Mathematics · Oct / Nov 2025 · Paper 61 · Question 6
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As printed in the paper
The question as text
6. The Venn diagram, where x, y and z are probabilities, represents the events A, B and C and their associated probabilities. B A C 0.36 0.24 x 0.09 y z Given that the events A and B are independent,
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Show the answer and mark scheme
6(a)[3]
P( )
A =
0.6
P( )
B
=
0.24
+
0.09
+
x
P(
A
∩
B
)
=
0.24
(
)
0.6 0.33
+
x
=
0.24
x = 0.07
0.24
+
"0.07"
31
(
)
6(b)[2]
P
B C
|
′=
=
+
+
0.6
"0.07"
z
75
z = 0.08
6(c)[2]
y = 1 – 0.36 – 0.24 – “0.07” – 0.09 – “0.08”
y = 0.16
(d)(i)
0.36
(ii)
0.67
Notes
for at least 2 of the 3 correct probabilities/probability
expressions
(a)
M1
seen or used (implied by 2nd M1)
+
+
may be implied by embedded working
0.24 0.09
x
e.g. x = “P(B)”– 0.24 – 0.09
for use of P(
A
∩
B
)
=
P( )
A
×
P( )
B
oe to form a correct
M1
equation in x
A1
oe correct answer scores 3/3
for a correct conditional probability equation ft “their x”
substituted or just x
(b)
M1
0.24
+
"0.07"
31
allow (
′=
)
P
B C
|
=
1 0.09
−
−
y
75
A1
oe correct answer scores 2 out of 2
for a correct expression to find y ft their x and their z
values substituted where x and z are probabilities
(c)
M1
do not allow blank/no answers for x and/or z as zero
may be implied by x + y + z = 0.31 with x, y, and z as
probabilities
ft their x and their z
A1ft
x + y + z = 0.31 where x, y, and z are probabilities
(d)(i)
B1
oe
0.67 oe or 0.6 + “their x”
(ii)
B1ft
ft “their x” where “their x” and 0.6 + “their x” are
probabilities
54.4
−
51
48.45
−
51
[
]
6(d)[2]
P( )
A =
0.6
P( )
B
=
0.24
+
0.09
+
x
P(
A
∩
B
)
=
0.24
(
)
0.6 0.33
+
x
=
0.24
x = 0.07
0.24
+
"0.07"
31
(
)
P
B C
|
′=
=
+
+
0.6
"0.07"
z
75
z = 0.08
y = 1 – 0.36 – 0.24 – “0.07” – 0.09 – “0.08”
y = 0.16
(d)(i)
0.36
(ii)
0.67
Notes
for at least 2 of the 3 correct probabilities/probability
expressions
(a)
M1
seen or used (implied by 2nd M1)
+
+
may be implied by embedded working
0.24 0.09
x
e.g. x = “P(B)”– 0.24 – 0.09
for use of P(
A
∩
B
)
=
P( )
A
×
P( )
B
oe to form a correct
M1
equation in x
A1
oe correct answer scores 3/3
for a correct conditional probability equation ft “their x”
substituted or just x
(b)
M1
0.24
+
"0.07"
31
allow (
′=
)
P
B C
|
=
1 0.09
−
−
y
75
A1
oe correct answer scores 2 out of 2
for a correct expression to find y ft their x and their z
values substituted where x and z are probabilities
(c)
M1
do not allow blank/no answers for x and/or z as zero
may be implied by x + y + z = 0.31 with x, y, and z as
probabilities
ft their x and their z
A1ft
x + y + z = 0.31 where x, y, and z are probabilities
(d)(i)
B1
oe
0.67 oe or 0.6 + “their x”
(ii)
B1ft
ft “their x” where “their x” and 0.6 + “their x” are
probabilities
54.4
−
51
48.45
−
51
[
]
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