Question

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 first, 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 ?

Answer 1

\(0.4+(1-0.4)0.3+(1-0.4)(1-0.3)0.2+(1-0.4)(1-0.3)(1-0.2)0.1=0.6976\)

Answer 2

Probability of the gun hitting the plane is equivalent of saying probability of at least 1 shell hitting the plane. The opposite of this case would then be no shells striking the plane. Thus, probability of a shell hitting the plane is equal to 1-(probability of no shells hitting the plane..
In this case, simply calculate the probabliility of no shells landing on the plane, which is the probability of the shell MISSING the plane on each shot.
1st shot missed: 1-(0.4) = 0.6
2nd shot: 1-(0.3) = 0.7
3rd shot: 1-0.2=0.8
4th shot: 1-0.1=0.9
For the probability of missing the plane completely:
P(shots hit=0) = 0.6*0.7*0.8*0.9
Subtract this result from 1 and you have the probability of at least 1 shell hitting the plane.

Answer 3

probability of hitting the first time: \(0.4\)
probability of missing the first yet hitting the second time: \((1-0.4)0.3\)
probability of missing the first two yet hitting the third time: \((1-0.4)(1-0.3)0.2\)
probability of missing the first three yet hitting the fourth time: \((1-0.4)(1-0.3)(1-0.2)0.1\)

Answer 4

Thank you sir