An anti-aircraft gun takes four shots at an e | vedclass

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(D) Let $E_1, E_2, E_3, E_4$ be the events of hitting the plane in the first,second,third,and fourth shots respectively.Given probabilities are $P(E_1

Vedclass · Accepted Answer

$0.6976$

Vedclass · Answer

(D) Let $E_1, E_2, E_3, E_4$ be the events of hitting the plane in the first,second,third,and fourth shots respectively.Given probabilities are $P(E_1) = 0.4, P(E_2) = 0.3, P(E_3) = 0.2, P(E_4) = 0.1$.The probability of not hitting the plane in any shot is $P(E_i^c) = 1 - P(E_i)$.$P(E_1^c) = 1 - 0.4 = 0.6$$P(E_2^c) = 1 - 0.3 = 0.7$$P(E_3^c) = 1 - 0.2 = 0.8$$P(E_4^c) = 1 - 0.1 = 0.9$The probability that the gun hits the plane at least once is $1 - P(	ext{not hitting the plane in any shot})$.Required probability $= 1 - (P(E_1^c) 	imes P(E_2^c) 	imes P(E_3^c) 	imes P(E_4^c))$$= 1 - (0.6 	imes 0.7 	imes 0.8 	imes 0.9)$$= 1 - 0.3024$$= 0.6976$

An anti-aircraft gun takes four 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. The probability that the gun hits the plane is:

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