Probability Ques 84
An anti-aircraft gun can take a maximum of four shots at an enemy plane moving away from it. The probabilities of hitting the plane at the first, second, third and fourth shot are $0.4,0.3,0.2$, and $0.1$ , respectively. What is the probability that the gun hits the plane?
(1981, 2M)
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Answer:
Correct Answer: 84.$(0.6976)$
Solution:
Formula:
Set theoretical notation of probability and some important results:
- Let $P\left(H _1\right)=0.4, P\left(H _2\right)=0.3, P\left(H _3\right)=0.2, P\left(H _4\right)=0.1$
$P$ (gun hits the plane)
$ \begin{aligned} & =1-P(\text { gun does not hit the plane }) \\ & =1-P\left(\bar{H} _1\right) \cdot P\left(\bar{H} _2\right) \cdot P\left(\bar{H} _3\right) \cdot P\left(\bar{H} _4\right) \\ & =1-(0.6)(0.7)(0.8)(0.9)=1-0.3024=0.6976 \end{aligned} $
BETA
