Mathematics · May / June 2023 · Paper 61 · Question 5

YMA01

As printed in the paper

The question as text

5. A discrete random variable Y has probability function k y y = 1, 2 ( 3 ) y = 3, 4, 5 k y ( 2 8 ) Y y P( ) y = 6 k otherwise 0 where k is a constant.

Read automatically from the paper; the printed page above is exact.

Show the answer and mark scheme

5(a)[2]
P(Y = y)
2k
k
k
8k
17k
k
2
k
+
k
+
k
+
8
k
+
17
k
+
k
= or
1
30
1
1
*
k
=
k
=
30
5(b)[2]
k
+
k
+
8
k
or 1 (2
−
k
+
17
k
+
k
)
= 1
3 oe awrt 0.333
5(c)[2]
(
1 2
×
) (
+
2
×
) (
+
3
×
) (
+
4 8
×
) (
+
5 17
×
) (
+
6
×
)
=
k
k
k
k
k
k
13
3 oe awrt 4.33
(
)
5(d)[3]
P
Y
15
−
2
Y
or [X =] 13 11 9 7 5 3 only or [Y =] 5 or 6 only
] = "17"
"1"
[ (
)
= (
)
(
)
P
Y
5
P
Y
=
5
+
P
Y
=
6
+
30
30
= 3
5 oe
(e) Var (X) = 4Var(Y)
302
[E(
Y
2
) ](1 2 )
=
×
k
+
(2
2
×
k
)
+
(3
2
×
k
)
+
(4
2
×
8 )
k
+
(5
2
×
17 )
k
+
(6
2
×
k
)
=
or awrt 20.1
15
(
)
( )
302
13
2
61
[Var
=
]"
"
−
"
"
=
or awrt 1.36
Y
15
3
45
[
]244
Var ( )
X =
oe awrt 5.42
45
ALT for 1st 3 marks
[E(X) =]
19
(13 2 )
×
k
+
(11
×
k
)
+
(9
×
k
)
+
(7 8 )
×
k
+
(5 17 )
×
k
+
(3
×
k
)
=
or awrt 6.33
3
683
[E(
X
2
) ](13
=
2
×
2 )
k
+
(11
2
×
k
)
+
(9
2
×
k
)
+
(7
2
×
8 )
k
+
(5
2
×
17 )
k
+
(3
2
×
k
)
=
or awrt 45.5
15
2
(
)
[Var
(
)
]"
683
"
"
190
"
X =
−
15
30
Notes
(a) M1 for finding the probabilities in terms of k. The individual probabilities must be seen either in a table
or in the calculation (but do not need to be simplified)
A1* Method mark must be awarded. For a correct equation which would lead to
k =
1 30
*
NB Verification ( ) ( ) ( )
( )
( ) ( )
2
1
8
17
1
+
1
+
1
+
1
+
1
+
1
= gains M1 A0
30
30
30
30
30
30
M1 for using P(
=
2)
+
P(
=
3)
+
P(
=
4)
or 1
−
P(
=
1)
+
P(
=
5)
+
P(
=
6)
Allow in terms of k or
(b)
Y
Y
Y
Y
Y
Y
with
1 30
subst or with their probabilities. Do not allow in terms of y
k =
A1 awrt 0.333
(c)
M1 for using
x
P( )
x
At least 3 terms given Allow with
k =
1 30
subst or ft their probabilities.
A1 awrt 4.33
(d) M1 forming correct inequality in Y or 13,11,9,7,5,3 seen anywhere or for 5 and 6 only. Implied by 2nd M1
(
)
(
)
or (
)
(
)
M1 finding
their P
Y
=
5
+
their P
Y
=
6
P
X
=
5
+
P
X
=
3
eg 17k + k
A1ft ft their probabilities
(e)
M1 written or used 4Var(Y) (may come at the end of the calculation) or written or used E(X) allow awrt
22 Var( )
6.33 NB condone
−
Y
if used 4Var(Y)
2
2
M1 Correct method, at least 3 products correct, to find
E(
Y
)
or
E(
X
)
condone incorrect labels
M1d dep on the 2nd M mark being awarded. For correct use of
