Question

An anti-aircraft gun can take a maximum of four shots at any plane moving away from it. The probabilities of hitting the plane at 1st, 2nd, 3rd and 4th shots are 0.4, 0.3, 0.2 and 0.1 respectively. What is the probability that at least one shot hits the plane ? 

(a) 0.0024 

(b) 0.3024 

(c) 0.6976 

(d) 0.9976

Answer 1

(c) 0.6976

Let A, B, C, D be the events that 1st, 2nd, 3rd and 4th shots hits the plane respectively. Then, 

P(A) = 0.4, P(B) = 0.3, P(C) = 0.2, P(D) = 0.1 

⇒ P(not A) = 1 – P(A),

= 1 – 0.4 = 0.6 

P(not B) = 1 – P(B)

= 1 – 0.3 = 0.7 

P(not C) = 1 – P(C), 

= 1 – 0.2 = 0.8 

P(not D) = 1 – P(D)

= 1 – 0.1 = 0.9 

∴ P(None of the shots hits the plane) 

= P(not A) × P(not B) × P(not C) × P(D)

= 0.6 × 0.7 × 0.8 × 0.9 = 0.3024 

(As A, B, C and D are independent events) 

P(At least one shot hits the plane) = 1 – P(None hits the plane) 

= 1 – 0.3024 = 0.6976