Question

An aircraft gun can take a maximum of four shots at an enemys plane moving away from it. The probability of hitting the plane at first, second, third & fourth shots are $0.4,0.3,0.2$ & $0.1$ respectively. What is the probability that the gun hits the plane?

• A. $0.6976$
• B. $0.64$
• C. $0.775$
• D. none of these

Answer 1

The correct option is A $0.6976$

Let the probability of hitting the plane at first shot, second shot, third shot and fourth be P(A), P(B), P(C), P(D) respectively.

Given, $P\left(A\right)=0.4,P\left(B\right)=0.3,P\left(C\right)=0.2,P\left(D\right)=0.1$

Probability of gunshot not hitting the plane

$=P\left(\overline{A}\right).P\left(\overline{B}\right).P\left(\overline{C}\right).P\left(\overline{D}\right)$

$=(1-0.4)\times (1-0.3)\times (1-0.2)\times (1-0.1)=0.6\times 0.7\times 0.8\times 0.9=0.3024$

$\therefore Probabilityofhittingtheplane=1-Probabilitynothittingtheplane=1-0.3024=0.6976$