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P2 Q5
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17 P2 Q5
2017 H2 Mathematics Paper 2 Question 5
Discrete Random Variables (DRVs)
Answers
(i)
t
{t}
t
2
{2}
2
3
{3}
3
4
{4}
4
5
{5}
5
P
(
T
=
t
)
{\mathrm{P}(T=t)}
P
(
T
=
t
)
5
12
{\frac{5}{12}}
12
5
5
14
{\frac{5}{14}}
14
5
5
28
{\frac{5}{28}}
28
5
1
21
{\frac{1}{21}}
21
1
(ii)
E
(
T
)
=
20
7
{\mathrm{E}(T)=\frac{20}{7}}
E
(
T
)
=
7
20
V
a
r
(
T
)
=
75
98
{\mathrm{Var}(T)=\frac{75}{98}}
Var
(
T
)
=
98
75
(iii)
0.238
{0.238}
0.238
Full solutions
(i)
P
(
T
=
2
)
=
6
9
(
5
8
)
=
5
12
■
P
(
T
=
3
)
=
6
9
(
3
8
)
(
5
7
)
×
2
!
=
5
14
■
P
(
T
=
4
)
=
6
9
(
3
8
)
(
2
7
)
(
5
6
)
×
3
!
2
!
=
5
28
■
P
(
T
=
5
)
=
6
9
(
3
8
)
(
2
7
)
(
1
6
)
(
5
5
)
×
4
!
3
!
=
1
21
■
\begin{align*} \mathrm{P}(T=2) &= \frac{6}{9} \left( \frac{5}{8} \right) \\ &= \frac{5}{12} \; \blacksquare \\ \mathrm{P}(T=3) &= \frac{6}{9} \left( \frac{3}{8} \right) \left( \frac{5}{7} \right) \times 2! \\ &= \frac{5}{14} \; \blacksquare \\ \mathrm{P}(T=4) &= \frac{6}{9} \left( \frac{3}{8} \right) \left( \frac{2}{7} \right) \left( \frac{5}{6} \right) \times \frac{3!}{2!} \\ &= \frac{5}{28} \; \blacksquare \\ \mathrm{P}(T=5) &= \frac{6}{9} \left( \frac{3}{8} \right) \left( \frac{2}{7} \right) \left( \frac{1}{6} \right) \left( \frac{5}{5} \right) \times \frac{4!}{3!} \\ &= \frac{1}{21} \; \blacksquare \\ \end{align*}
P
(
T
=
2
)
P
(
T
=
3
)
P
(
T
=
4
)
P
(
T
=
5
)
=
9
6
(
8
5
)
=
12
5
■
=
9
6
(
8
3
)
(
7
5
)
×
2
!
=
14
5
■
=
9
6
(
8
3
)
(
7
2
)
(
6
5
)
×
2
!
3
!
=
28
5
■
=
9
6
(
8
3
)
(
7
2
)
(
6
1
)
(
5
5
)
×
3
!
4
!
=
21
1
■
t
{t}
t
2
{2}
2
3
{3}
3
4
{4}
4
5
{5}
5
P
(
T
=
t
)
{\mathrm{P}(T=t)}
P
(
T
=
t
)
5
12
{\frac{5}{12}}
12
5
5
14
{\frac{5}{14}}
14
5
5
28
{\frac{5}{28}}
28
5
1
21
{\frac{1}{21}}
21
1
(ii)
E
(
T
)
=
∑
t
t
P
(
T
=
t
)
=
2
(
5
12
)
+
3
(
5
14
)
+
4
(
5
28
)
+
5
(
1
21
)
=
20
7
■
\begin{align*} \mathrm{E}(T) &= \sum_t t P(T=t) \\ &= 2\left(\frac{5}{12}\right) + 3\left(\frac{5}{14}\right) + 4\left(\frac{5}{28}\right) + 5\left(\frac{1}{21}\right) \\ &= \frac{20}{7} \; \blacksquare \end{align*}
E
(
T
)
=
t
∑
tP
(
T
=
t
)
=
2
(
12
5
)
+
3
(
14
5
)
+
4
(
28
5
)
+
5
(
21
1
)
=
7
20
■
E
(
T
2
)
=
∑
t
t
2
P
(
T
=
t
)
=
2
2
(
5
12
)
+
3
2
(
5
14
)
+
4
2
(
5
28
)
+
5
2
(
1
21
)
=
125
14
\begin{align*} \mathrm{E}(T^2) &= \sum_t t^2 P(T=t) \\ &= 2^2\left(\frac{5}{12}\right) + 3^2\left(\frac{5}{14}\right) + 4^2\left(\frac{5}{28}\right) + 5^2\left(\frac{1}{21}\right) \\ &= \frac{125}{14} \end{align*}
E
(
T
2
)
=
t
∑
t
2
P
(
T
=
t
)
=
2
2
(
12
5
)
+
3
2
(
14
5
)
+
4
2
(
28
5
)
+
5
2
(
21
1
)
=
14
125
V
a
r
(
T
)
=
E
(
T
2
)
−
(
E
(
T
)
)
2
=
125
14
−
(
20
7
)
2
=
75
98
■
\begin{align*} \mathrm{Var}(T) &= \mathrm{E}(T^2) - \left( \mathrm{E}(T) \right)^2 \\ &= \frac{125}{14} - \left( \frac{20}{7} \right)^2 \\ &= \frac{75}{98} \; \blacksquare \end{align*}
Var
(
T
)
=
E
(
T
2
)
−
(
E
(
T
)
)
2
=
14
125
−
(
7
20
)
2
=
98
75
■
(iii)
P
(
T
≥
4
)
=
P
(
T
=
4
)
+
P
(
T
=
5
)
=
5
28
+
1
21
=
19
84
\begin{align*} \mathrm{P}(T\geq 4) &= \mathrm{P}(T = 4) + \mathrm{P}(T = 5) \\ &= \frac{5}{28} + \frac{1}{21} \\ &= \frac{19}{84} \end{align*}
P
(
T
≥
4
)
=
P
(
T
=
4
)
+
P
(
T
=
5
)
=
28
5
+
21
1
=
84
19
Let
X
{X}
X
denote the r.v. of the number of times Lee takes at least 4 counters out of the bag in his
15
{15}
15
games
X
∼
B
(
15
,
19
84
)
X \sim \textrm{B}\left(15, \frac{19}{84}\right)
X
∼
B
(
15
,
84
19
)
P
(
X
≥
5
)
=
1
−
P
(
X
≤
4
)
=
0.238
■
\begin{align*} \mathrm{P}(X \geq 5) &= 1 - \mathrm{P}(X\leq 4) \\ &= 0.238 \; \blacksquare \end{align*}
P
(
X
≥
5
)
=
1
−
P
(
X
≤
4
)
=
0.238
■
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