Mathematics · May / June 2023 · Paper 61 · Question 6
YMA01
As printed in the paper
The question as text
6. Three events A, B and C are such that P (A) = 0.1 P (B½A) = 0.3 P (A ∪ B) = 0.25 P (C ) = 0.5 Given that A and C are mutually exclusive
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
6(a)[1]
P
(
)
6(b)[3]
A
∩
B
(
)
P
∩
=
0.1 0.3
×
or
0.3
=
A
B
0.1
( )
( )
(
)
0.25
=
0.1 P
+
B
−
"0.03"
or
0.25
=
0.1 P
+
B
−
P
A
∩
B
( )
P
B −
0.15
( )
0.25
=
0.1 P
+
B
−
0.03
or
0.3
=
P(B) = 0.18*
0.1
6(c)[5]
Notes
B1cao
(a)
(
)
(b)
P
A
∩
B
M1 for use of (
)
P
B A
|
=
with 0.1 and 0.3 substituted. Allow for 0.1 0.3
×
seen
P A
( )
( )
(
)
M1
0.25
=
0.1 P B
+
−
p
where 0
<
p
< or
1
p
=
P
A
∩
B
oe eg
0.25
−
0.1
+
p
=
P
any letter for P(B))
A1* P(B) = 0.18 depends on both previous M marks for a fully correct equation in terms of P(B)
(allow any letter for P(B)) followed by P(B) = 0.18
NB 0.03 used/stated with no working could get M0M1A0
Using (
P
∩
)
=
0.1 P( )
×
then they get M0M0A0
A
B
B
Verification could get M1M1A0
M1 for 0.1 0.3
×
0.18
−
0.15
(
M1 for 0.25
−
0.18
−
0.1
= −
0.03
or
0.3
=
or
0.25
=
0.1 0.18
+
−
P A
0.1
(c)
M1 for 3 circles as per either diagram. If using Diagram 2 we must see exactly 2 zeros in one of the
intersections (as shaded). (Do Not accept blank or dash instead of 0) Condone missing rectangle.
Ignore labels
M1 for 0.09 and 0.41 marked correctly in diagram – condone incorrect/no label but must be in the
left or right hand circles in 1st diagram or must have zeros (condone blank or dash) in the 2 other
regions of the circle if in 2nd diagram
B1ft their "0.03" in correct place on diagram. Correct label required
B1ft for 0.34 or ft 0.75 – "their 0.41" where their 0.41
≠
0.5
No other ft accepted. Do not allow
0.75
A1 fully correct Venn diagram including the rectangle and all 3 labels.
SC no labels could get M1M1B0B1A0 if using 3 intersecting circles must have blanks or 0 for the
2nd M1
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