Solution:
Solution
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, $$ P(A)=0.4,P(B)=0.3,P(C)=0.2,P(D)=0.1$$
Probability of gun shot not hitting the plane=$$P(\overline A).P(\overline B).P(\overline C).P(\overline D)$$=$$(1-0.4)\times(1-0.3)\times(1-0.2)\times(1-0.1)=0.6\times0.7\times0.8\times0.9=0.3024$$
$$\therefore\;Probability\;of\;hitting\;the\;plane=1-Probability\;not\;hitting\;the\;plane=1-0.3024=0.6976$$