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

Plane I Hit Probability is $$ P(I) = 0.3 $$

Plane I Miss Probability is $$ P(I') = 1 - 0.3 = 0.7 $$

Plane II Hit Probability is $$ P(II) = 0.2 $$

Plane II Miss Probability is $$ P(II') = 1 - 0.2 = 0.8 $$

Continuous Rounds Until Target is Hit

The process repeats indefinitely in turns until someone hits the target. Plane II can win on different turns.

Turn 1 value is $$ 0.7 \times 0.2 = 0.14 $$

Turn 2 value is $$ 0.7 \times 0.8 \times 0.7 \times 0.2 = 0.14 \times 0.56 $$

Turn 3 value is $$ 0.7 \times 0.8 \times 0.7 \times 0.8 \times 0.7 \times 0.2 = 0.14 \times (0.56)^2 $$

This forms an infinite geometric progression series where:

First term is $$ a = 0.14 $$

Common ratio is $$ r = 0.7 \times 0.8 = 0.56 $$

Formula for Infinite Sum is $$ S = \frac{a}{1 - r} $$

Calculation is $$ \frac{0.14}{1 - 0.56} = \frac{0.14}{0.44} = \frac{14}{44} = \frac{7}{22} $$

Final Answer

The probability that the target is hit by the second plane is $$ \frac{7}{22} $$