Mathematics · May / June 2022 · Paper 61 · Question 4
YMA01
As printed in the paper
The question as text
4. The events H and W are such that P(H) = 3 P(H ∪ W) = 3 8 4 Given that H and W are independent,
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
4(a)[4]
H
W
H
W
H
W
H
W
H
W
H
3
(
)
P(
∩
)
=
×
P
H
W
W
P(
H
'
∩
W
)
=
P(
H
') P(
×
W
)
8
3
3
3
3
5 P
(
)
(
)
(
)
=
+
P
−
P
=
W
W
W
4
8
8
8
8
3
3
(
)
P(
W =
)
*
P
W =
*
5
5
−
+
3
1
9
1
1
(
)
P N
′
|
=
8
15
or
40
12
or 1
−
15
4(b)[2]
H
M1
3
3
3
8
8
8
37
=
= awrt 0.822
A1
45
4(c)[5]
Notes
for use of P(
)
P(
)
(
)
P(
)
(a)
M1
H
∪
W
=
H
+
P W
−
H
∩
W
(with at least one value correctly substituted)
or use of P(
H
'
∩
W
)
=
P(
H
∪
W
)
−
P(
H
)
(with at least one value correctly substituted)
for use of P(
∩
)
=
P(
) P(
×
)
or use of P(
'
∩
)
=
P(
') P(
×
M1
H
W
H
W
H
W
H
a correct equation in P(W) (allow W or x here)
A1
A1cso* Correct solution ending with P(W)= 3
5 with no wrong working seen. Dep. on all previous marks.
or 3
3
9
A method which uses 3
×
=
can score maximum M1M1A0A0.
NB
5
5
8
40
(b)
3
x×
3
p where
For 3
0
<
p
<
use of independence is M0 e.g.
8
M1
8
3
8
8
awrt 0.822
A1
H′
(c)
3 circles labelled. Either N inside H or allow as intersecting circles with P(
N
∩
)
=
B1
not allow blank space to be considered 0. Condone missing box for this mark.
Allow all 3 circles overlapping with all zeros correctly labelled.
3
3
9
P(
)
For
H
∩
W
=
×
=
seen or correctly placed in Venn diagram.
M1
8
5
40
For their 3
9
3
−
"
"
=
" "
(may be implied by the regions in their P(W) adding to 0.6)
M1
5
40
8
For their 3
9
1
1
−
"
"
−
=
"
"
M1
8
40
15
12
Fully correct diagram with ¼ and box and correct probabilities (allow exact decimal equivalents)
A1
(
)
E
2
2
2
0.25
3
2
0.3
4
2
0.15
5
2
0.1 6
2
0.2
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