Mathematics · Oct / Nov 2021 · Paper 61 · Question 4

YMA01

As printed in the paper

The question as text

4. Three bags A, B and C each contain coloured balls. Bag A contains 4 red balls and 2 yellow balls only. Bag B contains 4 red balls and 1 yellow ball only. Bag C contains 6 red balls only. In a game Mike takes a ball at random from bag A, records the colour and places it in bag C. He then takes a ball at random from bag B, records the colour and places it in bag C. Finally, Mike takes a ball at random from bag C and records the colour.

Read automatically from the paper; the printed page above is exact.

Show the answer and mark scheme

4(a)[3]
(b) [Cases RYY or YRY or YYR] Prob = 2
1
1
1
4
1
1
1
6
×
×
+
×
×
+
×
×
3
5
8
3
5
8
3
5
8
1
] = 12
1
(
)
(
)
[=
2
+
4
+
6 or 0.0166...
+
0.033...
+
0.05
or
(*)
120
120
10
"
"
2
× ×
1
1
3
5
8
(c) [P(RYY | RYY or YRY or YYR) = ]
1
10
= 1
6
4(b)[3]
(b) [Cases RYY or YRY or YYR] Prob = 2
1
1
1
4
1
1
1
6
×
×
+
×
×
+
×
×
3
5
8
3
5
8
3
5
8
1
] = 12
1
(
)
(
)
[=
2
+
4
+
6 or 0.0166...
+
0.033...
+
0.05
or
(*)
120
120
10
"
"
2
× ×
1
1
3
5
8
(c) [P(RYY | RYY or YRY or YYR) = ]
1
10
= 1
6
x
0
1
2
3
P(X = x)
64
or
8
42
or
7
1
2
or
1
120
15
120
20
10
120
60
1
(
)
(e) [E(X) = ]
0
+
"42"
+
"2 12"
×
+
"3 2"
×
120
= 72
120 or 0.6
Notes
(a) 1st B1 completing the structure of branches: 2, 4 then 7 or 8 and suitable labels e.g. Y or R’
2nd B1 correct probabilities for at least bag A and bag B. Allow exact decimals
3rd B1 for a fully correct tree diagram. Condone missing 0 as probability. Allow exact decimals
(b) 1st M1 for at least one correct product of 3 probabilities (ft their tree diagram)
1st A1ft for all 3 products of 3 probabilities added ( no extras) (ft their tree diag.)
2nd A1*cso for fully correct solution with no incorrect statements seen
(c) M1 for a ratio of probabilities with denominator of 0.1 and numerator 1 oe
or the product
60
of 3 probabilities seen from their tree diagram representing P(RYY) provided num < 0.1
A1 for 1
or exact equivalent
6
(d) B1 for a correct sample space i.e. { 0, 1, 2, 3} Allow extras if they have a probability of 0.
M1 for at least 1 correct value of x and associated probability (excluding x = 2) [ft their tree]
A1 for a fully correct probability distribution
(e) M1 for attempt at a correct expression (at least 2 correct ft part(d) nonzero products)
A1 for 0.6 or any exact equivalent
5. (a) [By symmetry E(Y) ] = 0
19
(b)
q
+ +
r
u
=
30
(
2
)
1
q
+
r
+
u
= [and attempt to solve e.g. q + r = …]
8
4
u =
=
(*) A1*cso
30
15
(
)
(c)
(
)
2
(
)
2
E
Y
2
= −
9
×
q
+
−
5
×
r
+
5
2
×
r
+
9
2
q
or 162q + 50r
Var(Y) = 37 = (
)
E
Y
2
−
"0"
2
⇒
37
=
162
q
+
50
r
oe
11
55
Solving with
q
r
(162
50)
q
37
+
=
oe e.g.
