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2009
P2 Q11
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Binomial
09 P2 Q11
2009 H2 Mathematics Paper 2 Question 11
The Binomial Distribution
Answers
(i)
The probability of a car being red is the same for each car
Whether a car is red is independent of any other car
(ii)
0.346
{0.346}
0.346
(iii)
Out of syllabus
(iv)
Out of syllabus
(v)
0.142
{0.142}
0.142
Full solutions
(i)
The probability of a car being red is the same for each car
Whether a car is red is independent of any other car
(ii)
R
∼
B
(
20
,
0.15
)
R \sim \textrm{B}\left(20, 0.15\right)
R
∼
B
(
20
,
0.15
)
P
(
4
≤
R
<
8
)
=
P
(
R
≤
7
)
−
P
(
R
≤
3
)
=
0.346
(3 sf)
■
\begin{align*} & \mathrm{P}(4 \leq R < 8) \\ &= \mathrm{P}(R \leq 7) - \mathrm{P}(R \leq 3) \\ &= 0.346 \textrm{ (3 sf)} \; \blacksquare \end{align*}
P
(
4
≤
R
<
8
)
=
P
(
R
≤
7
)
−
P
(
R
≤
3
)
=
0.346
(3 sf)
■
(iii)
Out of syllabus
(iv)
Out of syllabus
(v)
R
∼
B
(
20
,
p
)
R \sim \mathrm{B}\left(20, p\right)
R
∼
B
(
20
,
p
)
P
(
R
=
0
or
1
)
=
0.2
(
20
0
)
p
0
(
1
−
p
)
20
−
0
+
(
20
1
)
p
1
(
1
−
p
)
20
−
1
=
0.2
\begin{gather*} \mathrm{P}(R = 0 \textrm{ or } 1) = 0.2 \\ {20 \choose 0} p^0 (1-p)^{20-0} + {20 \choose 1} p^1 (1-p)^{20-1} = 0.2 \\ \end{gather*}
P
(
R
=
0
or
1
)
=
0.2
(
0
20
)
p
0
(
1
−
p
)
20
−
0
+
(
1
20
)
p
1
(
1
−
p
)
20
−
1
=
0.2
(
1
−
p
)
20
+
20
p
(
1
−
p
)
19
=
0.2
■
(1-p)^{20} + 20p(1-p)^{19} = 0.2 \; \blacksquare
(
1
−
p
)
20
+
20
p
(
1
−
p
)
19
=
0.2
■
Solving with a GC, since
0
≤
p
≤
1
,
{0 \leq p \leq 1,}
0
≤
p
≤
1
,
p
=
0.142
■
p=0.142 \; \blacksquare
p
=
0.142
■
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