Two airplanes bomb the same target one after the other, each…

Two airplanes bomb the same target one after the other, each dropping one bomb per pass. Each airplane has three bombs. The probability that the first airplane hits the target with one bomb is $p_1=0.3$, for the second $p_2=0.4$. If one bomb hits the target, the target is destroyed and the bombing stops. Find the probability that the target will be destroyed before all the bombs of the airplanes are used up.

Step 1: Order of the bombs

Bombs are dropped alternately: plane 1, plane 2, plane 1, plane 2, plane 1, plane 2. Misses happen with probabilities $q_1=0.7$ and $q_2=0.6$, independently.

Step 2: What must happen

The target is destroyed before all bombs are used iff one of bombs $1$–$5$ hits. (If bombs $1$–$5$ miss, then either the sixth bomb is used, or the target survives; in both cases all bombs have been dropped.)

Step 3: The opposite event

Bombs $1$–$5$ are three bombs of plane 1 and two bombs of plane 2:

\[ P(\text{first five miss}) = q_1^3q_2^2 = 0.7^3\cdot0.6^2 = 0.343\cdot0.36 = 0.12348. \]

Step 4: The answer

\[ \boxed{P = 1-q_1^3q_2^2 = 0.87652\approx0.877.} \]

Step 5: Hit by hit

Step 6: Another reading

If the question is simply "the target is destroyed (by any of the six bombs)", the answer is $1-(q_1q_2)^3=1-0.42^3\approx0.926$. The words "before all the bombs are used" exclude destruction by the very last bomb, which gives $0.877$.