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2018
P2 Q8
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DRV
18 P2 Q8
2018 H2 Mathematics Paper 2 Question 8
Discrete Random Variables (DRVs)
Answers
(i)
s
{s}
s
6
{6}
6
7
{7}
7
8
{8}
8
9
{9}
9
10
{10}
10
P
(
S
=
s
)
{\mathrm{P}(S=s)}
P
(
S
=
s
)
2
(
n
+
5
)
(
n
+
4
)
{\frac{2}{(n+5)(n+4)}}
(
n
+
5
)
(
n
+
4
)
2
12
(
n
+
5
)
(
n
+
4
)
{\frac{12}{(n+5)(n+4)}}
(
n
+
5
)
(
n
+
4
)
12
4
n
+
6
(
n
+
5
)
(
n
+
4
)
{\frac{4n+6}{(n+5)(n+4)}}
(
n
+
5
)
(
n
+
4
)
4
n
+
6
6
n
(
n
+
5
)
(
n
+
4
)
{\frac{6n}{(n+5)(n+4)}}
(
n
+
5
)
(
n
+
4
)
6
n
n
(
n
−
1
)
(
n
+
5
)
(
n
+
4
)
{\frac{n(n-1)}{(n+5)(n+4)}}
(
n
+
5
)
(
n
+
4
)
n
(
n
−
1
)
(ii)
P
(
S
=
10
)
=
0
{\mathrm{P}(S=10)=0}
P
(
S
=
10
)
=
0
This is because if there is only
1
{1}
1
ball numbered
5
,
{5, }
5
,
then it is impossible to get
S
=
10
{S=10}
S
=
10
when taking two balls without replacement
(iii)
V
a
r
(
S
)
=
22
n
2
+
78
n
+
36
(
n
+
5
)
2
(
n
+
4
)
{\mathrm{Var}(S)=\frac{22 n^2 + 78 n + 36}{(n+5)^2(n+4)}}
Var
(
S
)
=
(
n
+
5
)
2
(
n
+
4
)
22
n
2
+
78
n
+
36
Full solutions
(i)
P
(
S
=
6
)
=
2
n
+
5
(
1
n
+
4
)
=
2
(
n
+
5
)
(
n
+
4
)
■
P
(
S
=
7
)
=
2
n
+
5
(
3
n
+
4
)
×
2
!
=
12
(
n
+
5
)
(
n
+
4
)
■
P
(
S
=
8
)
=
3
n
+
5
(
2
n
+
4
)
+
2
n
+
5
(
n
n
+
4
)
×
2
!
=
4
n
+
6
(
n
+
5
)
(
n
+
4
)
■
P
(
S
=
9
)
=
3
n
+
5
(
n
n
+
4
)
×
2
!
=
6
n
(
n
+
5
)
(
n
+
4
)
■
P
(
S
=
10
)
=
n
n
+
5
(
n
−
1
n
+
4
)
=
n
(
n
−
1
)
(
n
+
5
)
(
n
+
4
)
■
\begin{align*} \mathrm{P}(S=6) &= \frac{2}{n+5} \left( \frac{1}{n+4} \right) \\ &= \frac{2}{(n+5)(n+4)} \; \blacksquare \\ \mathrm{P}(S=7) &= \frac{2}{n+5} \left( \frac{3}{n+4} \right) \times 2! \\ &= \frac{12}{(n+5)(n+4)} \; \blacksquare \\ \mathrm{P}(S=8) &= \frac{3}{n+5} \left( \frac{2}{n+4} \right) + \frac{2}{n+5} \left( \frac{n}{n+4} \right)\times 2! \\ &= \frac{4n+6}{(n+5)(n+4)} \; \blacksquare \\ \mathrm{P}(S=9) &= \frac{3}{n+5} \left( \frac{n}{n+4} \right) \times 2! \\ &= \frac{6n}{(n+5)(n+4)} \; \blacksquare \\ \mathrm{P}(S=10) &= \frac{n}{n+5} \left( \frac{n-1}{n+4} \right) \\ &= \frac{n(n-1)}{(n+5)(n+4)} \; \blacksquare \\ \end{align*}
P
(
S
=
6
)
P
(
S
=
7
)
P
(
S
=
8
)
P
(
S
=
9
)
P
(
S
=
10
)
=
n
+
5
2
(
n
+
4
1
)
=
(
n
+
5
)
(
n
+
4
)
2
■
=
n
+
5
2
(
n
+
4
3
)
×
2
!
=
(
n
+
5
)
(
n
+
4
)
12
■
=
n
+
5
3
(
n
+
4
2
)
+
n
+
5
2
(
n
+
4
n
)
×
2
!
=
(
n
+
5
)
(
n
+
4
)
4
n
+
6
■
=
n
+
5
3
(
n
+
4
n
)
×
2
!