−
=
−
or
30
3
1
1
q
=
and
r
=
A1
6
5
(d)
Y
0
D
12 ,
D
1
2
2
Y
2
;
Y
5
D
13 or
Y
9
D
1
5
=
⇒
=
=
+
= ± ⇒
=
= ± ⇒
=
d
12
13
15
4
6
2
5
1
P(D = d)
or
or
15
15
5
15
3
Notes
(a) B1 for 0
(b) 1st M1 for a correct equation in q, r and u using F(0)
2nd M1 for a second equation clearly based on sum of probs = 1 and an attempt to solve these 2
equations
A1* cso correct value for u found with no incorrect working
(c)
1st M1 for an attempt at E(Y2) with at least 3 correct products seen. The negative numbers do not
need to be in brackets
2nd dM1 for attempt at correct equation in q and r using Var(Y) [ft their E(Y) and E(Y2)] Condone
missing subtraction of 02 if 0 in part(a)
3rd M1 using q + r = 11/30 (awrt 0.37) to attempt to solve two linear equations in q and r leading to
equation in one variable. May be implied by correct answers.
A1 for q = 1
6 and r = 1
5 or exact equivalents
(d) B1 for D = 12
1st M1 for use of Pythagoras to work out D = 13 or 15
1st A1 for D = 13 and 15
2nd M1 for a correct value of D and an associated probability. Allow two occurrences (for 15 and
13) which add to the appropriate probability.
2ndA1ft for two correct values of D and associated probs ft their +ve q and r if q + r = 11
two occurrences (for 15 and 13) which add to the appropriate probability.
3rd A1ft for a fully correct probability distribution ft their +ve q and r if q + r = 11
30
6. (a) H ~ N(25.1, 5.52)
30.8
−
25.1
P(H > 30.8) =
P
Z
or P(Z > 1.03636…)
>
5.5
= 1 – 0.8508
= 0.1492 or better (calc: 0.1500…)
y −
25.1
(b) [P(H < y) = 0.05 implies]
= −
1.6449
5.5
y = 16.053.… so range is awrt 16.1 ~ 30.8
(c)(i) P(H < d) = 0.05 + 0.2 + 0.3 [= 0.55]
d −
25.1
0.13
=
(Calc 0.12566)
5.5
d =
0.13 5.5
×
+
25.1
=
25.815
(25.791… calc)
(ii) P(H < m) = 0.05 + 0.2 [= 0.25]
m −
25.1
= −
0.67
(Calc 0.674489)
5.5
m = awrt 21.4
(d) Height = 2×"m" + 3×25.8 + 3×30.8 [+8]
= 220.6 awrt 221 (cm)
Notes
(a) 1st M1 for standardising 30.8 with 25.1 and 5.5 (allow +) Allow use of z = 1.04
2nd M1 for 1 – p (where 0.84 < p < 0.86)
A1cso for an answer of 0.1492… or better(calc: 0.1500…) with evidence of both M's seen
(b) M1 for standardising their letter y with 25.1 and 5.5 and setting equal to z value 1 < |z| < 2
B1 for use of z = + 1.6449 or better (calc 1.6448536…) with the correct standardisation.
A1 for awrt 16.1 (ISW)(calc 16.053305.…) or range [16.1, 30.8](Allow 30.8 – 16.1 = 14.7[5])
Ans only [awrt 16.05 scores 3/3 16.1 scores M1B0A1 unless 1.6449 or better is seen]
(c)(i) 1st M1 for a correct method to calculate P(H < d) implied by z = awrt 0.13
Allow 0.05 + awrt 0.200 + awrt 0.300[=0.5505]
2nd M1 for a correct standardisation = z where 0.125
„ „
z
0.13
A1cso both method marks awarded, no errors seen and awrt 25.82 or awrt 25.79
d =
awrt 0.13 5.5
×
+
25.1
or
= awrt 25.8
ALT Verification 2nd M1 allow 25.8
−
25.1
=
0.127...
or 0.13
5.5
A1 for 0.55 and 0.5517 (calc 0.5506 or better) seen
(ii) 1st M1 correct method for P(H < m) Allow 0.05 + awrt 0.200 implied by |z| = [0.67 − 0.68]
2nd M1 for standardising m with 25.1 and 5.5 and setting equal to z value (0.65 |z| 0.69)
3rd M1 for standardising m with 25.1 and 5.5 and setting equal to awrt −0.67 oe
A1 for m = awrt 21.4 (use of z = 0.67 gives 21.415… and z = 0.68 gives 21.36…)
No need for 3rd M1 to be awarded
Answer only 21.4 gets M1M1M0A1. 21.39 gets 4/4
(d) M1 for 2×"m" + 3×25.8 + 3×30.8 [ + n ] where n is an integer 0 Allow m or ft their m
A1 for awrt 221 (cm)
(c)(i) e.g. P(H > 25.8 | "16.1" < H < 30.8) or P(25.8
30.8)
ALT 1
<
H
<
1 (0.15
−
+
0.05)
= 0.8508
−
0.5517
(tables) or 0.299345...
