An anti-aircraft gun can take a maximum of 4 | vedclass

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(C) Let $A, B, C,$ and $D$ be the events that the plane is hit by the first,second,third,and fourth shots,respectively.The probabilities of hitting th

Vedclass · Accepted Answer

$0.6976$

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(C) Let $A, B, C,$ and $D$ be the events that the plane is hit by the first,second,third,and fourth shots,respectively.The probabilities of hitting the plane are $P(A) = 0.4, P(B) = 0.3, P(C) = 0.2,$ and $P(D) = 0.1$.The probabilities of missing the plane in each shot are:$P(\bar{A}) = 1 - 0.4 = 0.6$$P(\bar{B}) = 1 - 0.3 = 0.7$$P(\bar{C}) = 1 - 0.2 = 0.8$$P(\bar{D}) = 1 - 0.1 = 0.9$The gun hits the plane if at least one shot hits the target. This is the complement of the event that all shots miss the target.Required probability $= 1 - P(	ext{all shots miss})$$= 1 - [P(\bar{A}) 	imes P(\bar{B}) 	imes P(\bar{C}) 	imes P(\bar{D})]$$= 1 - [0.6 	imes 0.7 	imes 0.8 	imes 0.9]$$= 1 - 0.3024$$= 0.6976$

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 the gun hits the plane?

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