E(
Y
2
)
−
[E( )]
Y
2
or
E(
X
2
)
−
[E(
the ALT In addition to the 2nd M1 the 1st M1 must be awarded. Condone incorrect labelling
A1 awrt 5.42
0.6
5(e)[4]
P(Y = y)
2k
k
k
8k
17k
k
2
k
+
k
+
k
+
8
k
+
17
k
+
k
= or
1
30
1
1
*
k
=
k
=
30
k
+
k
+
8
k
or 1 (2
−
k
+
17
k
+
k
)
= 1
3 oe awrt 0.333
(
1 2
×
) (
+
2
×
) (
+
3
×
) (
+
4 8
×
) (
+
5 17
×
) (
+
6
×
)
=
k
k
k
k
k
k
13
3 oe awrt 4.33
(
)
P
Y
15
−
2
Y
or [X =] 13 11 9 7 5 3 only or [Y =] 5 or 6 only
] = "17"
"1"
[ (
)
= (
)
(
)
P
Y
5
P
Y
=
5
+
P
Y
=
6
+
30
30
= 3
5 oe
(e) Var (X) = 4Var(Y)
302
[E(
Y
2
) ](1 2 )
=
×
k
+
(2
2
×
k
)
+
(3
2
×
k
)
+
(4
2
×
8 )
k
+
(5
2
×
17 )
k
+
(6
2
×
k
)
=
or awrt 20.1
15
(
)
( )
302
13
2
61
[Var
=
]"
"
−
"
"
=
or awrt 1.36
Y
15
3
45
[
]244
Var ( )
X =
oe awrt 5.42
45
ALT for 1st 3 marks
[E(X) =]
19
(13 2 )
×
k
+
(11
×
k
)
+
(9
×
k
)
+
(7 8 )
×
k
+
(5 17 )
×
k
+
(3
×
k
)
=
or awrt 6.33
3
683
[E(
X
2
) ](13
=
2
×
2 )
k
+
(11
2
×
k
)
+
(9
2
×
k
)
+
(7
2
×
8 )
k
+
(5
2
×
17 )
k
+
(3
2
×
k
)
=
or awrt 45.5
15
2
(
)
[Var
(
)
]"
683
"
"
190
"
X =
−
15
30
Notes
(a) M1 for finding the probabilities in terms of k. The individual probabilities must be seen either in a table
or in the calculation (but do not need to be simplified)
A1* Method mark must be awarded. For a correct equation which would lead to
k =
1 30
*
NB Verification ( ) ( ) ( )
( )
( ) ( )
2
1
8
17
1
+
1
+
1
+
1
+
1
+
1
= gains M1 A0
30
30
30
30
30
30
M1 for using P(
=
2)
+
P(
=
3)
+
P(
=
4)
or 1
−
P(
=
1)
+
P(
=
5)
+
P(
=
6)
Allow in terms of k or
(b)
Y
Y
Y
Y
Y
Y
with
1 30
subst or with their probabilities. Do not allow in terms of y
k =
A1 awrt 0.333
(c)
M1 for using
x
P( )
x
At least 3 terms given Allow with
k =
1 30
subst or ft their probabilities.
A1 awrt 4.33
(d) M1 forming correct inequality in Y or 13,11,9,7,5,3 seen anywhere or for 5 and 6 only. Implied by 2nd M1
(
)
(
)
or (
)
(
)
M1 finding
their P
Y
=
5
+
their P
Y
=
6
P
X
=
5
+
P
X
=
3
eg 17k + k
A1ft ft their probabilities
(e)
M1 written or used 4Var(Y) (may come at the end of the calculation) or written or used E(X) allow awrt
22 Var( )
6.33 NB condone
−
Y
if used 4Var(Y)
2
2
M1 Correct method, at least 3 products correct, to find
E(
Y
)
or
E(
X
)
condone incorrect labels
M1d dep on the 2nd M mark being awarded. For correct use of
E(
Y
2
)
−
[E( )]
Y
2
or
E(
X
2
)
−
[E(
the ALT In addition to the 2nd M1 the 1st M1 must be awarded. Condone incorrect labelling
A1 awrt 5.42
0.6

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