3
(calc)
≈
0.8
0.8
8
Notes
(c)(i) 1st M1 for a correct conditional probability statement ft their answer to (b) i.e. their y
q where q = 0.3 to 1sf
2nd M1 for a ratio of probs of the form 0.8
A1 for probability of approx 3
8
Pearson Education Limited. Registered company number 872828
with its registered office at 80 Strand, London, WC2R 0RL, United Kingdom
4(c)[2]
(b) [Cases RYY or YRY or YYR] Prob = 2
1
1
1
4
1
1
1
6
×
×
+
×
×
+
×
×
3
5
8
3
5
8
3
5
8
1
] = 12
1
(
)
(
)
[=
2
+
4
+
6 or 0.0166...
+
0.033...
+
0.05
or
(*)
120
120
10
"
"
2
× ×
1
1
3
5
8
(c) [P(RYY | RYY or YRY or YYR) = ]
1
10
= 1
6
x
0
1
2
3
P(X = x)
64
or
8
42
or
7
1
2
or
1
120
15
120
20
10
120
60
1
(
)
(e) [E(X) = ]
0
+
"42"
+
"2 12"
×
+
"3 2"
×
120
= 72
120 or 0.6
Notes
(a) 1st B1 completing the structure of branches: 2, 4 then 7 or 8 and suitable labels e.g. Y or R’
2nd B1 correct probabilities for at least bag A and bag B. Allow exact decimals
3rd B1 for a fully correct tree diagram. Condone missing 0 as probability. Allow exact decimals
(b) 1st M1 for at least one correct product of 3 probabilities (ft their tree diagram)
1st A1ft for all 3 products of 3 probabilities added ( no extras) (ft their tree diag.)
2nd A1*cso for fully correct solution with no incorrect statements seen
(c) M1 for a ratio of probabilities with denominator of 0.1 and numerator 1 oe
or the product
60
of 3 probabilities seen from their tree diagram representing P(RYY) provided num < 0.1
A1 for 1
or exact equivalent
6
(d) B1 for a correct sample space i.e. { 0, 1, 2, 3} Allow extras if they have a probability of 0.
M1 for at least 1 correct value of x and associated probability (excluding x = 2) [ft their tree]
A1 for a fully correct probability distribution
(e) M1 for attempt at a correct expression (at least 2 correct ft part(d) nonzero products)
A1 for 0.6 or any exact equivalent
5. (a) [By symmetry E(Y) ] = 0
19
(b)
q
+ +
r
u
=
30
(
2
)
1
q
+
r
+
u
= [and attempt to solve e.g. q + r = …]
8
4
u =
=
(*) A1*cso
30
15
(
)
(c)
(
)
2
(
)
2
E
Y
2
= −
9
×
q
+
−
5
×
r
+
5
2
×
r
+
9
2
q
or 162q + 50r
Var(Y) = 37 = (
)
E
Y
2
−
"0"
2
⇒
37
=
162
q
+
50
r
oe
11
55
Solving with
q
r
(162
50)
q
37
+
=
oe e.g.
