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2018
P2 Q7
Topical
Probability
18 P2 Q7
2018 H2 Mathematics Paper 2 Question 7
Probability
Answers
(i)
1
−
a
−
b
+
a
b
{1-a-b+ab}
1
−
a
−
b
+
ab
(ii)
1
−
a
−
c
{1-a-c}
1
−
a
−
c
(iii)
Maximum
P
(
A
∩
B
)
=
1
3
{\textrm{P}\left(A \cap B\right) = \frac{1}{3}}
P
(
A
∩
B
)
=
3
1
Minimum
P
(
A
∩
B
)
=
2
15
{\textrm{P}\left(A \cap B\right) = \frac{2}{15}}
P
(
A
∩
B
)
=
15
2
Full solutions
(i)
Since
A
{A}
A
and
B
{B}
B
are independent,
P
(
A
∩
B
)
=
P
(
A
)
⋅
P
(
B
)
\textrm{P}\left(A \cap B\right) = \textrm{P}\left(A\right)\cdot\textrm{P}\left(B\right)
P
(
A
∩
B
)
=
P
(
A
)
⋅
P
(
B
)
P
(
A
′
∩
B
′
)
=
1
−
P
(
A
∪
B
)
=
1
−
(
P
(
A
)
+
P
(
B
)
−
P
(
A
∩
B
)
)
=
1
−
(
a
+
b
−
a
b
)
=
1
−
a
−
b
+
a
b
■
=
1
−
a
−
b
(
1
−
a
)
=
(
1
−
a
)
(
1
−
b
)
=
P
(
A
′
)
⋅
P
(
B
′
)
\begin{align*} & \textrm{P}\left(A' \cap B'\right) \\ &= 1 - \textrm{P}\left(A \cup B\right) \\ &= 1 - \Big( \textrm{P}\left(A\right) + \textrm{P}\left(B\right) - \textrm{P}\left(A \cap B\right) \Big) \\ &= 1 - (a+b-ab) \\ &= 1- a - b + ab \; \blacksquare \\ &= 1-a - b (1-a) \\ &= (1-a)(1-b) \\ &= \textrm{P}\left(A'\right) \cdot \textrm{P}\left(B'\right) \end{align*}
P
(
A
′
∩
B
′
)
=
1
−
P
(
A
∪
B
)
=
1
−
(
P
(
A
)
+
P
(
B
)
−
P
(
A
∩
B
)
)
=
1
−
(
a
+
b
−
ab
)
=
1
−
a
−
b
+
ab
■
=
1
−
a
−
b
(
1
−
a
)
=
(
1
−
a
)
(
1
−
b
)
=
P
(
A
′
)
⋅
P
(
B
′
)
Since
P
(
A
′
∩
B
′
)
=
P
(
A
′
)
⋅
P
(
B
′
)
,
\textrm{P}\left(A' \cap B'\right) = \textrm{P}\left(A'\right) \cdot \textrm{P}\left(B'\right),
P
(
A
′
∩
B
′
)
=
P
(
A
′
)
⋅
P
(
B
′
)
,
A
′
{A'}
A
′
and
B
′
{B'}
B
′
are independent
■
{\blacksquare}
■
(ii)
Since
A
{A}
A
and
B
{B}
B
are mutually exclusive,
P
(
A
∩
C
)
=
0
\textrm{P}\left(A \cap C\right) = 0
P
(
A
∩
C
)
=
0
P
(
A
′
∩
C
′
)
=
1
−
P
(
A
∪
C
)
=
1
−
(
P
(
A
)
+
P
(
C
)
−
P
(
A
∩
C
)
)
=
1
−
(
a
+
c
)
=
1
−
a
−
c
■
\begin{align*} & \textrm{P}\left(A' \cap C'\right) \\ &= 1 - \textrm{P}\left(A \cup C\right) \\ &= 1 - \Big( \textrm{P}\left(A\right) + \textrm{P}\left(C\right) - \textrm{P}\left(A \cap C\right) \Big) \\ &= 1 - (a+c) \\ &= 1- a - c \; \blacksquare \\ \end{align*}
P
(
A
′
∩
C
′
)
=
1
−
P
(
A
∪
C
)
=
1
−
(
P
(
A
)
+
P
(
C
)
−
P
(
A
∩
C
)
)
=
1
−
(
a
+
c
)
=
1
−
a
−
c
■
If
A
′
{A'}
A
′
and
C
′
{C'}
C
′
are also mutually exclusive,
P
(
A
′
∩
C
′
)
=
0
1
−
a
−
c
=
0
a
+
c
=
1
\begin{align*} \textrm{P}\left(A' \cap C'\right) &= 0 \\ 1- a - c &= 0 \\ a + c &= 1 \end{align*}
P
(
A
′
∩
C
′
)
1
−
a
−
c
a
+
c
=
0
=
0
=
1
Hence the regions
A
{A}
A
and
C
{C}
C
make up the universal set in our Venn diagram
(iii)
From the Venn diagram, since all probabilities are between
0
{0}
0
and
1
,
{1,}
1
,
3
5
b
−
1
5
≥
0
b
≥
1
3
\begin{align*} \frac{3}{5}b - \frac{1}{5} &\geq 0 \\ b &\geq \frac{1}{3} \end{align*}
5
3
b
−
5
1
b
≥
0
≥
3
1
1
2
−
3
5
b
≥
0
b
≤
5
6
\begin{align*} \frac{1}{2} - \frac{3}{5}b &\geq 0 \\ b &\leq \frac{5}{6} \end{align*}
2
1
−
5
3
b
b
≥
0
≤
6
5
P
(
A
∩
B
)
=
2
5
b
\textrm{P}\left(A \cap B\right) = \frac{2}{5}b
P
(
A
∩
B
)
=
5
2
b
1
3
≤
b
≤
5
6
2
15
≤
P
(
A
∩
B
)
≤
1
3
\begin{gather*} \frac{1}{3} \leq b \leq \frac{5}{6} \\ \frac{2}{15} \leq \textrm{P}\left(A \cap B\right) \leq \frac{1}{3} \end{gather*}
3
1
≤
b
≤
6
5
15
2
≤
P
(
A
∩
B
)
≤
3
1
Maximum P
(
A
∩
B
)
=
1
3
■
Minimum P
(
A
∩
B
)
=
2
15
■
\begin{align*} & \textrm{Maximum } \textrm{P}\left(A \cap B\right) \\ &\quad= \frac{1}{3} \; \blacksquare \\ & \textrm{Minimum } \textrm{P}\left(A \cap B\right) \\ &\quad= \frac{2}{15} \; \blacksquare \end{align*}
Maximum
P
(
A
∩
B
)
=
3
1
■
Minimum
P
(
A
∩
B
)
=
15
2
■
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