Mathematics · Oct / Nov 2023 · Paper 61 · Question 3
YMA01
As printed in the paper
Show the answer and mark scheme
3(i)(a)[4]
w =
0.15
x =
0.7
−
0.15
=
0.55
y =
0.65
−
0.55
=
0.1
z = −
1
0.15
−
0.55
−
0.1
=
0.2
3(i)(b)[1]
'0.15' '0.1'
+
=
'0.25'
'0.55'
[P( ) P( )]
×
=
'0.65' '0.7'
×
≠
'0.55'[ P(
=
∩
)]
or
[P(
|
) ]
'0.65'[
P( )]
oe M1
3(i)(c)[2]
C
O
C
O
C O
=
≠
=
C
'0.7'
'0.455'
≠
'0.55'
or '0.7857'...
≠
'0.65'
[So not independent]*
2
1
15
(
∪
)
=
+
=
3(ii)(a)[1]
P F
H
7
4
28
5
2
2
( )
( )
3(ii)(b)[3]
=
+
P
−
P
G
G
8
7
7
5
2
−
19
5
8
7
P
( )
G
=
=
÷
2
56
7
1
−
7
19
( )
P
G =
40
2
19
19
3(ii)(c)[1]
P F
(
∩
G
)
=
×
'
'
=
7
40
140
Notes
(i)(a)
w =
0.15
If answer is given in the script and the Venn diagram, then mark the script
B1
x =
0.55
B1
If answer is given in the script and the Venn diagram, then mark the script
0.1
y =
If answer is given in the script and the Venn diagram, then mark the script
B1
z =
0.2
B1
If answer is given in the script and the Venn diagram, then mark the script
For w + y = '0.25' follow through their w and their y (You will need to check for their values)
(b)
B1ft
provided this is a probability
' '
x
For '(
)' '(
)'
' '
(c)
x
+
y
×
w
+
x
≠
x
or
x ≠
'
x
+
y
'
ft their w, x and y
M1
'
w
+
'
A1*
A fully correct solution with values evaluated and no errors ft their w, x and y
For 15
(ii) (a)
B1
oe Allow awrt 0.536
28
For use of P(
∪
)=P( ) + P( )
−
P( ) P( )
×
(b)
M1
F
G
F
G
F
G
Dependent on M1. For a correct rearrangement to find P(G) e.g. 5
2
2
1
Allow
−
÷
−
8
7
7
dM1
19
5 P( )
May be implied by 19
=
G
56
7
40
For 19
oe
A1
40
For 19
oe or 2
evaluated correctly and where P( )
G is a probability
(c)
×
P( )
B1ft
G
140
7
1
1
1
1
1
3
1
2
2
= ×
*
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