−
=
−
or
30
3
1
1
q
=
and
r
=
A1
6
5
(d)
Y
0
D
12 ,
D
1
2
2
Y
2
;
Y
5
D
13 or
Y
9
D
1
5
=
⇒
=
=
+
= ± ⇒
=
= ± ⇒
=
d
12
13
15
4
6
2
5
1
P(D = d)
or
or
15
15
5
15
3
Notes
(a) B1 for 0
(b) 1st M1 for a correct equation in q, r and u using F(0)
2nd M1 for a second equation clearly based on sum of probs = 1 and an attempt to solve these 2
equations
A1* cso correct value for u found with no incorrect working
(c)
1st M1 for an attempt at E(Y2) with at least 3 correct products seen. The negative numbers do not
need to be in brackets
2nd dM1 for attempt at correct equation in q and r using Var(Y) [ft their E(Y) and E(Y2)] Condone
missing subtraction of 02 if 0 in part(a)
3rd M1 using q + r = 11/30 (awrt 0.37) to attempt to solve two linear equations in q and r leading to
equation in one variable. May be implied by correct answers.
A1 for q = 1
6 and r = 1
5 or exact equivalents
(d) B1 for D = 12
1st M1 for use of Pythagoras to work out D = 13 or 15
1st A1 for D = 13 and 15
2nd M1 for a correct value of D and an associated probability. Allow two occurrences (for 15 and
13) which add to the appropriate probability.
2ndA1ft for two correct values of D and associated probs ft their +ve q and r if q + r = 11
two occurrences (for 15 and 13) which add to the appropriate probability.
3rd A1ft for a fully correct probability distribution ft their +ve q and r if q + r = 11
30
6. (a) H ~ N(25.1, 5.52)
30.8
−
25.1
P(H > 30.8) =
P
Z
or P(Z > 1.03636…)
>
5.5
= 1 – 0.8508
= 0.1492 or better (calc: 0.1500…)
y −
25.1
(b) [P(H < y) = 0.05 implies]
= −
1.6449
5.5
y = 16.053.… so range is awrt 16.1 ~ 30.8
(c)(i) P(H < d) = 0.05 + 0.2 + 0.3 [= 0.55]
d −
25.1
0.13
=
(Calc 0.12566)
5.5
d =
0.13 5.5
×
+
25.1
=
25.815
(25.791… calc)
(ii) P(H < m) = 0.05 + 0.2 [= 0.25]
m −
25.1
= −
0.67
(Calc 0.674489)
5.5
m = awrt 21.4
(d) Height = 2×"m" + 3×25.8 + 3×30.8 [+8]
= 220.6 awrt 221 (cm)
Notes
(a) 1st M1 for standardising 30.8 with 25.1 and 5.5 (allow +) Allow use of z = 1.04
2nd M1 for 1 – p (where 0.84 < p < 0.86)
A1cso for an answer of 0.1492… or better(calc: 0.1500…) with evidence of both M's seen
(b) M1 for standardising their letter y with 25.1 and 5.5 and setting equal to z value 1 < |z| < 2
B1 for use of z = + 1.6449 or better (calc 1.6448536…) with the correct standardisation.
A1 for awrt 16.1 (ISW)(calc 16.053305.…) or range [16.1, 30.8](Allow 30.8 – 16.1 = 14.7[5])
Ans only [awrt 16.05 scores 3/3 16.1 scores M1B0A1 unless 1.6449 or better is seen]
(c)(i) 1st M1 for a correct method to calculate P(H < d) implied by z = awrt 0.13
Allow 0.05 + awrt 0.200 + awrt 0.300[=0.5505]
2nd M1 for a correct standardisation = z where 0.125
„ „
z
0.13
A1cso both method marks awarded, no errors seen and awrt 25.82 or awrt 25.79
d =
awrt 0.13 5.5
×
+
25.1
or
= awrt 25.8
ALT Verification 2nd M1 allow 25.8
−
25.1
=
0.127...
or 0.13
5.5
A1 for 0.55 and 0.5517 (calc 0.5506 or better) seen
(ii) 1st M1 correct method for P(H < m) Allow 0.05 + awrt 0.200 implied by |z| = [0.67 − 0.68]
2nd M1 for standardising m with 25.1 and 5.5 and setting equal to z value (0.65 |z| 0.69)
3rd M1 for standardising m with 25.1 and 5.5 and setting equal to awrt −0.67 oe
A1 for m = awrt 21.4 (use of z = 0.67 gives 21.415… and z = 0.68 gives 21.36…)
No need for 3rd M1 to be awarded
Answer only 21.4 gets M1M1M0A1. 21.39 gets 4/4
(d) M1 for 2×"m" + 3×25.8 + 3×30.8 [ + n ] where n is an integer 0 Allow m or ft their m
A1 for awrt 221 (cm)
(c)(i) e.g. P(H > 25.8 | "16.1" < H < 30.8) or P(25.8
30.8)
ALT 1
<
H
<
1 (0.15
−
+
0.05)
= 0.8508
−
0.5517
(tables) or 0.299345...
