NIMCET 2009 Question Paper with solutions

Question : 39 of 120

Marks: +1, -0

An anti aircraft gun can take a maximum of four shots at an enemy plane moving away from it. The probabilities of hitting the plane at first, second, third and fourth shot are

0.4,0.3,0.2

and

0.1

respectively. The probability that the gun hits the plane then is

$0.6972$
$0.6978$
$0.6976$
$0.6974$

Validate

Solution:

Required probability

1-0.6 × 0.7 × 0.8 × 0.9

=1-\;{3024}/{10000}

=\;{6976}/{10000}=0.6976

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