The probabilities of an anti-aircraft gun sho | vedclass

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Question

(B) Let $A, B$,and $C$ be the events of shooting down the plane in the first,second,and third shots respectively.Given probabilities are $P(A) = 0.6,

Vedclass · Accepted Answer

$0.892$

Vedclass · Answer

(B) Let $A, B$,and $C$ be the events of shooting down the plane in the first,second,and third shots respectively.Given probabilities are $P(A) = 0.6, P(B) = 0.7, P(C) = 0.1$.Since the events are independent,the probability of not hitting the plane in any of the three shots is $P(\overline{A}) 	imes P(\overline{B}) 	imes P(\overline{C})$.$P(\overline{A}) = 1 - 0.6 = 0.4$$P(\overline{B}) = 1 - 0.7 = 0.3$$P(\overline{C}) = 1 - 0.1 = 0.9$Probability of not hitting the plane = $0.4 	imes 0.3 	imes 0.9 = 0.108$.Therefore,the probability of shooting down the plane is $1 - 0.108 = 0.892$.

The probabilities of an anti-aircraft gun shooting down an enemy plane in the first,second,and third shots are $0.6, 0.7$,and $0.1$ respectively. What is the probability that the gun shoots down the plane?

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