3
(calc)
≈
0.8
0.8
8
Notes
(c)(i) 1st M1 for a correct conditional probability statement ft their answer to (b) i.e. their y
q where q = 0.3 to 1sf
2nd M1 for a ratio of probs of the form 0.8
A1 for probability of approx 3
8
Pearson Education Limited. Registered company number 872828
with its registered office at 80 Strand, London, WC2R 0RL, United Kingdom
4(d)[3]
x
0
1
2
3
P(X = x)
64
or
8
42
or
7
1
2
or
1
120
15
120
20
10
120
60
1
(
)
(e) [E(X) = ]
0
+
"42"
+
"2 12"
×
+
"3 2"
×
120
= 72
120 or 0.6
Notes
(a) 1st B1 completing the structure of branches: 2, 4 then 7 or 8 and suitable labels e.g. Y or R’
2nd B1 correct probabilities for at least bag A and bag B. Allow exact decimals
3rd B1 for a fully correct tree diagram. Condone missing 0 as probability. Allow exact decimals
(b) 1st M1 for at least one correct product of 3 probabilities (ft their tree diagram)
1st A1ft for all 3 products of 3 probabilities added ( no extras) (ft their tree diag.)
2nd A1*cso for fully correct solution with no incorrect statements seen
(c) M1 for a ratio of probabilities with denominator of 0.1 and numerator 1 oe
or the product
60
of 3 probabilities seen from their tree diagram representing P(RYY) provided num < 0.1
A1 for 1
or exact equivalent
6
(d) B1 for a correct sample space i.e. { 0, 1, 2, 3} Allow extras if they have a probability of 0.
M1 for at least 1 correct value of x and associated probability (excluding x = 2) [ft their tree]
A1 for a fully correct probability distribution
(e) M1 for attempt at a correct expression (at least 2 correct ft part(d) nonzero products)
A1 for 0.6 or any exact equivalent
5. (a) [By symmetry E(Y) ] = 0
19
(b)
q
+ +
r
u
=
30
(
2
)
1
q
+
r
+
u
= [and attempt to solve e.g. q + r = …]
8
4
u =
=
(*) A1*cso
30
15
(
)
(c)
(
)
2
(
)
2
E
Y
2
= −
9
×
q
+
−
5
×
r
+
5
2
×
r
+
9
2
q
or 162q + 50r
Var(Y) = 37 = (
)
E
Y
2
−
"0"
2
⇒
37
=
162
q
+
50
r
oe
11
55
Solving with
q
r
(162
50)
q
37
+
=
oe e.g.
−
=
−
or
30
3
1
1
q
=
and
r
=
A1
6
5
(d)
Y
0
D
12 ,
D
1
2
2
Y
2
;
Y
5
D
13 or
Y
9
D
1
5
=
⇒
=
=
+
= ± ⇒
=
= ± ⇒
=
d
12
13
15
4
6
2
5
1
P(D = d)
or
or
15
15
5
15
3
Notes
(a) B1 for 0
(b) 1st M1 for a correct equation in q, r and u using F(0)
2nd M1 for a second equation clearly based on sum of probs = 1 and an attempt to solve these 2
equations
A1* cso correct value for u found with no incorrect working
(c)
1st M1 for an attempt at E(Y2) with at least 3 correct products seen. The negative numbers do not
need to be in brackets
2nd dM1 for attempt at correct equation in q and r using Var(Y) [ft their E(Y) and E(Y2)] Condone
missing subtraction of 02 if 0 in part(a)
3rd M1 using q + r = 11/30 (awrt 0.37) to attempt to solve two linear equations in q and r leading to
equation in one variable. May be implied by correct answers.