=
(
n
+
5
)
(
n
+
4
)
6
n
■
=
n
+
5
n
(
n
+
4
n
−
1
)
=
(
n
+
5
)
(
n
+
4
)
n
(
n
−
1
)
■
s
{s}
s
6
{6}
6
7
{7}
7
8
{8}
8
9
{9}
9
10
{10}
10
P
(
S
=
s
)
{\mathrm{P}(S=s)}
P
(
S
=
s
)
2
(
n
+
5
)
(
n
+
4
)
{\frac{2}{(n+5)(n+4)}}
(
n
+
5
)
(
n
+
4
)
2
12
(
n
+
5
)
(
n
+
4
)
{\frac{12}{(n+5)(n+4)}}
(
n
+
5
)
(
n
+
4
)
12
4
n
+
6
(
n
+
5
)
(
n
+
4
)
{\frac{4n+6}{(n+5)(n+4)}}
(
n
+
5
)
(
n
+
4
)
4
n
+
6
6
n
(
n
+
5
)
(
n
+
4
)
{\frac{6n}{(n+5)(n+4)}}
(
n
+
5
)
(
n
+
4
)
6
n
n
(
n
−
1
)
(
n
+
5
)
(
n
+
4
)
{\frac{n(n-1)}{(n+5)(n+4)}}
(
n
+
5
)
(
n
+
4
)
n
(
n
−
1
)
(ii)
P
(
S
=
10
)
=
1
(
1
−
1
)
(
1
+
5
)
(
1
+
4
)
=
0
■
\begin{align*} \mathrm{P}(S=10) &= \frac{1(1-1)}{(1+5)(1+4)} \\ &= 0 \; \blacksquare \end{align*}
P
(
S
=
10
)
=
(
1
+
5
)
(
1
+
4
)
1
(
1
−
1
)
=
0
■
This is because if there is only
1
{1}
1
ball numbered
5
,
{5, }
5
,
then it is impossible to get
S
=
10
{S=10}
S
=
10
when taking two balls without replacement
(iii)
E
(
S
)
=
∑
s
s
P
(
S
=
s
)
=
12
+
84
+
8
(
4
n
+
6
)
+
54
n
+
10
n
(
n
−
1
)
(
n
+
5
)
(
n
+
4
)
=
10
n
2
+
76
n
+
144
(
n
+
5
)
(
n
+
4
)
=
2
(
n
+
4
)
(
5
n
+
18
)
(
n
+
5
)
(
n
+
4
)
=
10
n
+
36
n
+
5
■
\begin{align*} \mathrm{E}(S) &= \sum_s s \mathrm{P}(S=s) \\ &= \frac{12+84+8(4n+6)+54n+10n(n-1)}{(n+5)(n+4)} \\ &= \frac{10 n^2 + 76 n + 144}{(n+5)(n+4)} \\ &= \frac{2(n + 4)(5 n + 18)}{(n+5)(n+4)} \\ &= \frac{10 n + 36}{n+5} \; \blacksquare \end{align*}
E
(
S
)
=
s
∑
s
P
(
S
=
s
)
=
(
n
+
5
)
(
n
+
4
)
12
+
84
+
8
(
4
n
+
6
)
+
54
n
+
10
n
(
n
−
1
)
=
(
n
+
5
)
(
n
+
4
)
10
n
2
+
76
n
+
144
=
(
n
+
5
)
(
n
+
4
)
2
(
n
+
4
)
(
5
n
+
18
)
=
n
+
5
10
n
+
36
■
E
(
S
2
)
=
∑
s
s
2
P
(
S
=
s
)
=
72
+
588
+
64
(
4
n
+
6
)
+
486
n
+
100
n
(
n
−
1
)
(
n
+
5
)
(
n
+
4
)
=
100
n
2
+
642
n
+
1044
(
n
+
5
)
(
n
+
4
)
\begin{align*} \mathrm{E}(S^2) &= \sum_s s^2 \mathrm{P}(S=s) \\ &= \frac{72+588+64(4n+6)+486n+100n(n-1)}{(n+5)(n+4)} \\ &= \frac{100 n^2 + 642 n + 1044}{(n+5)(n+4)} \\ \end{align*}
E
(
S
2
)
=
s
∑
s
2
P
(
S
=
s
)
=
(
n
+
5
)
(
n
+
4
)
72
+
588
+
64
(
4
n
+
6
)
+
486
n
+
100
n
(
n
−
1
)
=
(
n
+
5
)
(
n
+
4
)
100
n
2
+
642
n
+
1044
V
a
r
(
S
)
=
E
(
S
2
)
−
(
E
(
S
)
)
2
=
100
n
2
+
642
n
+
1044
(
n
+
5
)
(
n
+
4
)
−
(
10
n
+
36
)
2
(
n
+
5
)
2
=
(
100
n
2
+
642
n
+
1044
)
(
n
+
5
)
−
(
10
n
+
36
)
2
(
n
+
4
)
(
n
+
5
)
2
(
n
+
4
)
=
22
n
2
+
78
n
+
36
(
n
+
5
)
2
(
n
+
4
)
■
\begin{align*} & \mathrm{Var}(S) \\ &= \mathrm{E}(S^2) - \left(\mathrm{E}(S)\right)^2 \\ &= \frac{100 n^2 + 642 n + 1044}{(n+5)(n+4)} - \frac{(10 n + 36)^2}{(n+5)^2} \\ &= \frac{(100 n^2 + 642 n + 1044)(n+5)-(10 n + 36)^2(n+4)}{(n+5)^2(n+4)} \\ &= \frac{22 n^2 + 78 n + 36}{(n+5)^2(n+4)} \; \blacksquare \end{align*}
Var
(
S
)
=
E
(
S
2
)
−
(
E
(
S
)
)
2
=
(
n
+
5
)
(
n
+
4
)
100
n
2
+
642
n
+
1044
−
(
n
+
5
)
2
(
10
n
+
36
)
2
=
(
n
+
5
)
2
(
n
+
4
)
(
100
n
2
+
642
n
+
1044
)
(
n
+
5
)
−
(
10
n
+
36
)
2
(
n
+
4
)
=
(
n
+
5
)
2
(
n
+
4
)
22
n
2
+
78
n
+
36
■
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