Mathematics · May / June 2024 · Paper 61 · Question 6
YMA01
As printed in the paper
The question as text
6. The Venn diagram shows the probabilities related to teenagers playing 3 particular board games. C is the event that a teenager plays Chess S is the event that a teenager plays Scrabble G is the event that a teenager plays Go where p and q are probabilities. C G p 0.04 0 0.15 0.12 0.1 q 0.23 S
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
6(a)[1]
0.16 oe
Mark parts (b) and (c) together
If values are given in the diagram and the script, then the script takes
precedence
(
)
[
]
6(b)[4]
P
( )
C =
0.04
+
0.15
+
0.12
+
[
=
0.31
+
]
P
′
=
0.1 0.23
+
+
=
0.33
+
p
p
C
q
q
( )
[
]
(
)
(
)[
P
S =
0.1 0.15
+
+
0.12
+
0.23
=
0.6
P
′
=
1
−
0.12
+
0.15
+
0.1 0.23
+
=
0.4
S
(
)
[
]
(
)
P
∩
C =
0.15
+
0.12
=
0.27
P
'
∩
'
=
S
S
C
q
(
)
(
)
"0.31"
+
×
"0.6"
=
"0.27"
oe
"0.33"
+
×
"0.4"
=
oe
p
q
q
0.14
oe
0.22
oe
p =
q =
6(c)[2]
q
= −
1 (0.04
+
0.12
+
0.15
+
0.1 0.23 " ")
+
+
p
p
= −
1 (0.04
+
0.12
+
0.15
+
0.1 0.23 " ")
+
+
q
q =
0.22
oe
p =
0.14
oe
(
)
(
)
(d)(i)
P
C
∪
S
∩
G
′
=
0.39
oe
0.15
(
)
(
)
(ii)
P
C
|
S
∩
G
=
0.15
+
0.1
= 0.6 oe
76
6(d)[3]
0.16 oe
Mark parts (b) and (c) together
If values are given in the diagram and the script, then the script takes
precedence
(
)
[
]
P
( )
C =
0.04
+
0.15
+
0.12
+
[
=
0.31
+
]
P
′
=
0.1 0.23
+
+
=
0.33
+
p
p
C
q
q
( )
[
]
(
)
(
)[
P
S =
0.1 0.15
+
+
0.12
+
0.23
=
0.6
P
′
=
1
−
0.12
+
0.15
+
0.1 0.23
+
=
0.4
S
(
)
[
]
(
)
P
∩
C =
0.15
+
0.12
=
0.27
P
'
∩
'
=
S
S
C
q
(
)
(
)
"0.31"
+
×
"0.6"
=
"0.27"
oe
"0.33"
+
×
"0.4"
=
oe
p
q
q
0.14
oe
0.22
oe
p =
q =
q
= −
1 (0.04
+
0.12
+
0.15
+
0.1 0.23 " ")
+
+
p
p
= −
1 (0.04
+
0.12
+
0.15
+
0.1 0.23 " ")
+
+
q
q =
0.22
oe
p =
0.14
oe
(
)
(
)
(d)(i)
P
C
∪
S
∩
G
′
=
0.39
oe
0.15
(
)
(
)
(ii)
P
C
|
S
∩
G
=
0.15
+
0.1
= 0.6 oe
76
Number of teenagers = 0.15 " "
oe
+
p
76
Number who don't play Scrabble
=
"
" 0.4
×
( 104.8...)
=
0.15
+
p
= 104.8… awrt 105 A1
Notes
(a)
correct probability
B1
(b)
for 2 correct probability expressions
M1
M1
all 3 correct probability expressions Allow P(C) = 0.45
dependent on the 1st M1 being awarded for use of (
)
( )
P
C
∩
S
=
P
C
×
M1d
(
′
′
)
(
′
)
(
′
)
P
C
∩
S
=
P
C
×
P
S
ft their probabilities if labelled clearly
for 0.14 or exact equivalent or 0.22 or exact equivalent
A1
for a correct expression for q ft their value of p or a correct expression for p ft their
(c)
M1
value of q May be implied by a correct value for q ft their p or a correct value for p
ft their q
for 0.22 or exact equivalent ft their value of p or 0.14 or exact equivalent ft their
A1ft
value of q (p + q = 0.36 provided p and q are probabilities)
(d)(i)
for 0.39 or exact equivalent do not allow 0.04 + 0.12 + 0.23
B1
0.15
for
(ii)
M1
0.15
+
0.1
A1
for 0.6 or exact equivalent
Relating 76 to their P(
∩
)
May be implied by awrt 262
(e)
M1
C
G
for number of teenagers × 0.4 ft their number of teenagers e.g. 0.4 × “262”
M1
provided that the number of teenagers is not 76
awrt 105 ISW
A1
Pearson Education Limited. Registered company number 872828
with its registered office at 80 Strand, London, WC2R 0RL, United Kingdom
6(e)[3]
Number of teenagers = 0.15 " "
oe
+
p
76
Number who don't play Scrabble
=
"
" 0.4
×
( 104.8...)
=
0.15
+
p
= 104.8… awrt 105 A1
Notes
(a)
correct probability
B1
(b)
for 2 correct probability expressions
M1
M1
all 3 correct probability expressions Allow P(C) = 0.45
dependent on the 1st M1 being awarded for use of (
)
( )
P
C
∩
S
=
P
C
×
M1d
(
′
′
)
(
′
)
(
′
)
P
C
∩
S
=
P
C
×
P
S
ft their probabilities if labelled clearly
for 0.14 or exact equivalent or 0.22 or exact equivalent
A1
for a correct expression for q ft their value of p or a correct expression for p ft their
(c)
M1
value of q May be implied by a correct value for q ft their p or a correct value for p
ft their q
for 0.22 or exact equivalent ft their value of p or 0.14 or exact equivalent ft their
A1ft
value of q (p + q = 0.36 provided p and q are probabilities)
(d)(i)
for 0.39 or exact equivalent do not allow 0.04 + 0.12 + 0.23
B1
0.15
for
(ii)
M1
0.15
+
0.1
A1
for 0.6 or exact equivalent
Relating 76 to their P(
∩
)
May be implied by awrt 262
(e)
M1
C
G
for number of teenagers × 0.4 ft their number of teenagers e.g. 0.4 × “262”
M1
provided that the number of teenagers is not 76
awrt 105 ISW
A1
Pearson Education Limited. Registered company number 872828
with its registered office at 80 Strand, London, WC2R 0RL, United Kingdom
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