NIMCET 2011 Question Paper with solutions

Question : 25 of 120

Marks: +1, -0

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(III)=0.2, P(IV)=0.1

⇒ \;\; P(\ov{I})=0.6, P(\ov{II})=0.7, P(\ov{III})=0.8 \and P(\ov{IV})=0.9

∴ Required probability

=P(I)+P(\ov{I}) P(II)+P\(\ov{I}) P(\ov{II}) P(III) +P(\ov{I}) P(\ov{II}) P(\ov{III}) 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

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