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 -Maths 9th

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 -Maths 9th
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1 Answer

Answer :

(c) 0.6976Let 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
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