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AP EAMCET Engineering 2018 Apr 22 Shift 1 Paper
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© examsnet.com
Question : 39
Total: 160
If the probability function of a random variable X is defined by
P
(
X
=
k
)
=
a
(
k
+
1
2
k
)
for
k
=
0
,
1
,
2
,
3
,
4
,
5
then the probability that X takes a prime value is
13/20
23/60
11/20
19/60
Validate
Solution:
It is given.
P
(
x
=
k
)
=
a
(
k
+
1
2
k
)
We know.
∑
P
(
x
=
k
)
=
1
P
(
x
=
0
)
+
P
(
x
=
1
)
+
P
(
x
=
2
)
+
P
(
x
=
3
)
+
P
(
x
=
4
)
+
P
(
x
=
5
)
=
1
Putting the values in above we get.
a
+
2
a
2
+
3
a
4
+
4
a
8
+
5
a
16
+
6
a
32
=
1
16
a
+
16
a
+
12
a
+
8
a
+
5
a
+
3
a
16
=
1
a
=
4
15
The required probability is given as.
P
(
x
=
2
)
+
P
(
x
=
3
)
+
P
(
x
=
4
)
=
3
a
4
+
a
2
+
3
a
16
=
12
a
+
8
a
+
3
a
16
=
23
a
16
=
23
(
4
15
)
16
=
23
60
© examsnet.com
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