The probabilities of an anti-aircraft gun sho | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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?

Similar Questions

Given $P(A \cup B)=0.6$ and $P(A \cap B)=0.2$,the probability that exactly one of the events occurs is:

The probability of getting the number $5$ when throwing a fair die is:

The probability that a number selected at random from the set of numbers $(1, 2, 3, \ldots, 100)$ is a cube,is

If three students $A, B, C$ independently solve a problem with probabilities $\frac{1}{3}, \frac{1}{4}$ and $\frac{1}{5}$ respectively,then the probability that the problem will be solved is

$A$ and $B$ are two events such that $P(A)=0.54$,$P(B)=0.69$ and $P(A \cap B)=0.35$. Find $P(B \cap A^{\prime})$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module