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AIEEE 2003 Math Solved Paper
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© examsnet.com
Question : 41
Total: 66
Events
A
,
B
,
C
are mutually exclusive events such that
P
(
A
)
=
3
x
+
1
3
,
P
(
B
)
=
1
−
x
4
and
P
(
C
)
=
1
−
2
x
2
The set of possible values of
x
are in the interval.
[
0
,
1
]
[
1
3
,
1
2
]
[
1
3
,
2
3
]
[
13
,
13
3
]
Validate
Solution:
Given
P
(
A
)
=
3
x
+
1
3
,
P
(
B
)
=
1
−
x
4
and
P
(
C
)
=
1
−
2
x
2
We know for any event
X
,
0
≤
P
(
X
)
≤
1
∴
0
≤
3
x
+
1
3
≤
1
⇒
−
1
≤
3
x
≤
2
⇒
−
1
3
≤
x
≤
2
3
0
≤
1
−
x
4
≤
1
⇒
−
3
≤
x
≤
1
0
≤
1
−
2
x
2
≤
1
⇒
−
1
≤
2
x
≤
1
⇒
−
1
2
≤
x
≤
1
2
In the question given that
A
,
B
and
C
are mutually exclusive
So
P
(
A
∪
B
∪
C
)
=
P
(
A
)
+
P
(
B
)
+
P
(
C
)
⇒
P
(
A
∪
B
∪
C
)
=
3
x
+
1
3
+
1
−
x
4
+
1
−
2
x
2
∴
0
≤
3
x
+
1
3
+
1
−
x
4
+
1
−
2
x
2
≤
1
0
≤
13
−
3
x
≤
12
⇒
1
3
≤
x
≤
13
3
From all those relations, we get
max
{
−
1
3
,
−
3
,
−
1
2
,
1
3
}
≤
x
≤
min
{
2
3
,
1
,
1
2
,
13
3
}
So,
1
3
≤
x
≤
1
2
⇒
⇒
x
∈
[
1
3
,
1
2
]
© examsnet.com
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