A1 for q = 1
6 and r = 1
5 or exact equivalents
(d) B1 for D = 12
1st M1 for use of Pythagoras to work out D = 13 or 15
1st A1 for D = 13 and 15
2nd M1 for a correct value of D and an associated probability. Allow two occurrences (for 15 and
13) which add to the appropriate probability.
2ndA1ft for two correct values of D and associated probs ft their +ve q and r if q + r = 11
two occurrences (for 15 and 13) which add to the appropriate probability.
3rd A1ft for a fully correct probability distribution ft their +ve q and r if q + r = 11
30
6. (a) H ~ N(25.1, 5.52)
30.8
−
25.1
P(H > 30.8) =
P
Z
or P(Z > 1.03636…)
>
5.5
= 1 – 0.8508
= 0.1492 or better (calc: 0.1500…)
y −
25.1
(b) [P(H < y) = 0.05 implies]
= −
1.6449
5.5
y = 16.053.… so range is awrt 16.1 ~ 30.8
(c)(i) P(H < d) = 0.05 + 0.2 + 0.3 [= 0.55]
d −
25.1
0.13
=
(Calc 0.12566)
5.5
d =
0.13 5.5
×
+
25.1
=
25.815
(25.791… calc)
(ii) P(H < m) = 0.05 + 0.2 [= 0.25]
m −
25.1
= −
0.67
(Calc 0.674489)
5.5
m = awrt 21.4
(d) Height = 2×"m" + 3×25.8 + 3×30.8 [+8]
= 220.6 awrt 221 (cm)
Notes
(a) 1st M1 for standardising 30.8 with 25.1 and 5.5 (allow +) Allow use of z = 1.04
2nd M1 for 1 – p (where 0.84 < p < 0.86)
A1cso for an answer of 0.1492… or better(calc: 0.1500…) with evidence of both M's seen
(b) M1 for standardising their letter y with 25.1 and 5.5 and setting equal to z value 1 < |z| < 2
B1 for use of z = + 1.6449 or better (calc 1.6448536…) with the correct standardisation.
A1 for awrt 16.1 (ISW)(calc 16.053305.…) or range [16.1, 30.8](Allow 30.8 – 16.1 = 14.7[5])
Ans only [awrt 16.05 scores 3/3 16.1 scores M1B0A1 unless 1.6449 or better is seen]
(c)(i) 1st M1 for a correct method to calculate P(H < d) implied by z = awrt 0.13
Allow 0.05 + awrt 0.200 + awrt 0.300[=0.5505]
2nd M1 for a correct standardisation = z where 0.125
„ „
z
0.13
A1cso both method marks awarded, no errors seen and awrt 25.82 or awrt 25.79
d =
awrt 0.13 5.5
×
+
25.1
or
= awrt 25.8
ALT Verification 2nd M1 allow 25.8
−
25.1
=
0.127...
or 0.13
5.5
A1 for 0.55 and 0.5517 (calc 0.5506 or better) seen
(ii) 1st M1 correct method for P(H < m) Allow 0.05 + awrt 0.200 implied by |z| = [0.67 − 0.68]
2nd M1 for standardising m with 25.1 and 5.5 and setting equal to z value (0.65 |z| 0.69)
3rd M1 for standardising m with 25.1 and 5.5 and setting equal to awrt −0.67 oe
A1 for m = awrt 21.4 (use of z = 0.67 gives 21.415… and z = 0.68 gives 21.36…)
No need for 3rd M1 to be awarded
Answer only 21.4 gets M1M1M0A1. 21.39 gets 4/4
(d) M1 for 2×"m" + 3×25.8 + 3×30.8 [ + n ] where n is an integer 0 Allow m or ft their m
A1 for awrt 221 (cm)
(c)(i) e.g. P(H > 25.8 | "16.1" < H < 30.8) or P(25.8
30.8)
ALT 1
<
H
<
1 (0.15
−
+
0.05)
= 0.8508
−
0.5517
(tables) or 0.299345...
