Two aeroplanes I and II bomb a target in succession. The probabilities of I and II scoring a hit correctly are 0.3 and 0.2, respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the second plane, is

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

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