A1
(3)
Notes
(a) 1st B1 for correctly placing 0.3 and 0.5
2nd B1 for correctly placing 0.7, 0.1 and 0.9

Apart from (d), a correct answer with no incorrect working scores full marks.
(b) M1 for a correct expression (ft from their tree diagram)
A1 for 0.9055 or exact equivalent e.g. 1811
2000 Accept 0.906 only

(c) M1 for a correct ratio of probs ft their 0.3 and their answer to (b)[if < 1]. Num > Den M0
A1ft for correct numerator and their part (b) on denominator
A1 for awrt 0.939 or accept exact fraction eg 17001811

(d) 1st M1 for a correct expression for P(pass) in terms of p[ condone p ( p 1)( p 0.2) etc]
2 dM1 dep. on 1st M1 for expanding brackets and forming an equation in p
nd

Allow one slip

A1cso correct processing leading to printed answer. No incorrect working seen.

(e) M1 for attempt to solve given equation, correct expression. Condone just + not +
st
1 A1 for correct expression and simplified square root 1.34 a 0.855
2nd A1 for p = 0.855 only (penalise any extra value > 1) Correct ans only scores 3/3
Ans. only For 101 11 6 or 0.855... score M1A1A0 (not to 3dp) but for 0.855 can score M1A1A1
 www.yesterdaysmathsexam.com

4. (a)(i) x+ 0.1 [P(x + 0.1) is B0] B1

(ii) P( B A) 0. 1 M1 A1
P(B | A) =
P( A) x 0.1 (3)
(b) x+y+ 0.1 (o.e.) [ P(x+y+ 0.1) is B0] B1
(1)
(c) x+y+ 0.1 + 0.32 = 1 or x+y+ 0.1 = 0.68 or “(b)” + 0.32 = 1 o.e. M1
x+ 0.1 = 2(y+ 0.1) M1
Eliminating x gives 3y = 0.48 M1
x = 0.42 y = 0.16 A1 A1
(5)
(9 marks)
Notes
(a)(ii) M1 for a correct ratio of probabilities formula with at least one correct probability
0.1
value (may ft their (a)(i) in the denominator) or a prob ratio of the form
(a)(i)
0.1
If num’ > denom’ score M0. NB P(A) = 0.68 – y and P(B|A) = is B0M1A0
0.68 y
0.1
A1 for as their final answer
x 0.1

(b) B1 for any correct expression in x and y e.g. 0.1 + x + 0.1 + y – 0.1
Condone x + y + 0.1 = 1 – 0.32 or 0.68 since LHS is a correct expression

(c) 1st M1 for using sum of probs. = 1 to form a “correct” linear equ’n in x and y [x + y = 0.58]
Ft their (b) and or their (a)(i) e.g. “(a)(i)” +0.32 + y = 1
2nd M1 for using P(A) = 2P(B) to form a “correct” linear equ’n in x and y[x – 2y = 0.1]
Ft their P(A) from part (a)
If they use 2P(A) = P(B) or swap x and y score 2nd M0 but allow access to 3rd M
3rd M1 for an attempt to solve their 2 linear equations. Implied by 1st 2 Ms and correct ans.
Requires correct algebraic steps leading to an equation in one variable.
If there are not 2 equations this cannot be scored (but see SC)
1 A1 for x = 0.42 (following correct working and dep. on 1st 2 Ms)
st

2nd A1 for y = 0.16 (following correct working and dep. on 1st 2 Ms)
Beware 0.42 = 0.32 + 0.1 so answer only does not score full marks

SC P(A) = 0.68 – y = 2(y + 0.1) score M2 (2nd and 3rd Ms) and 2nd A1 when y = 0.16 seen
Sight of x + y + 0.1 = 0.68 (o.e.)(scores 1st M1) and then 1st A1 if x = 0.42 follows.

or P(A) = x + 0.1 = 2(0.68 – x ) score M2 (2nd and 3rd Ms) and 1st A1 when x = 0.42 seen
Sight of x + y + 0.1 = 0.68 (o.e.)(scores 1st M1) and then 2nd A1 if y = 0.16 follows.
 www.yesterdaysmathsexam.com

Question
Scheme Marks
Number
5. (a)
B1
B1
B1
B1

(4)
(b) F and S or R and S B1
(1)
33
(c) P F R S = or 0.33 B1
100
(1)
30 12 21
(d) P(R)= or 0.42 B1
100 50
(1)
30 25 11
(e) P F S = or 0.55 B1
100 20
(1)
P F R "0.30"
(f) P F|R M1
P( R) "0.42"
30 5
= or (o.e.) A1
42 7
(2)
Total 10
Notes
(a) In the diagram do not treat a blank space as zero. Allow probabilities or integers
1st B1 for 3 labelled loops and a box. The 33 is not required for any marks in (a)
2nd B1 for F R or indicated by zeros
3rd B1 for 30 and 12 correctly placed and n(F) = 30 and n F R 12
4th B1 for S a separate loop, or indicated by zeros, and the 25

(b) B1 for a correct pair. If there is more than one pair then each pair must be correct.
Do not allow P(F) etc or e.g. P( R S ) = 0

B1 cao for each answer. Accept any exact equivalent (fractions or decimals) for the
(c),(d),(e)
probabilities

M1 f he 30 a d he a e (d). F a c ec a f he bab e a
(f)
correct ratio expression and at least one correct probability. If num > den score M0
5 0.3
A1 for or any exact equivalent. Must be proper fraction not
7 0.42
0.3
NB = 0.714 is A0 since it is not a proper fraction and the answer is not exact
0.42
Condone P(R|F) = 30 42 and allow M1A1 for the correct answer
P( R F ) 0.30 30
but P(R | F ) is M0A0
P( F ) 0.42 42
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Question
Scheme Marks
Number
P( M L) 3
5
1
5
7. (a) [P(M | L) =] M1
P( L) 3
10
= 0.40 (o.e) A1
(2)
P(L F ) 3
10
3
5
1
5 3 1 3 3
(b) x P( L | F ) or 1 x M1
P(F ) 1 3
5 5 5 5 10
0.3 0.12
x or 0.4 x 0.3 0.12 M1
0.40
x = 0.45 (o.e.) A1
(3)
(c) P(M R) 0.6 P(M L) or 0.6 1 0.2 M1
= 0.48 (o.e.) A1
(2)
3 7 21
(d) P(one is left handed and the other right handed) = 2 , or 0.42 M1, A1
10 10 50
(2)
Total 9
Notes
0.12
(a) M1 for a fully correct ratio e.g. or a correct ratio expression and one correct prob.
0.30
If numerator > denominator then M0
A1 for 0.40 or any exact equivalent

st
(b) 1 M1 for an equation for x with at least 2 of : 3
5
1
5 or 3
10 or (1 3
5 ) correct
2 3
5 10
BUT 2
is M0 or allow M1 for P( L F) 0.18
5
2nd M1 for a fully correct expression for x = … or 0.4x = …
A1 for 0.45 or any exact equivalent

(c) M1 for a correct expression with 0.6 follow through their P(M L) = 0.12
A1 for 0.48 or any exact equivalent

(d) M1 for a fully correct expression including the 2. Allow 1 0.3 instead of 0.7
A1 for 0.42 or any exact equivalent

NB You may see Venn or tree diagram drawn but marks are given when values are
used in correct expressions as above