3
(calc)
≈
0.8
0.8
8
Notes
(c)(i) 1st M1 for a correct conditional probability statement ft their answer to (b) i.e. their y
q where q = 0.3 to 1sf
2nd M1 for a ratio of probs of the form 0.8
A1 for probability of approx 3
8
Pearson Education Limited. Registered company number 872828
with its registered office at 80 Strand, London, WC2R 0RL, United Kingdom
4(e)[2]
(b) [Cases RYY or YRY or YYR] Prob = 2
1
1
1
4
1
1
1
6
×
×
+
×
×
+
×
×
3
5
8
3
5
8
3
5
8
1
] = 12
1
(
)
(
)
[=
2
+
4
+
6 or 0.0166...
+
0.033...
+
0.05
or
(*)
120
120
10
"
"
2
× ×
1
1
3
5
8
(c) [P(RYY | RYY or YRY or YYR) = ]
1
10
= 1
6
x
0
1
2
3
P(X = x)
64
or
8
42
or
7
1
2
or
1
120
15
120
20
10
120
60
1
(
)
(e) [E(X) = ]
0
+
"42"
+
"2 12"
×
+
"3 2"
×
120
= 72
120 or 0.6
Notes
(a) 1st B1 completing the structure of branches: 2, 4 then 7 or 8 and suitable labels e.g. Y or R’
2nd B1 correct probabilities for at least bag A and bag B. Allow exact decimals
3rd B1 for a fully correct tree diagram. Condone missing 0 as probability. Allow exact decimals
(b) 1st M1 for at least one correct product of 3 probabilities (ft their tree diagram)
1st A1ft for all 3 products of 3 probabilities added ( no extras) (ft their tree diag.)
2nd A1*cso for fully correct solution with no incorrect statements seen
(c) M1 for a ratio of probabilities with denominator of 0.1 and numerator 1 oe
or the product
60
of 3 probabilities seen from their tree diagram representing P(RYY) provided num < 0.1
A1 for 1
or exact equivalent
6
(d) B1 for a correct sample space i.e. { 0, 1, 2, 3} Allow extras if they have a probability of 0.
M1 for at least 1 correct value of x and associated probability (excluding x = 2) [ft their tree]
A1 for a fully correct probability distribution
(e) M1 for attempt at a correct expression (at least 2 correct ft part(d) nonzero products)
A1 for 0.6 or any exact equivalent
5. (a) [By symmetry E(Y) ] = 0
19
(b)
q
+ +
r
u
=
30
(
2
)
1
q
+
r
+
u
= [and attempt to solve e.g. q + r = …]
8
4
u =
=
(*) A1*cso
30
15
(
)
(c)
(
)
2
(
)
2
E
Y
2
= −
9
×
q
+
−
5
×
r
+
5
2
×
r
+
9
2
q
or 162q + 50r
Var(Y) = 37 = (
)
E
Y
2
−
"0"
2
⇒
37
=
162
q
+
50
r
oe
11
55
Solving with
q
r
(162
50)
q
37
+
=
oe e.g.
−
=
−
or
30
3
1
1
q
=
and
r
=
A1
6
5
(d)
Y
0
D
12 ,
D
1
2
2
Y
2
;
Y
5
D
13 or
Y
9
D
1
5
=
⇒
=
=
+
= ± ⇒
=
= ± ⇒
=
d
12
13
15
4
6
2
5
1
P(D = d)
or
or
15
15
5
15
3
Notes
(a) B1 for 0
(b) 1st M1 for a correct equation in q, r and u using F(0)
2nd M1 for a second equation clearly based on sum of probs = 1 and an attempt to solve these 2
equations
A1* cso correct value for u found with no incorrect working
(c)
1st M1 for an attempt at E(Y2) with at least 3 correct products seen. The negative numbers do not
need to be in brackets
2nd dM1 for attempt at correct equation in q and r using Var(Y) [ft their E(Y) and E(Y2)] Condone
missing subtraction of 02 if 0 in part(a)
3rd M1 using q + r = 11/30 (awrt 0.37) to attempt to solve two linear equations in q and r leading to
equation in one variable. May be implied by correct answers.
A1 for q = 1
6 and r = 1
5 or exact equivalents
(d) B1 for D = 12
1st M1 for use of Pythagoras to work out D = 13 or 15
1st A1 for D = 13 and 15
2nd M1 for a correct value of D and an associated probability. Allow two occurrences (for 15 and
13) which add to the appropriate probability.
