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TS EAMCET 20-July-2022 Shift 1 Solved Paper
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© examsnet.com
Question : 39
Total: 160
If the probability distribution of a random variable
X
is given by
X=x
0
2
4
6
8
10
P(X=x)
0
k
2k
5
k
2
2
k
2
3k
then the mean of
X
is
384
121
60
13
163
25
326
49
Validate
Solution:
👈: Video Solution
We know that sum of total probability is 1 .
⇒
0
+
k
+
2
k
+
5
k
2
+
2
k
2
+
3
k
=
1
⇒
7
k
2
+
6
k
−
1
=
0
⇒
7
k
2
+
7
k
−
k
−
1
=
0
⇒
(
7
k
−
1
)
(
k
+
1
)
=
0
⇒
k
=
1
7
,
k
≠
−
1
Mean of
X
=
0
+
2
k
+
8
k
+
30
k
2
+
16
k
2
+
30
k
=
40
k
+
46
k
2
=
k
(
40
+
46
k
)
=
1
7
(
40
+
46
7
)
=
1
7
×
326
7
=
326
49
© examsnet.com
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