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AP EAMCET Engineering 2015 Code C Paper
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© examsnet.com
Question : 6
Total: 160
The probability distribution of a random variable is given below
X
=
x
0
1
2
3
4
5
6
7
P
(
X
=
x
)
0
K
2K
2K
3
K
K
2
2
K
2
7
K
2
+
K
Then
P
(
0
<
X
<
5
)
=
1
10
3
10
8
10
7
10
12
Validate
Solution:
∑
P
(
x
=
x
i
)
=
1
10
K
2
+
9
K
−
1
=
0
10
K
2
+
10
K
−
K
−
1
=
0
10
K
+
(
K
+
1
)
−
1
(
K
+
1
)
=
0
K
=
−
1
,
1
10
,
1
But
K
≠
−
1
P
(
0
<
X
<
5
)
=
P
(
X
=
1
)
+
P
(
X
=
2
)
+
P
(
X
=
3
)
+
P
(
X
=
4
)
=
K
+
2
K
+
2
K
+
3
K
=
8
K
=
0.8
=
8
10
© examsnet.com
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