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2012
P2 Q9
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Binomial
12 P2 Q9
2012 H2 Mathematics Paper 2 Question 9
The Binomial Distribution
Answers
(i)
The probability of a voter supporting the Alliance Party is the same for each voter
Whether a voter supports the Alliance Party is independent of any other voter
(ii)
0.373
{0.373}
0.373
(iii)
Out of syllabus
(iv)
k
=
0.23790
{k=0.23790}
k
=
0.23790
p
=
0.39
{p=0.39}
p
=
0.39
Full solutions
(i)
The probability of a voter supporting the Alliance Party is the same for each voter
Whether a voter supports the Alliance Party is independent of any other voter
(ii)
A
∼
B
(
30
,
0.15
)
A \sim \textrm{B}\left(30, 0.15\right)
A
∼
B
(
30
,
0.15
)
P
(
A
=
3
or
4
)
=
P
(
A
=
3
)
+
P
(
A
=
4
)
=
0.17026
+
0.20281
=
0.373
(3 sf)
■
\begin{align*} & \mathrm{P}(A=3 \textrm{ or } 4) \\ & = \mathrm{P}(A=3) + \mathrm{P}(A=4) \\ & = 0.17026 + 0.20281 \\ & = 0.373 \textrm{ (3 sf)} \; \blacksquare \end{align*}
P
(
A
=
3
or
4
)
=
P
(
A
=
3
)
+
P
(
A
=
4
)
=
0.17026
+
0.20281
=
0.373
(3 sf)
■
(iii)
Out of syllabus
(iv)
P
(
A
=
15
)
=
0.06864
(
30
15
)
3
0
15
(
1
−
p
)
30
−
15
=
0.06864
155
,
117
,
520
p
15
(
1
−
p
)
15
=
0.06864
p
(
1
−
p
)
=
(
0.06864
155
,
117
,
520
)
1
15
\begin{gather*} \mathrm{P}(A=15) = 0.06864 \\ {30 \choose 15} 30^{15} (1-p)^{30-15} = 0.06864 \\ 155,117,520 p^{15} (1-p)^{15} = 0.06864 \\ p(1-p) = \left( \frac{0.06864}{155,117,520} \right)^{\frac{1}{15}} \\ \end{gather*}
P
(
A
=
15
)
=
0.06864
(
15
30
)
3
0
15
(
1
−
p
)
30
−
15
=
0.06864
155
,
117
,
520
p
15
(
1
−
p
)
15
=
0.06864
p
(
1
−
p
)
=
(
155
,
117
,
520
0.06864
)
15
1
p
(
1
−
p
)
=
0.23790
■
p(1-p) = 0.23790 \; \blacksquare
p
(
1
−
p
)
=
0.23790
■
p
2
−
p
+
0.23790
=
0
p^2 - p + 0.23790 = 0
p
2
−
p
+
0.23790
=
0
Since
p
<
0.5
,
{p<0.5, }
p
<
0.5
,
using a GC,
p
=
0.39
(3 sf)
■
p=0.39 \textrm{ (3 sf)} \; \blacksquare
p
=
0.39
(3 sf)
■
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