An anti-aircraft gun can take a maximum of 4 shots at an enemy plane moving away from it. The probabilities of hitting the plane at the first, second, third and fourth shots are 0.4, 0.3, 0.2 and 0.1 respectively. What is the probability that at least one shot hits the plane?
Given:Let A ,B , C and Dbe first second third and fourth shots whose probability of hitting the plane is given i.e, P(A) = 0.4 ,P(B) = 0.3 , P(C) = 0.2 and P(D) = 0.1 respectively
P( 
) = 0.6 and P( 
) = 0.7 and P( 
) = 0.8 and P( 
) = 0.9
To Find: The probability that atleast one shot hits the plane .
Here , P(atleast one shot hits the plane) = 1 – P(none of the shots hit the plane )
= 1- P( 
)
= 1 – [P( 
) 
P( 
) 
P( 
) 
P(
) ]
= 1- [0.6
0.7 
0.8
0.9 ]
= 1 – 0.3024
= 0.6976
Therefore, The probability that atleast one shot hits the plane is 0.6976.