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NIMCET 2011 Question Paper with solutions
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© examsnet.com
Question : 25
Total: 120
An anti-aircraft gun can take a maximum of four slots at an enemy plane moving away from it. The probabilities of hitting the plane at the first, second, third and fourth slots are
0.4
,
0.3
,
0.2
and
0.1
respectively. The probability that the gun hits the plane, then is
0.5
0.7235
0.6976
1.0
Validate
Solution:
Given,
P
(
I
)
=
0.4
,
P
(
I
)
=
0.3
,
P
(
I
I
I
)
=
0.2
,
P
(
I
V
)
=
0.1
⇒
P
(
I
)
=
0.6
,
P
(
I
I
)
=
0.7
,
P
(
I
I
I
)
=
0.8
and
P
(
I
V
)
=
0.9
∴ Required probability
=
P
(
I
)
+
P
(
I
)
P
(
I
I
)
+
P
(
I
)
P
(
I
I
)
P
(
I
I
I
)
+
P
(
I
)
P
(
I
I
)
P
(
I
I
I
)
P
(
I
V
)
=
0.4
+
(
0.6
)
(
0.3
)
+
(
0.6
)
(
0.7
)
(
02
)
+
(
0.6
)
(
0.7
)
(
0.8
)
(
0.1
)
=
0.4
+
0.18
+
0.084
+
0.0336
=
0.6976
© examsnet.com
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