2ndA1ft for two correct values of D and associated probs ft their +ve q and r if q + r = 11
two occurrences (for 15 and 13) which add to the appropriate probability.
3rd A1ft for a fully correct probability distribution ft their +ve q and r if q + r = 11
30
6. (a) H ~ N(25.1, 5.52)
30.8
−
25.1
P(H > 30.8) =
P
Z
or P(Z > 1.03636…)
>
5.5
= 1 – 0.8508
= 0.1492 or better (calc: 0.1500…)
y −
25.1
(b) [P(H < y) = 0.05 implies]
= −
1.6449
5.5
y = 16.053.… so range is awrt 16.1 ~ 30.8
(c)(i) P(H < d) = 0.05 + 0.2 + 0.3 [= 0.55]
d −
25.1
0.13
=
(Calc 0.12566)
5.5
d =
0.13 5.5
×
+
25.1
=
25.815
(25.791… calc)
(ii) P(H < m) = 0.05 + 0.2 [= 0.25]
m −
25.1
= −
0.67
(Calc 0.674489)
5.5
m = awrt 21.4
(d) Height = 2×"m" + 3×25.8 + 3×30.8 [+8]
= 220.6 awrt 221 (cm)
Notes
(a) 1st M1 for standardising 30.8 with 25.1 and 5.5 (allow +) Allow use of z = 1.04
2nd M1 for 1 – p (where 0.84 < p < 0.86)
A1cso for an answer of 0.1492… or better(calc: 0.1500…) with evidence of both M's seen
(b) M1 for standardising their letter y with 25.1 and 5.5 and setting equal to z value 1 < |z| < 2
B1 for use of z = + 1.6449 or better (calc 1.6448536…) with the correct standardisation.
A1 for awrt 16.1 (ISW)(calc 16.053305.…) or range [16.1, 30.8](Allow 30.8 – 16.1 = 14.7[5])
Ans only [awrt 16.05 scores 3/3 16.1 scores M1B0A1 unless 1.6449 or better is seen]
(c)(i) 1st M1 for a correct method to calculate P(H < d) implied by z = awrt 0.13
Allow 0.05 + awrt 0.200 + awrt 0.300[=0.5505]
2nd M1 for a correct standardisation = z where 0.125
„ „
z
0.13
A1cso both method marks awarded, no errors seen and awrt 25.82 or awrt 25.79
d =
awrt 0.13 5.5
×
+
25.1
or
= awrt 25.8
ALT Verification 2nd M1 allow 25.8
−
25.1
=
0.127...
or 0.13
5.5
A1 for 0.55 and 0.5517 (calc 0.5506 or better) seen
(ii) 1st M1 correct method for P(H < m) Allow 0.05 + awrt 0.200 implied by |z| = [0.67 − 0.68]
2nd M1 for standardising m with 25.1 and 5.5 and setting equal to z value (0.65 |z| 0.69)
3rd M1 for standardising m with 25.1 and 5.5 and setting equal to awrt −0.67 oe
A1 for m = awrt 21.4 (use of z = 0.67 gives 21.415… and z = 0.68 gives 21.36…)
No need for 3rd M1 to be awarded
Answer only 21.4 gets M1M1M0A1. 21.39 gets 4/4
(d) M1 for 2×"m" + 3×25.8 + 3×30.8 [ + n ] where n is an integer 0 Allow m or ft their m
A1 for awrt 221 (cm)
(c)(i) e.g. P(H > 25.8 | "16.1" < H < 30.8) or P(25.8
30.8)
ALT 1
<
H
<
1 (0.15
−
+
0.05)
= 0.8508
−
0.5517
(tables) or 0.299345...
3
(calc)
≈
0.8
0.8
8
Notes
(c)(i) 1st M1 for a correct conditional probability statement ft their answer to (b) i.e. their y
q where q = 0.3 to 1sf
2nd M1 for a ratio of probs of the form 0.8
A1 for probability of approx 3
8
Pearson Education Limited. Registered company number 872828
with its registered office at 80 Strand, London, WC2R 0RL, United Kingdom

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