Mathematics · May / June 2024 · Paper 61 · Question 2
YMA01
As printed in the paper
The question as text
2. A spinner can land on the numbers 2, 4, 5, 7 or 8 only. The random variable X represents the number that this spinner lands on when it is spun once. The probability distribution of X is given in the table below. x 2 4 5 7 8 P(X = x) 0.25 0.3 0.2 0.1 0.15
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
2(a)[1]
P 2
X
– 3
>
5
(
)
[
]
2(b)[3]
E
2
=
2
2
×
0.25
+
4
2
×
0.3
+
5
2
×
0.2
+
7
2
×
0.1 8
+
2
×
0.15
=
25.3
X
Var(
X =
)
2
2
×
0.25
+
4
2
×
0.3 5
+
2
×
0.2
+
7
2
×
0.1 8
+
2
×
0.15
−
4.6
2
or
Var(
X =
)
"25.3" 4.6
−
2
=
25.3
−
4.6
2
=
4.14
*
[
]
E( )
=
13.4
=
×
4.6
−
2(c)[4]
Y
a
b
[
]
Var( )
=
2
×
4.14
=
66.2
4
Y
a
a =
4
b =
5
Sam throws 8 and Alex throws 2, 4 or 5
2(d)[4]
Sam throws 7 and Alex throws 2 or 4
Sam throws 5 or 4 or 2 and Alex throws 2
0.15 (0.25
×
+
0.3
+
0.2)
or 0.1 (0.25
×
+
0.3)
or (0.2
+
0.3
+
0.25) 0.25
×
0.15 (0.25
×
+
0.3
+
0.2)
+ 0.1 (0.25
×
+
0.3)
+ (0.2
+
0.3
+
0.25) 0.25
×
71
= 0.355
=
A1
200
Alex throws 2
ALT
Alex throws 4 and Sam throws 7 or 8
Alex throws 5 and Sam throws 8
(
)
0.25 or 0.3
×
0.1 0.15 or 0.2 0.15
+
×
(
)
0.25
+
0.3
×
0.1 0.15
+
+
0.2 0.15
×
71
= 0.355
=
A1
200
Notes
(a)
B1
0.45 oe
for a correct method to find (
)
E X
2
At least 3 terms correct and added This is
(b)
M1
not implied by 25.3 on its own
for use of correct equation ft their (
)
E X
2
M1
for a correct expression, with all terms seen, leading to the given answer
A1*
(c)
for writing or using a correct equation for E(Y)
M1
M1
for writing or using a correct method for Var(Y)
for a = 4 may be seen as part of the expression 4
...
A1
X ±
for b = 5 may be seen as part of the expression …X – 5
A1
for recognising all the required combinations – implied by 3 correct
(d)
M1
probabilities Ignore any repeats but do not ignore any incorrect combinations
M1
for any one correct calculation from the 3 given
for any 2 correct calculations from the 3 given
M1
0.355 oe
A1
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