## Solution :

Let the events,

A = Ist aeroplane hit the target

B = 2nd aeroplane hit the target

And their corresponding probabilities are

P(A) = 0.3 and P(B) = 0.2

\(\implies\) P(A’) = 0.7 and P(B’) = 0.8

\(\therefore\) Required Probability = P(A’)P(B) + P(A’)P(B’)P(A’)P(B) + …….

= (0.7)(0.2) + (0.7)(0.8)(0.7)(0.2) + …….

= 0.14 [ 1 + 0.56 + \((0.56)^2\) + ….. ]

= 0.14 \(({1\over {1 – 0.56}})\)

= \(0.44\over 0.44\) = \(7\over 22\) = 0.32